Binomial Theorem
360 Questions
Start JEE Mains Test
2021
Q251
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
Q252
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
Q253
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
Q254
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
Q255
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
Q256
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
Q257
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
Q258
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
Q259
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
Q260
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
Q261
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
Q262
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
Q263
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
Q264
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
Q265
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
Q266
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
Q267
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
Q268
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
Q269
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
Q270
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
Q271
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
Q272
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
Q273
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
Q274
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.
2021
Q275
BITSAT
MCQ
11 Jun 2026
${{{C_1}} \over {{C_0}}} + 2{{{C_2}} \over {{C_1}}} + 3{{{C_3}} \over {{C_2}}} + 4{{{C_4}} \over {{C_3}}} + ....20{{{C_{20}}} \over {{C_{19}}}} = $
A.
120
B.
260
C.
210
D.
180
2020
Q276
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
Q277
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
Q278
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
Q279
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
Q280
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
Q281
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
Q282
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
Q283
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
Q284
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
Q285
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
Q286
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
Q287
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
Q288
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
Q289
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
Q290
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
Q291
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
Q292
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
Q293
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
2020
Q294
TS-EAMCET
MCQ
20 May 2026
If ${ }^n C_0,{ }^n C_1,{ }^n C_2, \ldots,{ }^n C_n$ respectively are the binomial coefficients in the expansion of $(1+x)^n$, then when $n=10, \sum_{r=1}^{10}{ }^n C_r \cdot r(r-4)=$
A.
5120
B.
7680
C.
20480
D.
28160
2020
Q295
TS-EAMCET
MCQ
20 May 2026
If sum of the coefficients of $x^r(r=0,1,2, \ldots, 2 n)$ in the expansion of $\left(1+3 x-2 x^2\right)^n$ is 128 , then $\sum_{r=1}^{2 n} r \frac{(2 n)_{C_r}}{(2 n)_{C_{r-1}}}=$
A.
120
B.
135
C.
90
D.
105
2020
Q296
TS-EAMCET
MCQ
20 May 2026
The approximate value of $\left(3 \sqrt{126}+\sin 61^{\circ}\right)$ correct to three decimal places, obtained by taking $1^{\circ}=0.0174$ radians, is
A.
5.772
B.
5.765
C.
5.806
D.
5.888
2020
Q297
TS-EAMCET
MCQ
20 May 2026
If $x$ is so small that all terms containing $x^2$ and higher powers of $x$ can be neglected, then the approximate value of $\frac{\left(1+\frac{2 x}{3}\right)^{-4}(4+5 x)^{1 / 2}}{(9+x)^{3 / 2}}$, when $x=\frac{6}{371}$, is
A.
$\frac{1}{27}$
B.
$\frac{29}{378}$
C.
$\frac{3}{27}$
D.
$\frac{1}{14}$
2020
Q298
TS-EAMCET
MCQ
20 May 2026
The sum of the coefficients of $x^{-3 / 2}$ and $x^3$ in the expansion of $\sqrt{3+x}+\sqrt{5+x}$ when $3 < x< 5$, is
A.
$ =\frac{-18+3(5)^{-5 / 2}}{8} $
B.
$\frac{5^{-5 / 2}-18}{16}$
C.
$\frac{-6+\sqrt{5}}{6}$
D.
$\frac{5-\sqrt{6}}{6}$
2020
Q299
TS-EAMCET
MCQ
20 May 2026
If the 9th and 10th terms are the numerically greatest terms in the expansion of $(5 x-6 y)^n$ when $x=2 / 5$ and $y=1 / 2$, then the absolute value of the middle terms of that expansion is
A.
$14 C_8 6^7$
B.
$14 C_7 6^7$
C.
$15 C_7 6^7$
D.
$15 C_8 6^8$
2020
Q300
TS-EAMCET
MCQ
20 May 2026
$ 1-\frac{3}{16}+\frac{1 \cdot 4}{1 \cdot 2}\left(\frac{3}{16}\right)^2-\frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3}\left(\frac{3}{16}\right)^3+\ldots $
A.
$\left(\frac{15}{6}\right)^{3 / 8}$
B.
$\left(\frac{4}{5}\right)^{2 / 3}$
C.
$\left(\frac{7}{4}\right)^{1 / 16}$
D.
$\left(\frac{4}{15}\right)^{-2 / 5}$