Binomial Theorem
252 Questions
Start JEE Mains Test
2021
Q151
JEE Mains
MCQ
14 Mar 2026
The value of $\sum\limits_{r = 0}^6 {\left( {{}^6{C_r}\,.\,{}^6{C_{6 - r}}} \right)} $ is equal to :
A.
924
B.
1024
C.
1124
D.
1324
2021
Q152
JEE Mains
MCQ
14 Mar 2026
If the fourth term in the expansion of ${(x + {x^{{{\log }_2}x}})^7}$ is 4480, then the value of x where x$\in$N is equal to :
A.
3
B.
1
C.
4
D.
2
2021
Q153
JEE Mains
MCQ
14 Mar 2026
If n is the number of irrational terms in the
expansion of ${\left( {{3^{1/4}} + {5^{1/8}}} \right)^{60}}$, then (n $-$ 1) is divisible by :
expansion of ${\left( {{3^{1/4}} + {5^{1/8}}} \right)^{60}}$, then (n $-$ 1) is divisible by :
A.
30
B.
8
C.
7
D.
26
2021
Q154
JEE Mains
MCQ
14 Mar 2026
Let [ x ] denote greatest integer less than or equal to x. If for n$\in$N,
${(1 - x + {x^3})^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}} $,
then $\sum\limits_{j = 0}^{\left[ {{{3n} \over 2}} \right]} {{a_{2j}} + 4} \sum\limits_{j = 0}^{\left[ {{{3n - 1} \over 2}} \right]} {{a_{2j}} + 1} $ is equal to :
${(1 - x + {x^3})^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}} $,
then $\sum\limits_{j = 0}^{\left[ {{{3n} \over 2}} \right]} {{a_{2j}} + 4} \sum\limits_{j = 0}^{\left[ {{{3n - 1} \over 2}} \right]} {{a_{2j}} + 1} $ is equal to :
A.
2n $-$ 1
B.
n
C.
2
D.
1
2021
Q155
JEE Mains
MCQ
14 Mar 2026
The maximum value of the term independent of 't' in the expansion
of ${\left( {t{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{10}}$ where x$\in$(0, 1) is :
of ${\left( {t{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{10}}$ where x$\in$(0, 1) is :
A.
${{10!} \over {\sqrt 3 {{(5!)}^2}}}$
B.
${{2.10!} \over {3\sqrt 3 {{(5!)}^2}}}$
C.
${{10!} \over {3{{(5!)}^2}}}$
D.
${{2.10!} \over {3{{(5!)}^2}}}$
2021
Q156
JEE Mains
MCQ
14 Mar 2026
If $n \ge 2$ is a positive integer, then the sum of the series ${}^{n + 1}{C_2} + 2\left( {{}^2{C_2} + {}^3{C_2} + {}^4{C_2} + ... + {}^n{C_2}} \right)$ is :
A.
${{n(2n + 1)(3n + 1)} \over 6}$
B.
${{n(n + 1)(2n + 1)} \over 6}$
C.
${{n{{(n + 1)}^2}(n + 2)} \over {12}}$
D.
${{n(n - 1)(2n + 1)} \over 6}$
2021
Q157
JEE Mains
MCQ
14 Mar 2026
The value of
-15C1 + 2.15C2 – 3.15C3 + ... - 15.15C15 + 14C1 + 14C3 + 14C5 + ...+ 14C11 is :
-15C1 + 2.15C2 – 3.15C3 + ... - 15.15C15 + 14C1 + 14C3 + 14C5 + ...+ 14C11 is :
A.
213 - 13
B.
216 - 1
C.
214
D.
213 - 14
2021
Q158
JEE Mains
Numerical
14 Mar 2026
If the sum of the coefficients in the expansion of (x + y)n is 4096, then the greatest coefficient in the expansion is _____________.
Correct Answer: 924
Explanation:
(x + y)n $\Rightarrow$ 2n = 4096
210 = 1024 $\times$ 2
$\Rightarrow$ 2n = 212
211 = 2048
n = 12
212 = 4096
${}^{12}{C_6}={{12 \times 11 \times 10 \times 9 \times 8 \times 7} \over {6 \times 5 \times 4 \times 3 \times 2 \times 1}}$
$ = 11 \times 3 \times 4 \times 7$
$ = 924$
210 = 1024 $\times$ 2
$\Rightarrow$ 2n = 212
211 = 2048
n = 12
212 = 4096
${}^{12}{C_6}={{12 \times 11 \times 10 \times 9 \times 8 \times 7} \over {6 \times 5 \times 4 \times 3 \times 2 \times 1}}$
$ = 11 \times 3 \times 4 \times 7$
$ = 924$
2021
Q159
JEE Mains
Numerical
14 Mar 2026
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
Q160
JEE Mains
Numerical
14 Mar 2026
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
Q161
JEE Mains
Numerical
14 Mar 2026
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
Q162
JEE Mains
Numerical
14 Mar 2026
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
Q163
JEE Mains
Numerical
14 Mar 2026
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
Q164
JEE Mains
Numerical
14 Mar 2026
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
Q165
JEE Mains
Numerical
14 Mar 2026
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
Q166
JEE Mains
Numerical
14 Mar 2026
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
Q167
JEE Mains
Numerical
14 Mar 2026
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
Q168
JEE Mains
Numerical
14 Mar 2026
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
Q169
JEE Mains
Numerical
14 Mar 2026
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
Q170
JEE Mains
Numerical
14 Mar 2026
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
Q171
JEE Mains
Numerical
14 Mar 2026
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
Q172
JEE Mains
Numerical
14 Mar 2026
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
Q173
JEE Mains
Numerical
14 Mar 2026
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
Q174
JEE Mains
Numerical
14 Mar 2026
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
Q175
JEE Mains
Numerical
14 Mar 2026
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
Q176
JEE Mains
Numerical
14 Mar 2026
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
Q177
JEE Mains
Numerical
14 Mar 2026
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
Q178
JEE Mains
Numerical
14 Mar 2026
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
Q179
JEE Mains
MCQ
14 Mar 2026
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
Q180
JEE Mains
MCQ
14 Mar 2026
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
Q181
JEE Mains
MCQ
14 Mar 2026
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
Q182
JEE Mains
MCQ
14 Mar 2026
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
Q183
JEE Mains
MCQ
14 Mar 2026
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
Q184
JEE Mains
MCQ
14 Mar 2026
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
Q185
JEE Mains
MCQ
14 Mar 2026
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
Q186
JEE Mains
MCQ
14 Mar 2026
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
Q187
JEE Mains
MCQ
14 Mar 2026
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
Q188
JEE Mains
MCQ
14 Mar 2026
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
Q189
JEE Mains
MCQ
14 Mar 2026
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
Q190
JEE Mains
Numerical
14 Mar 2026
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
Q191
JEE Mains
Numerical
14 Mar 2026
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
Q192
JEE Mains
Numerical
14 Mar 2026
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
Q193
JEE Mains
Numerical
14 Mar 2026
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
Q194
JEE Mains
Numerical
14 Mar 2026
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
Q195
JEE Mains
Numerical
14 Mar 2026
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
Q196
JEE Mains
Numerical
14 Mar 2026
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
Q197
JEE Mains
MCQ
14 Mar 2026
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
Q198
JEE Mains
MCQ
14 Mar 2026
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
Q199
JEE Mains
MCQ
14 Mar 2026
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
Q200
JEE Mains
MCQ
14 Mar 2026
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