Binomial Theorem
244 Questions
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
If the coefficient of a7b8 in the expansion of (a + 2b + 4ab)10 is K.216, then K is equal to _____________.
Correct Answer: 315
Explanation:
${{10!} \over {\alpha !\beta !\gamma !}}{a^\alpha }{(2b)^\beta }.{(4ab)^\gamma }$
${{10!} \over {\alpha !\beta !\gamma !}}{a^{\alpha + \gamma }}.\,{b^{\beta + \gamma }}\,.\,{2^\beta }\,.\,{4^\gamma }$
$\alpha + \beta + \gamma = 10$ ..... (1)
$\alpha + \gamma = 7$ .... (2)
$\beta + \gamma = 8$ ..... (3)
$(2) + (3) - (1) \Rightarrow \gamma = 5$
$\alpha = 2$
$\beta = 3$
so coefficients = ${{10!} \over {2!3!5!}}{2^3}{.2^{10}}$
$ = {{10 \times 9 \times 8 \times 7 \times 6 \times 5} \over {2 \times 3 \times 2 \times 5!}} \times {2^{13}}$
$ = 315 \times {2^{16}} \Rightarrow k = 315$
${{10!} \over {\alpha !\beta !\gamma !}}{a^{\alpha + \gamma }}.\,{b^{\beta + \gamma }}\,.\,{2^\beta }\,.\,{4^\gamma }$
$\alpha + \beta + \gamma = 10$ ..... (1)
$\alpha + \gamma = 7$ .... (2)
$\beta + \gamma = 8$ ..... (3)
$(2) + (3) - (1) \Rightarrow \gamma = 5$
$\alpha = 2$
$\beta = 3$
so coefficients = ${{10!} \over {2!3!5!}}{2^3}{.2^{10}}$
$ = {{10 \times 9 \times 8 \times 7 \times 6 \times 5} \over {2 \times 3 \times 2 \times 5!}} \times {2^{13}}$
$ = 315 \times {2^{16}} \Rightarrow k = 315$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
If $\left( {{{{3^6}} \over {{4^4}}}} \right)k$ is the term, independent of x, in the binomial expansion of ${\left( {{x \over 4} - {{12} \over {{x^2}}}} \right)^{12}}$, then k is equal to ___________.
Correct Answer: 55
Explanation:
${\left( {{x \over 4} - {{12} \over {{x^2}}}} \right)^{12}}$
${T_{r + 1}} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{x \over 4}} \right)^{12 - r}}{\left( {{{12} \over {{x^2}}}} \right)^r}$
${T_{r + 1}} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{1 \over 4}} \right)^{12 - r}}{\left( {12} \right)^r}\,.\,{(x)^{12 - 3r}}$
Term independent of x $\Rightarrow$ 12 $-$ 3r = 0 $\Rightarrow$ r = 4
${T_5} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{1 \over 4}} \right)^8}{\left( {12} \right)^4} = {{{3^6}} \over {{4^4}}}.\,k$
$\Rightarrow$ k = 55
${T_{r + 1}} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{x \over 4}} \right)^{12 - r}}{\left( {{{12} \over {{x^2}}}} \right)^r}$
${T_{r + 1}} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{1 \over 4}} \right)^{12 - r}}{\left( {12} \right)^r}\,.\,{(x)^{12 - 3r}}$
Term independent of x $\Rightarrow$ 12 $-$ 3r = 0 $\Rightarrow$ r = 4
${T_5} = {( - 1)^r}\,.\,{}^{12}{C_r}{\left( {{1 \over 4}} \right)^8}{\left( {12} \right)^4} = {{{3^6}} \over {{4^4}}}.\,k$
$\Rightarrow$ k = 55
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Evening Shift
3 $\times$ 722 + 2 $\times$ 1022 $-$ 44 when divided by 18 leaves the remainder __________.
Correct Answer: 15
Explanation:
3(1 + 6)22 + 2 . (1 + 9)22 $-$ 44 = (3 + 2 $-$ 44) = 18 . I
= $-$ 39 + 18 . I
= (54 $-$ 39) + 18(I $-$ 3)
= 15 + 18I1
$\Rightarrow$ Remainder = 15
= $-$ 39 + 18 . I
= (54 $-$ 39) + 18(I $-$ 3)
= 15 + 18I1
$\Rightarrow$ Remainder = 15
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
Let $\left( {\matrix{
n \cr
k \cr
} } \right)$ denotes ${}^n{C_k}$ and $\left[ {\matrix{
n \cr
k \cr
} } \right] = \left\{ {\matrix{
{\left( {\matrix{
n \cr
k \cr
} } \right),} & {if\,0 \le k \le n} \cr
{0,} & {otherwise} \cr
} } \right.$
If ${A_k} = \sum\limits_{i = 0}^9 {\left( {\matrix{ 9 \cr i \cr } } \right)\left[ {\matrix{ {12} \cr {12 - k + i} \cr } } \right] + } \sum\limits_{i = 0}^8 {\left( {\matrix{ 8 \cr i \cr } } \right)\left[ {\matrix{ {13} \cr {13 - k + i} \cr } } \right]} $ and A4 $-$ A3 = 190 p, then p is equal to :
If ${A_k} = \sum\limits_{i = 0}^9 {\left( {\matrix{ 9 \cr i \cr } } \right)\left[ {\matrix{ {12} \cr {12 - k + i} \cr } } \right] + } \sum\limits_{i = 0}^8 {\left( {\matrix{ 8 \cr i \cr } } \right)\left[ {\matrix{ {13} \cr {13 - k + i} \cr } } \right]} $ and A4 $-$ A3 = 190 p, then p is equal to :
Correct Answer: 49
Explanation:
${A_k} = \sum\limits_{i = 0}^9 {{}^9{C_i}} {}^{12}{C_{k - i}} + \sum\limits_{i = 0}^8 {{}^8{C_i}} {}^{13}{C_{k - i}}$
${A_k} = {}^{21}{C_k} + {}^{21}{C_k} = 2.{}^{21}{C_k}$
${A_4} - {A_3} = 2\left( {{}^{21}{C_4} - {}^{21}{C_3}} \right) = 2(5985 - 1330)$
$190p = 2(5985 - 1330) \Rightarrow p = 49$
${A_k} = {}^{21}{C_k} + {}^{21}{C_k} = 2.{}^{21}{C_k}$
${A_4} - {A_3} = 2\left( {{}^{21}{C_4} - {}^{21}{C_3}} \right) = 2(5985 - 1330)$
$190p = 2(5985 - 1330) \Rightarrow p = 49$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
Let n$\in$N and [x] denote the greatest integer less than or equal to x. If the sum of (n + 1) terms ${}^n{C_0},3.{}^n{C_1},5.{}^n{C_2},7.{}^n{C_3},.....$ is equal to 2100 . 101, then $2\left[ {{{n - 1} \over 2}} \right]$ is equal to _______________.
Correct Answer: 98
Explanation:
1. ${}^n{C_0} + 3.{}^n{C_1} + 5.{}^n{C_2} + ... + (2n + 1).{}^n{C_n}$
${T_r} = (2r + 1){}^n{C_r}$
$S = \sum {{T_r}} $
$S = \sum {(2r + 1){}^n{C_r}} = \sum {2r{}^n{C_r} + \sum {{}^n{C_r}} } $
$S = 2(n{.2^{n - 1}}) + {2^n} = {2^n}(n + 1)$
${2^n}(n + 1) = {2^{100}}.101 \Rightarrow n = 100$
$2\left[ {{{n - 1} \over 2}} \right] = 2\left[ {{{99} \over 2}} \right] = 98$
${T_r} = (2r + 1){}^n{C_r}$
$S = \sum {{T_r}} $
$S = \sum {(2r + 1){}^n{C_r}} = \sum {2r{}^n{C_r} + \sum {{}^n{C_r}} } $
$S = 2(n{.2^{n - 1}}) + {2^n} = {2^n}(n + 1)$
${2^n}(n + 1) = {2^{100}}.101 \Rightarrow n = 100$
$2\left[ {{{n - 1} \over 2}} \right] = 2\left[ {{{99} \over 2}} \right] = 98$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
If the co-efficient of x7 and x8 in the expansion of ${\left( {2 + {x \over 3}} \right)^n}$ are equal, then the value of n is equal to _____________.
Correct Answer: 55
Explanation:
${}^n{C_7}{2^{n - 7}}{1 \over {{3^7}}} = {}^n{C_8}{2^{n - 8}}{1 \over {{3^8}}}$
$\Rightarrow$ n $-$ 7 = 48 $\Rightarrow$ n = 55
$\Rightarrow$ n $-$ 7 = 48 $\Rightarrow$ n = 55
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
The ratio of the coefficient of the middle term in the expansion of (1 + x)20 and the sum of the coefficients of two middle terms in expansion of (1 + x)19 is _____________.
Correct Answer: 1
Explanation:
Coeff. of middle term in (1 + x)20 = ${}^{20}{C_{10}}$ & Sum of coeff. of two middle terms in (1 + x)19 = ${}^{19}{C_{9}}$ + ${}^{19}{C_{10}}$
So required ratio = ${{{}^{20}{C_{10}}} \over {^{19}{C_9}{ + ^{19}}{C_{10}}}} = {{^{20}{C_{10}}} \over {^{20}{C_{10}}}} = 1$
So required ratio = ${{{}^{20}{C_{10}}} \over {^{19}{C_9}{ + ^{19}}{C_{10}}}} = {{^{20}{C_{10}}} \over {^{20}{C_{10}}}} = 1$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
The term independent of 'x' in the expansion of
${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$, where x $\ne$ 0, 1 is equal to ______________.
${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$, where x $\ne$ 0, 1 is equal to ______________.
Correct Answer: 210
Explanation:
${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$
= ${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$
= ${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$
= ${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$
[Note:
For ${\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}$ the $\left( {r + 1} \right)$th term with power m of x is
$r = {{n\alpha - m} \over {\alpha + \beta }}$]
Here $\alpha = {1 \over 3}$, $\beta = {1 \over 2}$ and m = 0
then $r = {{10 \times {1 \over 3} - 0} \over {{1 \over 3} + {1 \over 2}}}$ = ${{10} \over 3} \times {6 \over 5}$ = 4
$\therefore$ T5 is the term independent of x.
$\therefore$ T5 = ${}^{10}{C_4}$ = 210
= ${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$
= ${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$
= ${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$
[Note:
For ${\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}$ the $\left( {r + 1} \right)$th term with power m of x is
$r = {{n\alpha - m} \over {\alpha + \beta }}$]
Here $\alpha = {1 \over 3}$, $\beta = {1 \over 2}$ and m = 0
then $r = {{10 \times {1 \over 3} - 0} \over {{1 \over 3} + {1 \over 2}}}$ = ${{10} \over 3} \times {6 \over 5}$ = 4
$\therefore$ T5 is the term independent of x.
$\therefore$ T5 = ${}^{10}{C_4}$ = 210
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 22th July Evening Shift
If the constant term, in binomial expansion of ${\left( {2{x^r} + {1 \over {{x^2}}}} \right)^{10}}$ is 180, then r is equal to __________________.
Correct Answer: 8
Explanation:
${\left( {2{x^r} + {1 \over {{x^2}}}} \right)^{10}}$
General term $ = {}^{10}{C_R}{(2{x^2})^{10 - R}}{x^{ - 2R}}$
$ \Rightarrow {2^{10 - R}}{}^{10}{C_R} = 180$ ....... (1)
& (10 $-$ R)r $-$ 2R = 0
$r = {{2R} \over {10 - R}}$
$r = {{2(R - 10)} \over {10 - R}} + {{20} \over {10 - R}}$
$ \Rightarrow r = - 2 + {{20} \over {10 - R}}$ ....... (2)
R = 8 or 5 reject equation (1) not satisfied
At R = 8
$ \Rightarrow {2^{10 - R}}\times{}^{10}{C_R} = 180 \Rightarrow r = 8$
General term $ = {}^{10}{C_R}{(2{x^2})^{10 - R}}{x^{ - 2R}}$
$ \Rightarrow {2^{10 - R}}{}^{10}{C_R} = 180$ ....... (1)
& (10 $-$ R)r $-$ 2R = 0
$r = {{2R} \over {10 - R}}$
$r = {{2(R - 10)} \over {10 - R}} + {{20} \over {10 - R}}$
$ \Rightarrow r = - 2 + {{20} \over {10 - R}}$ ....... (2)
R = 8 or 5 reject equation (1) not satisfied
At R = 8
$ \Rightarrow {2^{10 - R}}\times{}^{10}{C_R} = 180 \Rightarrow r = 8$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 22th July Evening Shift
The number of elements in the set {n $\in$ {1, 2, 3, ......., 100} | (11)n > (10)n + (9)n} is ______________.
Correct Answer: 96
Explanation:
${11^n} > {10^n} + {9^n}$
$ \Rightarrow {11^n} - {9^n} > {10^n}$
$ \Rightarrow {(10 + 1)^n} - {(10 - 1)^n} > {10^n}$
$ \Rightarrow 2\{ {}^n{C_1}{.10^{n - 1}} + {}^n{C_3}{10^{n - 10}} + {}^n{C_5}{10^{n - 5}} + .....\} > {10^n}$
$ \Rightarrow $ ${1 \over 5}\left[ {{}^n{C_1}{{10}^n} + {}^n{C_3}{{10}^{n - 2}} + {}^n{C_5}{{10}^{n - 4}} + .....} \right] > {10^n}$
$ \Rightarrow $ ${1 \over 5}\left[ {{}^n{C_1} + {}^n{C_3}{{10}^{ - 2}} + {}^n{C_5}{{10}^{ - 4}} + .....} \right] > 1$
Clearly the above inequality is true for n $ \ge $ 5
For n = 4, we have ${1 \over 5}\left[ {4 + {4 \over {{{10}^2}}}} \right] = {4 \over 5}\left( {{{101} \over {100}}} \right) < 1$
$\Rightarrow$ Inequality does not hold good for n = 1, 2, 3, 4
So, required number of elements ={5, 6, 7, ......., 100} = 96
$ \Rightarrow {11^n} - {9^n} > {10^n}$
$ \Rightarrow {(10 + 1)^n} - {(10 - 1)^n} > {10^n}$
$ \Rightarrow 2\{ {}^n{C_1}{.10^{n - 1}} + {}^n{C_3}{10^{n - 10}} + {}^n{C_5}{10^{n - 5}} + .....\} > {10^n}$
$ \Rightarrow $ ${1 \over 5}\left[ {{}^n{C_1}{{10}^n} + {}^n{C_3}{{10}^{n - 2}} + {}^n{C_5}{{10}^{n - 4}} + .....} \right] > {10^n}$
$ \Rightarrow $ ${1 \over 5}\left[ {{}^n{C_1} + {}^n{C_3}{{10}^{ - 2}} + {}^n{C_5}{{10}^{ - 4}} + .....} \right] > 1$
Clearly the above inequality is true for n $ \ge $ 5
For n = 4, we have ${1 \over 5}\left[ {4 + {4 \over {{{10}^2}}}} \right] = {4 \over 5}\left( {{{101} \over {100}}} \right) < 1$
$\Rightarrow$ Inequality does not hold good for n = 1, 2, 3, 4
So, required number of elements ={5, 6, 7, ......., 100} = 96
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Morning Shift
The number of rational terms in the binomial expansion of ${\left( {{4^{{1 \over 4}}} + {5^{{1 \over 6}}}} \right)^{120}}$ is _______________.
Correct Answer: 21
Explanation:
${\left( {{4^{{1 \over 4}}} + {5^{{1 \over 6}}}} \right)^{120}}$
${T_{r + 1}} = {}^{120}{C_r}{({2^{1/2}})^{120 - r}}{(5)^{r/6}}$
for rational terms r = 6$\lambda$
0 $\le$ r $\le$ 120
So total no of terms are 21.
${T_{r + 1}} = {}^{120}{C_r}{({2^{1/2}})^{120 - r}}{(5)^{r/6}}$
for rational terms r = 6$\lambda$
0 $\le$ r $\le$ 120
So total no of terms are 21.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
The term independent of x in the expansion of
${\left[ {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right]^{10}}$, x $\ne$ 1, is equal to ____________.
${\left[ {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right]^{10}}$, x $\ne$ 1, is equal to ____________.
Correct Answer: 210
Explanation:
${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$
= ${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$
= ${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$
= ${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$
${T_{r + 1}} = {}^{10}{C_r}{\left( {{x^{{1 \over 3}}}} \right)^{(10 - r)}}{\left( {{x^{ - {1 \over 2}}}} \right)^r}$
For being independent of $x:{{10 - r} \over 3} - {r \over 2} = 0 \Rightarrow r = 4$
Term independent of $x = {}^{10}{C_4} = 210$
= ${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$
= ${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$
= ${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$
= ${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$
${T_{r + 1}} = {}^{10}{C_r}{\left( {{x^{{1 \over 3}}}} \right)^{(10 - r)}}{\left( {{x^{ - {1 \over 2}}}} \right)^r}$
For being independent of $x:{{10 - r} \over 3} - {r \over 2} = 0 \Rightarrow r = 4$
Term independent of $x = {}^{10}{C_4} = 210$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
Let ${}^n{C_r}$ denote the binomial coefficient of xr in the expansion of (1 + x)n.
If $\sum\limits_{k = 0}^{10} {({2^2} + 3k)} {}^{10}{C_k} = \alpha {.3^{10}} + \beta {.2^{10}},\alpha ,\beta \in R$, then $\alpha$ + $\beta$ is equal to ___________.
Correct Answer: 19
Explanation:
$\sum\limits_{k = 0}^{10} {({2^2} + 3k){}^{10}{C_k}} $
$ = 4\sum\limits_{k = 0}^{10} {{}^{10}{C_k}} + 3\sum\limits_{k = 0}^{10} {k.{}^{10}{C_k}} $
$ = 4({2^{10}}) + 3\sum\limits_{k = 0}^{10} {k.{{10} \over k}.{}^9{C_{k - 1}}} $
= $4({2^{10}}) + 3.10({2^9})$
$ = 4({2^{10}}) + {3.5.2^{10}}$
$ = {2^{10}}(19)$
According to question,
$19({2^{10}}) = \alpha {.3^{10}} + \beta {.2^{10}}$
$ \therefore $ $\alpha = 0,\beta = 19$
$ \Rightarrow \alpha + \beta = 19$
$ = 4\sum\limits_{k = 0}^{10} {{}^{10}{C_k}} + 3\sum\limits_{k = 0}^{10} {k.{}^{10}{C_k}} $
$ = 4({2^{10}}) + 3\sum\limits_{k = 0}^{10} {k.{{10} \over k}.{}^9{C_{k - 1}}} $
= $4({2^{10}}) + 3.10({2^9})$
$ = 4({2^{10}}) + {3.5.2^{10}}$
$ = {2^{10}}(19)$
According to question,
$19({2^{10}}) = \alpha {.3^{10}} + \beta {.2^{10}}$
$ \therefore $ $\alpha = 0,\beta = 19$
$ \Rightarrow \alpha + \beta = 19$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Evening Shift
Let the coefficients of third, fourth and fifth terms in the expansion of ${\left( {x + {a \over {{x^2}}}} \right)^n},x \ne 0$, be in the ratio 12 : 8 : 3. Then the term independent of x in the expansion, is equal to ___________.
Correct Answer: 4
Explanation:
${T_{r + 1}} = {n_{C_r}}{x^{n - r}}.{\left( {{a \over {{x^2}}}} \right)^r}$
$ = {}^n{C_r}{a^r}{x^{n - 3r}}$
${T_3} = {}^n{C_2}{a^2}{x^{n - 6}}$, ${T_4} = {}^n{C_3}{a^3}{x^{n - 9}}$, ${T_5} = {}^n{C_4}{a^4}{x^{n - 12}}$
Now, ${{coefficient\,of\,{T_3}} \over {coefficient\,of\,{T_4}}} = {{{}^n{C_4}.{a^2}} \over {{}^n{C_3}.{a^3}}} = {3 \over {a(n - 2)}} = {3 \over 2}$
$ \Rightarrow a(n - 2) = 2$ .......... (i)
and ${{coefficient\,of\,{T_4}} \over {coefficient\,of\,{T_5}}} = {{{}^n{C_3}.{a^3}} \over {{}^n{C_4}.{a^4}}} = {4 \over {a(n - 3)}} = {8 \over 3}$
$ \Rightarrow a(n - 3) = {3 \over 2}$ ........ (ii)
by (i) and (ii) $n = 6,\,a = {1 \over 2}$
for term independent of 'x'
$n - 3r = 0 \Rightarrow r = {n \over 3} \Rightarrow r = {6 \over 3} = 2$
${T_3} = {}^6{C_2}{\left( {{1 \over 2}} \right)^2}{x^0} = {{15} \over 4} = 3.75 \approx 4$
$ = {}^n{C_r}{a^r}{x^{n - 3r}}$
${T_3} = {}^n{C_2}{a^2}{x^{n - 6}}$, ${T_4} = {}^n{C_3}{a^3}{x^{n - 9}}$, ${T_5} = {}^n{C_4}{a^4}{x^{n - 12}}$
Now, ${{coefficient\,of\,{T_3}} \over {coefficient\,of\,{T_4}}} = {{{}^n{C_4}.{a^2}} \over {{}^n{C_3}.{a^3}}} = {3 \over {a(n - 2)}} = {3 \over 2}$
$ \Rightarrow a(n - 2) = 2$ .......... (i)
and ${{coefficient\,of\,{T_4}} \over {coefficient\,of\,{T_5}}} = {{{}^n{C_3}.{a^3}} \over {{}^n{C_4}.{a^4}}} = {4 \over {a(n - 3)}} = {8 \over 3}$
$ \Rightarrow a(n - 3) = {3 \over 2}$ ........ (ii)
by (i) and (ii) $n = 6,\,a = {1 \over 2}$
for term independent of 'x'
$n - 3r = 0 \Rightarrow r = {n \over 3} \Rightarrow r = {6 \over 3} = 2$
${T_3} = {}^6{C_2}{\left( {{1 \over 2}} \right)^2}{x^0} = {{15} \over 4} = 3.75 \approx 4$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Morning Shift
If (2021)3762 is divided by 17, then the remainder is __________.
Correct Answer: 4
Explanation:
2021 = 17m - 2
(2021)3762 = (17m $-$ 2)3762 = multiple of 17 + 23762
= 17$\lambda$ + 22 (24)940
= 17$\lambda$ + 4 (17 $-$ 1)940
= 17$\lambda$ + 4 (17$\mu$ + 1)
= 17k + 4; (k $\in$ I)
$ \therefore $ Remainder = 4
(2021)3762 = (17m $-$ 2)3762 = multiple of 17 + 23762
= 17$\lambda$ + 22 (24)940
= 17$\lambda$ + 4 (17 $-$ 1)940
= 17$\lambda$ + 4 (17$\mu$ + 1)
= 17k + 4; (k $\in$ I)
$ \therefore $ Remainder = 4
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
Let n be a positive integer. Let
$A = \sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}\left[ {{{\left( {{1 \over 2}} \right)}^k} + {{\left( {{3 \over 4}} \right)}^k} + {{\left( {{7 \over 8}} \right)}^k} + {{\left( {{{15} \over {16}}} \right)}^k} + {{\left( {{{31} \over {32}}} \right)}^k}} \right]} $. If
$63A = 1 - {1 \over {{2^{30}}}}$, then n is equal to _____________.
$A = \sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}\left[ {{{\left( {{1 \over 2}} \right)}^k} + {{\left( {{3 \over 4}} \right)}^k} + {{\left( {{7 \over 8}} \right)}^k} + {{\left( {{{15} \over {16}}} \right)}^k} + {{\left( {{{31} \over {32}}} \right)}^k}} \right]} $. If
$63A = 1 - {1 \over {{2^{30}}}}$, then n is equal to _____________.
Correct Answer: 6
Explanation:
$A = \sum {{{( - 1)}^k}{}^n{C_k}{{\left( {{1 \over 2}} \right)}^k}} + \sum {{{( - 1)}^k}{}^n{C_k}{{\left( {{3 \over 4}} \right)}^k}} + .....$
$ = {\left( {1 - {1 \over 2}} \right)^n} + {\left( {1 - {3 \over 4}} \right)^n} + ..... + {\left( {1 - {{31} \over {32}}} \right)^n}$
$ = {\left( {{1 \over 2}} \right)^n} + {\left( {{1 \over 2}} \right)^{2n}} + {\left( {{1 \over 2}} \right)^{3n}} + ..... + {\left( {{1 \over 2}} \right)^{5n}}$
$ = {\left( {{1 \over 2}} \right)^n}\left( {{{1 - {{\left( {{1 \over 2}} \right)}^{5n}}} \over {1 - {{\left( {{1 \over 2}} \right)}^n}}}} \right) = {{{2^{5n}} - 1} \over {{2^{5n}}({2^n} - 1)}}$
$ \therefore $ $63A = {{63\left( {{2^{5n}} - 1} \right)} \over {{2^{5n}}\left( {{2^n} - 1} \right)}}$ = ${{63} \over {\left( {{2^n} - 1} \right)}}\left( {1 - {1 \over {{2^{5n}}}}} \right)$
Given, $63A = 1 - {1 \over {{2^{30}}}}$
$ \therefore $ ${{63} \over {\left( {{2^n} - 1} \right)}}\left( {1 - {1 \over {{2^{5n}}}}} \right)$ = $1 - {1 \over {{2^{30}}}}$
For n = 6, L.H.S = R.H.S
$ \therefore $ n = 6
$ = {\left( {1 - {1 \over 2}} \right)^n} + {\left( {1 - {3 \over 4}} \right)^n} + ..... + {\left( {1 - {{31} \over {32}}} \right)^n}$
$ = {\left( {{1 \over 2}} \right)^n} + {\left( {{1 \over 2}} \right)^{2n}} + {\left( {{1 \over 2}} \right)^{3n}} + ..... + {\left( {{1 \over 2}} \right)^{5n}}$
$ = {\left( {{1 \over 2}} \right)^n}\left( {{{1 - {{\left( {{1 \over 2}} \right)}^{5n}}} \over {1 - {{\left( {{1 \over 2}} \right)}^n}}}} \right) = {{{2^{5n}} - 1} \over {{2^{5n}}({2^n} - 1)}}$
$ \therefore $ $63A = {{63\left( {{2^{5n}} - 1} \right)} \over {{2^{5n}}\left( {{2^n} - 1} \right)}}$ = ${{63} \over {\left( {{2^n} - 1} \right)}}\left( {1 - {1 \over {{2^{5n}}}}} \right)$
Given, $63A = 1 - {1 \over {{2^{30}}}}$
$ \therefore $ ${{63} \over {\left( {{2^n} - 1} \right)}}\left( {1 - {1 \over {{2^{5n}}}}} \right)$ = $1 - {1 \over {{2^{30}}}}$
For n = 6, L.H.S = R.H.S
$ \therefore $ n = 6
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
Let m, n$\in$N and gcd (2, n) = 1. If $30\left( {\matrix{
{30} \cr
0 \cr
} } \right) + 29\left( {\matrix{
{30} \cr
1 \cr
} } \right) + ...... + 2\left( {\matrix{
{30} \cr
{28} \cr
} } \right) + 1\left( {\matrix{
{30} \cr
{29} \cr
} } \right) = n{.2^m}$, then n + m is equal to __________.
(Here $\left( {\matrix{ n \cr k \cr } } \right) = {}^n{C_k}$)
(Here $\left( {\matrix{ n \cr k \cr } } \right) = {}^n{C_k}$)
Correct Answer: 45
Explanation:
$30({}^{30}{C_0}) + 29({}^{30}{C_1}) + .... + 2({}^{30}{C_{28}}) + 1({}^{30}{C_{29}})$
$ = 30({}^{30}{C_{30}}) + 29({}^{30}{C_{29}}) + ...... + 2({}^{30}{C_2}) + 1({}^{30}{C_1})$
$ = \sum\limits_{r = 1}^{30} {r({}^{30}{C_r})} $
$ = \sum\limits_{r = 1}^{30} {r\left( {{{30} \over r}} \right)({}^{29}{C_{r - 1}}} )$
$ = 30\sum\limits_{r = 1}^{30} {{}^{29}{C_{r - 1}}} $
$ = 30({}^{29}{C_0} + {}^{29}{C_1} + {}^{29}{C_2} + ..... + {}^{29}{C_{29}})$
$ = 30({2^{29}}) = 15{(2)^{30}} = n{(2)^m}$
$ \therefore $ n = 15, m = 30
$ \Rightarrow $ n + m = 45
$ = 30({}^{30}{C_{30}}) + 29({}^{30}{C_{29}}) + ...... + 2({}^{30}{C_2}) + 1({}^{30}{C_1})$
$ = \sum\limits_{r = 1}^{30} {r({}^{30}{C_r})} $
$ = \sum\limits_{r = 1}^{30} {r\left( {{{30} \over r}} \right)({}^{29}{C_{r - 1}}} )$
$ = 30\sum\limits_{r = 1}^{30} {{}^{29}{C_{r - 1}}} $
$ = 30({}^{29}{C_0} + {}^{29}{C_1} + {}^{29}{C_2} + ..... + {}^{29}{C_{29}})$
$ = 30({2^{29}}) = 15{(2)^{30}} = n{(2)^m}$
$ \therefore $ n = 15, m = 30
$ \Rightarrow $ n + m = 45
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
If the remainder when x is divided by 4 is 3, then the remainder when (2020 + x)2022 is divided by 8 is __________.
Correct Answer: 1
Explanation:
Let x = 4k + 3
(2020 + x)2022
= (2020 + 4k + 3)2022
= (4(505) + 4k + 3)2022
= (4P + 3)2022
= (4P + 4 $-$ 1)2022
= (4A $-$ 1)2022
2022C0(4A)0($-$1)2022 + 2022C1(4A)1($-$1)2021 + ......
= 1 + 2022(4A)(-1) + .....
= 1 + 8$\lambda$
$ \therefore $ Reminder is 1.
(2020 + x)2022
= (2020 + 4k + 3)2022
= (4(505) + 4k + 3)2022
= (4P + 3)2022
= (4P + 4 $-$ 1)2022
= (4A $-$ 1)2022
2022C0(4A)0($-$1)2022 + 2022C1(4A)1($-$1)2021 + ......
= 1 + 2022(4A)(-1) + .....
= 1 + 8$\lambda$
$ \therefore $ Reminder is 1.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
The total number of two digit numbers 'n', such that 3n + 7n is a multiple of 10, is __________.
Correct Answer: 45
Explanation:
$ \because $ ${7^n} = {(10 - 3)^n} = 10k + {( - 3)^n}$
${7^n} + {3^n} = 10k + {( - 3)^n} + {3^n}$
$ \therefore $ 3n = 32t = (10 $-$ 1)t
= 10p + ($-$1)t
= 10p $\pm$ 1
$ \therefore $ if n = even then 7n + 3n will not be multiply of 10
So if n is odd then only 7n + 3n will be multiply of 10
$ \therefore $ n = 11, 13, 15, ..........., 99
$ \therefore $ Ans : 45
${7^n} + {3^n} = 10k + {( - 3)^n} + {3^n}$
$ \therefore $ 3n = 32t = (10 $-$ 1)t
= 10p + ($-$1)t
= 10p $\pm$ 1
$ \therefore $ if n = even then 7n + 3n will not be multiply of 10
So if n is odd then only 7n + 3n will be multiply of 10
$ \therefore $ n = 11, 13, 15, ..........., 99
$ \therefore $ Ans : 45
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
For integers n and r, let $\left( {\matrix{
n \cr
r \cr
} } \right) = \left\{ {\matrix{
{{}^n{C_r},} & {if\,n \ge r \ge 0} \cr
{0,} & {otherwise} \cr
} } \right.$ The maximum value of k for which the sum $\sum\limits_{i = 0}^k {\left( {\matrix{
{10} \cr
i \cr
} } \right)\left( {\matrix{
{15} \cr
{k - i} \cr
} } \right)} + \sum\limits_{i = 0}^{k + 1} {\left( {\matrix{
{12} \cr
i \cr
} } \right)\left( {\matrix{
{13} \cr
{k + 1 - i} \cr
} } \right)} $ exists, is equal to _________.
Correct Answer: 12
Explanation:
As k is unbounded so maximum value is not defined.
Question will be BONUS.
Question will be BONUS.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
If the constant term in the binomial expansion
of
${\left( {\sqrt x - {k \over {{x^2}}}} \right)^{10}}$ is 405, then |k| equals :
${\left( {\sqrt x - {k \over {{x^2}}}} \right)^{10}}$ is 405, then |k| equals :
A.
3
B.
9
C.
1
D.
2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
If {p} denotes the fractional part of the number p, then
$\left\{ {{{{3^{200}}} \over 8}} \right\}$, is equal to :
$\left\{ {{{{3^{200}}} \over 8}} \right\}$, is equal to :
A.
${5 \over 8}$
B.
${7 \over 8}$
C.
${1 \over 8}$
D.
${3 \over 8}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
If for some positive integer n, the coefficients
of three consecutive terms in the binomial
expansion of (1 + x)n + 5 are in the ratio
5 : 10 : 14, then the largest coefficient in this expansion is :
of three consecutive terms in the binomial
expansion of (1 + x)n + 5 are in the ratio
5 : 10 : 14, then the largest coefficient in this expansion is :
A.
330
B.
792
C.
252
D.
462
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
The value of $\sum\limits_{r = 0}^{20} {{}^{50 - r}{C_6}} $ is equal to:
A.
${}^{50}{C_6} - {}^{30}{C_6}$
B.
${}^{51}{C_7} - {}^{30}{C_7}$
C.
${}^{50}{C_7} - {}^{30}{C_7}$
D.
${}^{51}{C_7} + {}^{30}{C_7}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
If the term independent of x in the expansion of
${\left( {{3 \over 2}{x^2} - {1 \over {3x}}} \right)^9}$ is k, then 18 k is equal to :
${\left( {{3 \over 2}{x^2} - {1 \over {3x}}} \right)^9}$ is k, then 18 k is equal to :
A.
5
B.
9
C.
7
D.
11
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
If the number of integral terms in the expansion
of (31/2 + 51/8)n is exactly 33, then the least value of n is :
of (31/2 + 51/8)n is exactly 33, then the least value of n is :
A.
264
B.
256
C.
128
D.
248
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
Let
$\alpha $ > 0,
$\beta $ > 0 be such that
$\alpha $3 + $\beta $2 = 4. If the maximum value of the term independent of x in
the binomial expansion of ${\left( {\alpha {x^{{1 \over 9}}} + \beta {x^{ - {1 \over 6}}}} \right)^{10}}$ is 10k,
then k is equal to :
$\alpha $3 + $\beta $2 = 4. If the maximum value of the term independent of x in
the binomial expansion of ${\left( {\alpha {x^{{1 \over 9}}} + \beta {x^{ - {1 \over 6}}}} \right)^{10}}$ is 10k,
then k is equal to :
A.
176
B.
336
C.
352
D.
84
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
In the expansion of ${\left( {{x \over {\cos \theta }} + {1 \over {x\sin \theta }}} \right)^{16}}$, if ${\ell _1}$ is
the least value of the term independent of x
when ${\pi \over 8} \le \theta \le {\pi \over 4}$ and ${\ell _2}$ is the least value of the
term independent of x when ${\pi \over {16}} \le \theta \le {\pi \over 8}$, then
the ratio ${\ell _2}$ : ${\ell _1}$ is equal to :
A.
8 : 1
B.
16 : 1
C.
1 : 8
D.
1 : 16
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
If $\alpha $ and $\beta $ be the coefficients of x4 and x2
respectively in the expansion of
${\left( {x + \sqrt {{x^2} - 1} } \right)^6} + {\left( {x - \sqrt {{x^2} - 1} } \right)^6}$, then
${\left( {x + \sqrt {{x^2} - 1} } \right)^6} + {\left( {x - \sqrt {{x^2} - 1} } \right)^6}$, then
A.
$\alpha + \beta = 60$
B.
$\alpha - \beta = 60$
C.
$\alpha + \beta = -30$
D.
$\alpha - \beta = -132$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
The coefficient of x7
in the expression
(1 + x)10 + x(1 + x)9 + x2(1 + x)8 + ......+ x10 is:
(1 + x)10 + x(1 + x)9 + x2(1 + x)8 + ......+ x10 is:
A.
120
B.
330
C.
420
D.
210
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
The greatest positive integer k, for which 49k + 1 is a factor of the sum
49125 + 49124 + ..... + 492 + 49 + 1, is:
49125 + 49124 + ..... + 492 + 49 + 1, is:
A.
32
B.
60
C.
63
D.
65
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 5th September Evening Slot
The coefficient of x4
in the expansion of
(1 + x + x2 + x3)6 in powers of x, is ______.
(1 + x + x2 + x3)6 in powers of x, is ______.
Correct Answer: 120
Explanation:
(1 + x + x2
+ x3)6
= ((1 + x) (1 + x2))6
= (1 + x)6(1 + x2)6
= $\sum\limits_{r = 0}^6 {{}^6{C_r}.{x^r}} $ $\sum\limits_{r = 0}^6 {{}^6{C_r}.{x^{2r}}} $
Coefficient of x4 = 6C0 6C2 + 6C2 6C1 + 6C4 6C0 = 120
= ((1 + x) (1 + x2))6
= (1 + x)6(1 + x2)6
= $\sum\limits_{r = 0}^6 {{}^6{C_r}.{x^r}} $ $\sum\limits_{r = 0}^6 {{}^6{C_r}.{x^{2r}}} $
Coefficient of x4 = 6C0 6C2 + 6C2 6C1 + 6C4 6C0 = 120
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 5th September Morning Slot
The natural number m, for which the coefficient of x in the binomial expansion of
${\left( {{x^m} + {1 \over {{x^2}}}} \right)^{22}}$ is 1540, is .............
${\left( {{x^m} + {1 \over {{x^2}}}} \right)^{22}}$ is 1540, is .............
Correct Answer: 13
Explanation:
General term,
${T_{r + 1}} = {}^{22}{C_r}{({x^m})^{22 - r}}{\left( {{1 \over {{x^2}}}} \right)^r} = {}^{22}{C_r}{x^{22m - mr - 2r}}$
$ \because $ ${}^{22}{C_3} = {}^{22}{C_{19}} = 1540$
$ \therefore $ $r = 3\,or\,19$
$22m - mr - 2r = 1$
$m = {{2r + 1} \over {22 - 5}}$
When $r = 3$, $m = {7 \over {19}} \notin N$
When $r = 19$, $m = {{38 + 1} \over {22 - 19}} = {{39} \over 3} = 13$
$ \therefore $ $m = 13$
${T_{r + 1}} = {}^{22}{C_r}{({x^m})^{22 - r}}{\left( {{1 \over {{x^2}}}} \right)^r} = {}^{22}{C_r}{x^{22m - mr - 2r}}$
$ \because $ ${}^{22}{C_3} = {}^{22}{C_{19}} = 1540$
$ \therefore $ $r = 3\,or\,19$
$22m - mr - 2r = 1$
$m = {{2r + 1} \over {22 - 5}}$
When $r = 3$, $m = {7 \over {19}} \notin N$
When $r = 19$, $m = {{38 + 1} \over {22 - 19}} = {{39} \over 3} = 13$
$ \therefore $ $m = 13$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Morning Slot
Let ${\left( {2{x^2} + 3x + 4} \right)^{10}} = \sum\limits_{r = 0}^{20} {{a_r}{x^r}} $
Then ${{{a_7}} \over {{a_{13}}}}$ is equal to ______.
Then ${{{a_7}} \over {{a_{13}}}}$ is equal to ______.
Correct Answer: 8
Explanation:
Note : Multinomial Theorem :
The general term of ${\left( {{x_1} + {x_2} + ... + {x_n}} \right)^n}$ the expansion is
${{n!} \over {{n_1}!{n_2}!...{n_n}!}}x_1^{{n_1}}x_2^{{n_2}}...x_n^{{n_n}}$
where n1 + n2 + ..... + nn = n
Here, in ${(2{x^2} + 3x + 4)^{10}}$ general term is
$ = {{10!} \over {{n_1}!{n_2}!{n_3}!}}{(2{x^2})^{{n_1}}}{(3x)^{{n_2}}}{(4)^{{n_3}}}$
$ = {{10!} \over {{n_1}!{n_2}!{n_3}!}}{.2^{{n_1}}}{.3^{{n_2}}}{.4^{{n_3}}}.{x^{2{n_1} + {n_2}}}$
$ \therefore $ Coefficient of $ {x^{2{n_1} + {n_2}}}$ is
${{10!} \over {{n_1}!{n_2}!{n_3}!}}{.2^{{n_1}}}{.3^{{n_2}}}{.4^{{n_3}}}$
where ${n_1} + {n_2} + {n_3} = 10$
For, Coefficient of x7 :
2n1 + n2 = 7
Possible values of n1, n2 and n3 are
$ \therefore $ Coefficient of x7
$ = {{10!} \over {3!1!6!}}{(2)^3}{(3)^1}{(4)^6} + {{10!} \over {3!1!6!}}{(2)^2}{(3)^3}{(4)^5} + {{10!} \over {1!5!4!}}{(2)^1}{(3)^5}{(4)^4} + {{10!} \over {0!7!3!}}{(2)^0}{(3)^7}{(4)^3}$
Coefficient of x13 = a13
Here 2n1 + n2 = 13
possible values of n1, n2 and n3 are
$ \therefore $ Coefficient of x13
$ = {{10!} \over {6!1!3!}}{(2)^6}{(3)^1}{(4)^3} + {{10!} \over {5!3!2!}}{(2)^5}{(3)^3}{(4)^2} + {{10!} \over {4!5!1!}}{(2)^4}{(3)^5}{(4)^1} + {{10!} \over {3!7!0!}}{(2)^3}{(3)^7}{(4)^0}$
$ \therefore $ ${{{a_7}} \over {{a_{13}}}} = 8$
The general term of ${\left( {{x_1} + {x_2} + ... + {x_n}} \right)^n}$ the expansion is
${{n!} \over {{n_1}!{n_2}!...{n_n}!}}x_1^{{n_1}}x_2^{{n_2}}...x_n^{{n_n}}$
where n1 + n2 + ..... + nn = n
Here, in ${(2{x^2} + 3x + 4)^{10}}$ general term is
$ = {{10!} \over {{n_1}!{n_2}!{n_3}!}}{(2{x^2})^{{n_1}}}{(3x)^{{n_2}}}{(4)^{{n_3}}}$
$ = {{10!} \over {{n_1}!{n_2}!{n_3}!}}{.2^{{n_1}}}{.3^{{n_2}}}{.4^{{n_3}}}.{x^{2{n_1} + {n_2}}}$
$ \therefore $ Coefficient of $ {x^{2{n_1} + {n_2}}}$ is
${{10!} \over {{n_1}!{n_2}!{n_3}!}}{.2^{{n_1}}}{.3^{{n_2}}}{.4^{{n_3}}}$
where ${n_1} + {n_2} + {n_3} = 10$
For, Coefficient of x7 :
2n1 + n2 = 7
Possible values of n1, n2 and n3 are
| ${n_1}$ | ${n_2}$ | ${n_3}$ |
|---|---|---|
| 3 | 1 | 6 |
| 2 | 3 | 5 |
| 1 | 5 | 4 |
| 0 | 7 | 3 |
$ \therefore $ Coefficient of x7
$ = {{10!} \over {3!1!6!}}{(2)^3}{(3)^1}{(4)^6} + {{10!} \over {3!1!6!}}{(2)^2}{(3)^3}{(4)^5} + {{10!} \over {1!5!4!}}{(2)^1}{(3)^5}{(4)^4} + {{10!} \over {0!7!3!}}{(2)^0}{(3)^7}{(4)^3}$
Coefficient of x13 = a13
Here 2n1 + n2 = 13
possible values of n1, n2 and n3 are
| ${n_1}$ | ${n_2}$ | ${n_3}$ |
|---|---|---|
| 6 | 1 | 3 |
| 5 | 3 | 2 |
| 4 | 5 | 1 |
| 3 | 7 | 0 |
$ \therefore $ Coefficient of x13
$ = {{10!} \over {6!1!3!}}{(2)^6}{(3)^1}{(4)^3} + {{10!} \over {5!3!2!}}{(2)^5}{(3)^3}{(4)^2} + {{10!} \over {4!5!1!}}{(2)^4}{(3)^5}{(4)^1} + {{10!} \over {3!7!0!}}{(2)^3}{(3)^7}{(4)^0}$
$ \therefore $ ${{{a_7}} \over {{a_{13}}}} = 8$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 2nd September Evening Slot
For a positive integer n,
${\left( {1 + {1 \over x}} \right)^n}$ is expanded
in increasing powers of x. If three consecutive
coefficients in this expansion are in the ratio,
2 : 5 : 12, then n is equal to________.
${\left( {1 + {1 \over x}} \right)^n}$ is expanded
in increasing powers of x. If three consecutive
coefficients in this expansion are in the ratio,
2 : 5 : 12, then n is equal to________.
Correct Answer: 118
Explanation:
Let, three consecutive coefficients are
${}^n{C_{r - 1}},{}^n{C_r},{}^n{C_{r + 1}}$
${}^n{C_{r - 1}}:{}^n{C_r}:{}^n{C_{r + 1}} = 2:5:12$
Now, ${{{}^n{C_{r - 1}}} \over {{}^n{C_r}}} = {2 \over 5}$
$ \Rightarrow 7r = 2n + 2$ ...(i)
${{{}^n{C_r}} \over {{}^n{C_{r + 1}}}} = {5 \over {12}}$
$ \Rightarrow 7r = 5n - 12$ ...(ii)
On solving (i) and (ii) we get n = 118
${}^n{C_{r - 1}},{}^n{C_r},{}^n{C_{r + 1}}$
${}^n{C_{r - 1}}:{}^n{C_r}:{}^n{C_{r + 1}} = 2:5:12$
Now, ${{{}^n{C_{r - 1}}} \over {{}^n{C_r}}} = {2 \over 5}$
$ \Rightarrow 7r = 2n + 2$ ...(i)
${{{}^n{C_r}} \over {{}^n{C_{r + 1}}}} = {5 \over {12}}$
$ \Rightarrow 7r = 5n - 12$ ...(ii)
On solving (i) and (ii) we get n = 118
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Evening Slot
If Cr $ \equiv $ 25Cr and
C0 + 5.C1 + 9.C2 + .... + (101).C25 = 225.k, then k is equal to _____.
C0 + 5.C1 + 9.C2 + .... + (101).C25 = 225.k, then k is equal to _____.
Correct Answer: 51
Explanation:
S = 1.25C0 + 5.25C1 + 9.25C2 + .... + (101)25C25
S = (101).25C25 + (97).25C24 + .......... + (1).25C0
_________________________________________
2S = 102{25C0 + 25C1 + ......+ 25C25}
$ \Rightarrow $ S = 51 $ \times $ 225 = k.225
$ \Rightarrow $ k = 51
S = (101).25C25 + (97).25C24 + .......... + (1).25C0
_________________________________________
2S = 102{25C0 + 25C1 + ......+ 25C25}
$ \Rightarrow $ S = 51 $ \times $ 225 = k.225
$ \Rightarrow $ k = 51
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Morning Slot
The coefficient of x4 is the expansion of
(1 + x + x2)10 is _____.
Correct Answer: 615
Explanation:
(1 + x + x2)10
= 10C0(1 + x)10 + 10C1(1 + x)9.x2 + 10C2(1 + x)8.x4+ .....
Coefficient of x4
= 10C0.10C4 + 10C1.9C2 + 10C2.8C0
= 210 + 360 + 45
= 615
= 10C0(1 + x)10 + 10C1(1 + x)9.x2 + 10C2(1 + x)8.x4+ .....
Coefficient of x4
= 10C0.10C4 + 10C1.9C2 + 10C2.8C0
= 210 + 360 + 45
= 615
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Morning Slot
If the sum of the coefficients of all even powers of x in the product
(1 + x + x2 + ....+ x2n)(1 - x + x2 - x3 + ...... + x2n) is 61, then n is equal to _______.
(1 + x + x2 + ....+ x2n)(1 - x + x2 - x3 + ...... + x2n) is 61, then n is equal to _______.
Correct Answer: 30
Explanation:
(1 + x + x2
+ ....+ x2n)(1 - x + x2 - x3 + ...... + x2n)
= a0 + a1x + a2x2 + …..
put x = 1
(2n + 1)$ \times $1 = a0 + a1 + a2 + …… (1)
put x = –1
1$ \times $(2n + 1) = a0 – a1 + a2+ …….. (2)
Adding (1) and (2)
4n + 2 = 2(a0 + a2 + ….. )
$ \Rightarrow $ 4n + 2 = 2 $ \times $ 61
$ \Rightarrow $ n = 30
= a0 + a1x + a2x2 + …..
put x = 1
(2n + 1)$ \times $1 = a0 + a1 + a2 + …… (1)
put x = –1
1$ \times $(2n + 1) = a0 – a1 + a2+ …….. (2)
Adding (1) and (2)
4n + 2 = 2(a0 + a2 + ….. )
$ \Rightarrow $ 4n + 2 = 2 $ \times $ 61
$ \Rightarrow $ n = 30
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
If 20C1 + (22) 20C2 + (32) 20C3 + ..... + (202
)
20C20 = A(2$\beta $), then the ordered pair (A, $\beta $) is equal to :
A.
(420, 19)
B.
(420, 18)
C.
(380, 18)
D.
(380, 19)
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
The term independent of x in the expansion of
$\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$ is equal to :
$\left( {{1 \over {60}} - {{{x^8}} \over {81}}} \right).{\left( {2{x^2} - {3 \over {{x^2}}}} \right)^6}$ is equal to :
A.
36
B.
- 108
C.
- 36
D.
- 72
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
The coefficient of x18 in the product
(1 + x) (1 – x)10 (1 + x + x2)9 is :
(1 + x) (1 – x)10 (1 + x + x2)9 is :
A.
126
B.
- 84
C.
- 126
D.
84
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
The smallest natural number n, such that the coefficient of x in the expansion of ${\left( {{x^2} + {1 \over {{x^3}}}} \right)^n}$ is nC23, is :
A.
23
B.
58
C.
38
D.
35
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
If the coefficients of x2
and x3
are both zero, in the expansion of the expression (1 + ax + bx2
) (1 – 3x)15 in
powers of x, then the ordered pair (a,b) is equal to :
A.
(28, 861)
B.
(28, 315)
C.
(–21, 714)
D.
(–54, 315)
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
If some three consecutive in the binomial
expansion of (x + 1)n is powers of x are in the ratio
2 : 15 : 70, then the average of these three
coefficient is :-
A.
625
B.
227
C.
964
D.
232
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
If the fourth term in the binomial expansion of ${\left( {{2 \over x} + {x^{{{\log }_8}x}}} \right)^6}$
(x > 0) is 20 × 87, then a value of
x is :
A.
8–2
B.
82
C.
83
D.
8
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
If the fourth term in the binomial expansion of
${\left( {\sqrt {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} + {x^{{1 \over {12}}}}} \right)^6}$ is equal to 200, and x > 1, then the value of x is :
${\left( {\sqrt {{x^{\left( {{1 \over {1 + {{\log }_{10}}x}}} \right)}}} + {x^{{1 \over {12}}}}} \right)^6}$ is equal to 200, and x > 1, then the value of x is :
A.
100
B.
103
C.
10
D.
104
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
The sum of the co-efficients of all even
degree terms in x in the expansion of
${\left( {x + \sqrt {{x^3} - 1} } \right)^6}$ + ${\left( {x - \sqrt {{x^3} - 1} } \right)^6}$, (x > 1) is equal to:
${\left( {x + \sqrt {{x^3} - 1} } \right)^6}$ + ${\left( {x - \sqrt {{x^3} - 1} } \right)^6}$, (x > 1) is equal to:
A.
32
B.
26
C.
29
D.
24
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
The sum of the series
2.20C0 + 5.20C1 + 8.20C2 + 11.20C3 + ... +62.20C20 is equal to :
2.20C0 + 5.20C1 + 8.20C2 + 11.20C3 + ... +62.20C20 is equal to :
A.
225
B.
224
C.
226
D.
223
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
The total number of irrational terms in the binomial expansion of (71/5 – 31/10)60 is :
A.
54
B.
55
C.
49
D.
48
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Morning Slot
A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of ${\left( {{2^{1/3}} + {1 \over {2{{\left( 3 \right)}^{1/3}}}}} \right)^{10}}$ is :
A.
1 : 2(6)1/3
B.
1 : 4(6)1/3
C.
2(36)1/3 : 1
D.
4(36)1/3 : 1