Mathematical Induction and Binomial Theorem
54 Questions
2025
JEE Advanced
Numerical
JEE Advanced 2025 Paper 2 Online
Let $a_0, a_1, \ldots, a_{23}$ be real numbers such that
$ \left(1+\frac{2}{5} x\right)^{23}=\sum\limits_{i=0}^{23} a_i x^i $
for every real number $x$. Let $a_r$ be the largest among the numbers $a_j$ for $0 \leq j \leq 23$. Then the value of $r$ is ____________.
Correct Answer: 6
Explanation:
For $\mathrm{x}=1$
$ \left(1+\frac{2}{5}\right)^{23}=a_0+a_1+a_2+\ldots+a_{23} $
for numerically greatest term
$ \begin{aligned} & \frac{\mathrm{n}+1}{1+\left|\frac{\mathrm{a}}{\mathrm{~b}}\right|}=\frac{23+1}{1+\frac{5}{2}}=\frac{48}{7} \\ & \Rightarrow\left[\frac{48}{7}\right]=6=\mathrm{m} \quad \text { (where [.] greatest integer function) } \end{aligned} $
so, $\mathrm{T}_7$ is numerical greatest term
Hence $\mathrm{r}=6$
2023
JEE Advanced
Numerical
JEE Advanced 2023 Paper 1 Online
Let $a$ and $b$ be two nonzero real numbers. If the coefficient of $x^5$ in the expansion of $\left(a x^2+\frac{70}{27 b x}\right)^4$ is equal to the coefficient of $x^{-5}$ in the expansion of $\left(a x-\frac{1}{b x^2}\right)^7$, then the value of $2 b$ is :
Correct Answer: 3
Explanation:
$
\begin{aligned}
& \mathrm{T}_{\mathrm{r}+1}={ }^4 \mathrm{C}_{\mathrm{r}}\left(\mathrm{a} \cdot \mathrm{x}^2\right)^{4-\mathrm{r}} \cdot\left(\frac{70}{27 \mathrm{bx}}\right)^{\mathrm{r}} \\\\
& ={ }^4 \mathrm{C}_{\mathrm{r}} \cdot \mathrm{a}^{4-\mathrm{r}} \cdot \frac{70^{\mathrm{r}}}{(27 b)^{\mathrm{r}}} \cdot x^{8-3 \mathrm{r}}
\end{aligned}
$
$ \begin{aligned} & \text { here } 8-3 r=5 \\\\ & 8-5=3 r \Rightarrow r=1 \\\\ & \therefore \text { coeff. }=4 \cdot a^3 \cdot \frac{70}{27 b} \end{aligned} $
$ \begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^7 \mathrm{C}_{\mathrm{r}}(\mathrm{ax})^{7-\mathrm{r}}\left(\frac{-1}{\mathrm{bx}^2}\right)^{\mathrm{r}} \\\\ & ={ }^7 \mathrm{C}_{\mathrm{r}} \cdot a^{7-\mathrm{r}}\left(\frac{-1}{\mathrm{~b}}\right)^{\mathrm{r}} \cdot \mathrm{x}^{7-3 \mathrm{r}} \\\\ & 7-3 \mathrm{r}=-5 \Rightarrow 12=3 \mathrm{r} \Rightarrow \mathrm{r}=4 \end{aligned} $
$ \begin{aligned} & \text { now } \frac{35 \mathrm{a}^3}{\mathrm{~b}^4}=\frac{280 \mathrm{a}^3}{27 \mathrm{~b}} \\\\ & \mathrm{~b}^3=\frac{35 \times 27}{280}=\mathrm{b}=\frac{3}{2} \Rightarrow 2 \mathrm{~b}=3 \end{aligned} $
$ \begin{aligned} & \text { here } 8-3 r=5 \\\\ & 8-5=3 r \Rightarrow r=1 \\\\ & \therefore \text { coeff. }=4 \cdot a^3 \cdot \frac{70}{27 b} \end{aligned} $
$ \begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^7 \mathrm{C}_{\mathrm{r}}(\mathrm{ax})^{7-\mathrm{r}}\left(\frac{-1}{\mathrm{bx}^2}\right)^{\mathrm{r}} \\\\ & ={ }^7 \mathrm{C}_{\mathrm{r}} \cdot a^{7-\mathrm{r}}\left(\frac{-1}{\mathrm{~b}}\right)^{\mathrm{r}} \cdot \mathrm{x}^{7-3 \mathrm{r}} \\\\ & 7-3 \mathrm{r}=-5 \Rightarrow 12=3 \mathrm{r} \Rightarrow \mathrm{r}=4 \end{aligned} $
$ \begin{aligned} & \text { now } \frac{35 \mathrm{a}^3}{\mathrm{~b}^4}=\frac{280 \mathrm{a}^3}{27 \mathrm{~b}} \\\\ & \mathrm{~b}^3=\frac{35 \times 27}{280}=\mathrm{b}=\frac{3}{2} \Rightarrow 2 \mathrm{~b}=3 \end{aligned} $
2018
JEE Advanced
Numerical
JEE Advanced 2018 Paper 2 Offline
Let $X = {({}^{10}{C_1})^2} + 2{({}^{10}{C_2})^2} + 3{({}^{10}{C_3})^2} + ... + 10{({}^{10}{C_{10}})^2}$,
where ${}^{10}{C_r}$, r $ \in ${1, 2, ..., 10} denote binomial coefficients. Then, the value of ${1 \over {1430}}X$ is ..........
where ${}^{10}{C_r}$, r $ \in ${1, 2, ..., 10} denote binomial coefficients. Then, the value of ${1 \over {1430}}X$ is ..........
Correct Answer: 646
Explanation:
We have,
$X = {({}^{10}{C_1})^2} + 2{({}^{10}{C_2})^2} + 3{({}^{10}{C_3})^2} + ... + 10{({}^{10}{C_{10}})^2}$
$ \Rightarrow X = \sum\limits_{r = 1}^{10} r {({}^{10}{C_r})^2}$
$ \Rightarrow X = \sum\limits_{r = 1}^{10} r\times {}^{10}{C_r}\times{}^{10}{C_r}$
$ \Rightarrow X = \sum\limits_{r = 1}^{10} r \times {}^{10}{C_r}\times {{10} \over r}\times{}^9{C_{r - 1}}$ $\left[ \,\because{{}^{n}{C_r} = {n \over r}{}^{n - 1}{C_{r - 1}}} \right]$
$ \Rightarrow X = 10\sum\limits_{r = 1}^{10} {{}^9{C_{r - 1}}} {}^{10}{C_r}$
$ \Rightarrow X = 10\sum\limits_{r = 1}^{10} {{}^9{C_{r - 1}}} {}^{10}{C_{10 - r}}$ [ $ \because $ ${}^n{C_r} = {}^n{C_{n - r}}$]
$ \Rightarrow X = 10 \times {}^{19}{C_9}$
[ $ \because $ ${}^{n - 1}{C_{r - 1}}{}^n{C_{n - r}} = {}^{2n - 1}{C_{n - 1}}$]
Now, ${1 \over {1430}}X = {{10 \times {}^{19}{C_9}} \over {1430}} = {{{}^{19}{C_9}} \over {143}} = {{{}^{19}{C_9}} \over {11 \times 13}}$
$ = {{19 \times 17 \times 16} \over 8} = 19 \times 34 = 646$
$X = {({}^{10}{C_1})^2} + 2{({}^{10}{C_2})^2} + 3{({}^{10}{C_3})^2} + ... + 10{({}^{10}{C_{10}})^2}$
$ \Rightarrow X = \sum\limits_{r = 1}^{10} r {({}^{10}{C_r})^2}$
$ \Rightarrow X = \sum\limits_{r = 1}^{10} r\times {}^{10}{C_r}\times{}^{10}{C_r}$
$ \Rightarrow X = \sum\limits_{r = 1}^{10} r \times {}^{10}{C_r}\times {{10} \over r}\times{}^9{C_{r - 1}}$ $\left[ \,\because{{}^{n}{C_r} = {n \over r}{}^{n - 1}{C_{r - 1}}} \right]$
$ \Rightarrow X = 10\sum\limits_{r = 1}^{10} {{}^9{C_{r - 1}}} {}^{10}{C_r}$
$ \Rightarrow X = 10\sum\limits_{r = 1}^{10} {{}^9{C_{r - 1}}} {}^{10}{C_{10 - r}}$ [ $ \because $ ${}^n{C_r} = {}^n{C_{n - r}}$]
$ \Rightarrow X = 10 \times {}^{19}{C_9}$
[ $ \because $ ${}^{n - 1}{C_{r - 1}}{}^n{C_{n - r}} = {}^{2n - 1}{C_{n - 1}}$]
Now, ${1 \over {1430}}X = {{10 \times {}^{19}{C_9}} \over {1430}} = {{{}^{19}{C_9}} \over {143}} = {{{}^{19}{C_9}} \over {11 \times 13}}$
$ = {{19 \times 17 \times 16} \over 8} = 19 \times 34 = 646$
2016
JEE Advanced
Numerical
JEE Advanced 2016 Paper 1 Offline
Let $m$ be the smallest positive integer such that the coefficient of ${x^2}$ in the expansion of ${\left( {1 + x} \right)^2} + {\left( {1 + x} \right)^3} + ........ + {\left( {1 + x} \right)^{49}} + {\left( {1 + mx} \right)^{50}}\,\,$ is $\left( {3n + 1} \right)\,{}^{51}{C_3}$ for some positive integer $n$. Then the value of $n$ is
Correct Answer: 5
Explanation:
It is given that the coefficient of $x^2$ in $(1+x)^2+(1+x)^3+$ $\cdots+(1+x)^{49}+(1+m x)^{50}$ is $(3 n+1){ }^{51} C_3$.
Now,
$ \begin{aligned} & { }^2 C_2+{ }^3 C_2+{ }^4 C_2+\cdots+{ }^{49} C_2+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^3 C_3+{ }^3 C_2+{ }^4 C_2+\cdots+{ }^{49} C_2+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3 \\\\ & \quad \quad\left(\text { as } n C r+n C r-1=n^{+1} C r\right) \\\\ & \Rightarrow{ }^4 C_3+{ }^4 C_2+\cdots+{ }^{49} C_2+m^{250} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^{49} C_3+{ }^{49} C_2+m{ }^{250} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^{50} C_3+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^{50} C_3+{ }^{50} C_2+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3+{ }^{50} C_2 \\\\ & \Rightarrow{ }^{51} C_3+m^2{ }^{50} C_2=3 n{ }^{51} C_3+{ }^{51} C_3+{ }^{50} C_2 \\\\ & \Rightarrow{ }^{-50} C_2+m^{250} C_2=3 n{ }^{51} C_3 \\\\ & \Rightarrow{ }^{50} C_2\left(m^2-1\right)=3 n \cdot \frac{51}{3}{ }^{50} C_2\left({ }^n C_r=\frac{n}{r} \quad{ }^{n-1} C_{r-1}\right) \\\\ & \Rightarrow 2\left(m^2-1\right)=51 \end{aligned} $
Therefore, $m^2=51 n+1$.
$ \begin{gathered} \Rightarrow n=\frac{m^2-1}{51} \\\\ \Rightarrow m=16 \Rightarrow n=5\left(m, n \in \mathrm{I}^{+}\right) \end{gathered} $
Now,
$ \begin{aligned} & { }^2 C_2+{ }^3 C_2+{ }^4 C_2+\cdots+{ }^{49} C_2+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^3 C_3+{ }^3 C_2+{ }^4 C_2+\cdots+{ }^{49} C_2+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3 \\\\ & \quad \quad\left(\text { as } n C r+n C r-1=n^{+1} C r\right) \\\\ & \Rightarrow{ }^4 C_3+{ }^4 C_2+\cdots+{ }^{49} C_2+m^{250} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^{49} C_3+{ }^{49} C_2+m{ }^{250} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^{50} C_3+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3 \\\\ & \Rightarrow{ }^{50} C_3+{ }^{50} C_2+m^2{ }^{50} C_2=(3 n+1){ }^{51} C_3+{ }^{50} C_2 \\\\ & \Rightarrow{ }^{51} C_3+m^2{ }^{50} C_2=3 n{ }^{51} C_3+{ }^{51} C_3+{ }^{50} C_2 \\\\ & \Rightarrow{ }^{-50} C_2+m^{250} C_2=3 n{ }^{51} C_3 \\\\ & \Rightarrow{ }^{50} C_2\left(m^2-1\right)=3 n \cdot \frac{51}{3}{ }^{50} C_2\left({ }^n C_r=\frac{n}{r} \quad{ }^{n-1} C_{r-1}\right) \\\\ & \Rightarrow 2\left(m^2-1\right)=51 \end{aligned} $
Therefore, $m^2=51 n+1$.
$ \begin{gathered} \Rightarrow n=\frac{m^2-1}{51} \\\\ \Rightarrow m=16 \Rightarrow n=5\left(m, n \in \mathrm{I}^{+}\right) \end{gathered} $
2013
JEE Advanced
Numerical
JEE Advanced 2013 Paper 1 Offline
The coefficient of three consecutive terms of ${\left( {1 + x} \right)^{n + 5}}$ are in the ratio $5:10:14.$ Then $n$ =
Correct Answer: 6
Explanation:
We know $(1+x)^{n+5}=\sum_\limits{r=0}^{n+5}{ }^{n+5} \mathrm{C}_r \cdot x^r$
Given, the coefficients of three consecutive terms of $(1+x)^{n+5}$ are in the ratio $5: 10: 14$.
$\begin{aligned} & \Rightarrow{ }^{n+5} \mathrm{C}_r:{ }^{n+5} \mathrm{C}_{r+1}:{ }^{n+5} \mathrm{C}_{r+2}=5: 10: 14 \\ & \Rightarrow \frac{{ }^{n+5} \mathrm{C}_{r+1}}{{ }^{n+5} \mathrm{C}_r}=\frac{10}{5} \text { and } \frac{{ }^{n+5} \mathrm{C}_{r+2}}{{ }^{n+5} \mathrm{C}_{r+1}}=\frac{14}{10} \\ & \Rightarrow{ }^{n+5} \mathrm{C}_{r+1}=2 .{ }^{n+5} \mathrm{C}_r \text { and } 5 .{ }^{n+5} \mathrm{C}_{r+2}=7 .{ }^{n+5} \mathrm{C}_{r+1} \end{aligned}$
$ \Rightarrow {{\left| \!{\underline {\, {n + 5} \,}} \right. } \over {\left| \!{\underline {\, {r + 1} \,}} \right. \,\left| \!{\underline {\, {n - r + 4} \,}} \right. }} = 2{{\left| \!{\underline {\, {n + 5} \,}} \right. } \over {\left| \!{\underline {\, r \,}} \right. \,\left| \!{\underline {\, {n - r + 5} \,}} \right. }}$ and
$\,\,\,\,\,\,\,{{5\left| \!{\underline {\, {n + 5} \,}} \right. } \over {\left| \!{\underline {\, {r + 2} \,}} \right. \,\left| \!{\underline {\, {n - r + 3} \,}} \right. }} = {{7\left| \!{\underline {\, {n + 5} \,}} \right. } \over {\left| \!{\underline {\, {r + 1} \,}} \right. \,\left| \!{\underline {\, {n - r + 4} \,}} \right. }}$
$\begin{aligned} \Rightarrow \quad & \frac{1}{r+1}=\frac{2}{n-r+5} \text { and } \frac{5}{r+2}=\frac{7}{n-r+4} \\ \Rightarrow \quad & n-r+5=2 r+2 \text { and } \\ & 5 n-5 r+20=7 r+14 \\ \Rightarrow \quad & n=3 r-3=\frac{12 r-6}{5} \\ \Rightarrow \quad & n=3(r-1) \text { and } 15 r-15=12 r-6 \\ \Rightarrow \quad & r=3 \text { and } n=6 \\ \Rightarrow \quad & n=6 \end{aligned}$
Hints :
(i) Recall $(1+x)^m=\sum_\limits{r=1}^m{ }^m C_r \cdot x^r$
(ii) The coefficients of three consecutive terms $(1+x)^m$ are ${ }^m \mathrm{C}_r,{ }^m \mathrm{C}_{r+1},{ }^m \mathrm{C}_{r+2}$.
2020
JEE Advanced
MSQ
JEE Advanced 2020 Paper 2 Offline
For non-negative integers s and r, let
$\left( {\matrix{ s \cr r \cr } } \right) = \left\{ {\matrix{ {{{s!} \over {r!(s - r)!}}} & {if\,r \le \,s,} \cr 0 & {if\,r\, > \,s} \cr } } \right.$
For positive integers m and n, let
$g(m,\,n) = \sum\limits_{p = 0}^{m + n} {{{f(m,n,p)} \over {\left( {\matrix{ {n + p} \cr p \cr } } \right)}}} $
where for any non-negative integer p,
$f(m,n,p) = \sum\limits_{i = 0}^p {\left( {\matrix{ m \cr i \cr } } \right)\left( {\matrix{ {n + i} \cr p \cr } } \right)\left( {\matrix{ {p + n} \cr {p - i} \cr } } \right)} $
Then which of the following statements is/are TRUE?
$\left( {\matrix{ s \cr r \cr } } \right) = \left\{ {\matrix{ {{{s!} \over {r!(s - r)!}}} & {if\,r \le \,s,} \cr 0 & {if\,r\, > \,s} \cr } } \right.$
For positive integers m and n, let
$g(m,\,n) = \sum\limits_{p = 0}^{m + n} {{{f(m,n,p)} \over {\left( {\matrix{ {n + p} \cr p \cr } } \right)}}} $
where for any non-negative integer p,
$f(m,n,p) = \sum\limits_{i = 0}^p {\left( {\matrix{ m \cr i \cr } } \right)\left( {\matrix{ {n + i} \cr p \cr } } \right)\left( {\matrix{ {p + n} \cr {p - i} \cr } } \right)} $
Then which of the following statements is/are TRUE?
A.
g(m, n) = g(n, m) for all positive integers m, n
B.
g(m, n + 1) = g(m + 1, n) for all positive integers m, n
C.
g(2m, 2n) = 2g(m, n) for all positive integers m, n
D.
g(2m, 2n) = (g(m, n))2 for all positive integers m, n
2014
JEE Advanced
MCQ
JEE Advanced 2014 Paper 2 Offline
Coefficient of ${x^{11}}$ in the expansion of ${\left( {1 + {x^2}} \right)^4}{\left( {1 + {x^3}} \right)^7}{\left( {1 + {x^4}} \right)^{12}}$ is
A.
1051
B.
1106
C.
1113
D.
1120
2010
JEE Advanced
MCQ
IIT-JEE 2010 Paper 2 Offline
For $r = 0,\,1,....,$ let ${A_r},\,{B_r}$ and ${C_r}$ denote, respectively, the coefficient of ${X^r}$ in the expansions of ${\left( {1 + x} \right)^{10}},$ ${\left( {1 + x} \right)^{20}}$ and ${\left( {1 + x} \right)^{30}}.$
Then $\sum\limits_{r = 1}^{10} {{A_r}\left( {{B_{10}}{B_r} - {C_{10}}{A_r}} \right)} $ is equal to
Then $\sum\limits_{r = 1}^{10} {{A_r}\left( {{B_{10}}{B_r} - {C_{10}}{A_r}} \right)} $ is equal to
A.
$\left( {{B_{10}} - {C_{10}}} \right)$
B.
${A_{10}}\left( {{B^2}_{10}{C_{10}}{A_{10}}} \right)$
C.
$0$
D.
${{C_{10}} - {B_{10}}}$
2005
JEE Advanced
MCQ
IIT-JEE 2005 Screening
The value of $$\left( {\matrix{
{30} \cr
0 \cr
} } \right)\left( {\matrix{
{30} \cr
{10} \cr
} } \right) - \left( {\matrix{
{30} \cr
1 \cr
} } \right)\left( {\matrix{
{30} \cr
{11} \cr
} } \right) + \left( {\matrix{
{30} \cr
2 \cr
} } \right)\left( {\matrix{
{30} \cr
{12} \cr
} } \right)....... + \left( {\matrix{
{30} \cr
{20} \cr
} } \right)\left( {\matrix{
{30} \cr
{30} \cr
} } \right)$$
is where $\left( {\matrix{ n \cr r \cr } } \right) = {}^n{C_r}$
is where $\left( {\matrix{ n \cr r \cr } } \right) = {}^n{C_r}$
A.
$\left( {\matrix{
{30} \cr
{10} \cr
} } \right)$
B.
$\left( {\matrix{
{30} \cr
{15} \cr
} } \right)$
C.
$\left( {\matrix{
{60} \cr
{30} \cr
} } \right)$
D.
$\left( {\matrix{
{31} \cr
{10} \cr
} } \right)$
2004
JEE Advanced
MCQ
IIT-JEE 2004 Screening
If ${}^{n - 1}{C_r} = \left( {{k^2} - 3} \right)\,{}^n{C_{r + 1,}}$ then $k \in $
A.
$\left( { - \infty , - 2} \right)$
B.
$\left[ {2,\infty } \right)$
C.
$\left[ { - \sqrt 3 ,\sqrt 3 } \right]$
D.
$\left( {\sqrt 3 ,2} \right]$
2003
JEE Advanced
MCQ
IIT-JEE 2003 Screening
Coefficient of ${t^{24}}$ in ${\left( {1 + {t^2}} \right)^{12}}\left( {1 + {t^{12}}} \right)\left( {1 + {t^{24}}} \right)$ is
A.
${}^{12}{C_6} + 3$
B.
${}^{12}{C_6} + 1$
C.
${}^{12}{C_6}$
D.
${}^{12}{C_6} + 2$
2002
JEE Advanced
MCQ
IIT-JEE 2002 Screening
The sum $\sum\limits_{i = 0}^m {\left( {\matrix{
{10} \cr
i \cr
} } \right)\left( {\matrix{
{20} \cr
{m - i} \cr
} } \right),\,\left( {where\left( {\matrix{
p \cr
q \cr
} } \right) = 0\,\,if\,\,p < q} \right)} $ is maximum when $m$ is
A.
5
B.
10
C.
15
D.
20
2001
JEE Advanced
MCQ
IIT-JEE 2001 Screening
In the binomial expansion of ${\left( {a - b} \right)^n},\,n \ge 5,$ the sum of the ${5^{th}}$ and ${6^{th}}$ terms is zero. Then $a/b$ equals
A.
$\left( {n - 5} \right)/6$
B.
$\left( {n - 4} \right)/5$
C.
$5/\left( {n - 4} \right)$
D.
$6/\left( {n - 5} \right)$
2000
JEE Advanced
MCQ
IIT-JEE 2000 Screening
For $2 \le r \le n,\,\,\,\,\left( {\matrix{
n \cr
r \cr
} } \right) + 2\left( {\matrix{
n \cr
{r - 1} \cr
} } \right) + \left( {\matrix{
n \cr
{r - 2} \cr
} } \right) = $
A.
$\left( {\matrix{
{n + 1} \cr
{r - 1} \cr
} } \right)$
B.
$2\left( {\matrix{
{n + 1} \cr
{r + 1} \cr
} } \right)$
C.
$2\left( {\matrix{
{n + 2} \cr
r \cr
} } \right)$
D.
$\left( {\matrix{
{n + 2} \cr
r \cr
} } \right)$
1999
JEE Advanced
MCQ
IIT-JEE 1999
If in the expansion of ${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$ the coefficients of $x$ and ${x^2}$ are $3$ and $-6$ respectively, then $m$ is
A.
6
B.
9
C.
12
D.
24
1998
JEE Advanced
MCQ
IIT-JEE 1998
If ${a_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}},\,\,\,then\,\,\,\sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}}} } $ equals
A.
$\left( {n - 1} \right){a_n}$
B.
$n{a_n}$
C.
${1 \over 2}n{a_n}$
D.
None of the above
1992
JEE Advanced
MCQ
IIT-JEE 1992
The expansion ${\left( {x + {{\left( {{x^3} - 1} \right)}^{{1 \over 2}}}} \right)^5} + {\left( {x - {{\left( {{x^3} - 1} \right)}^{{1 \over 2}}}} \right)^5}$ is a polynomial of degree
A.
5
B.
6
C.
7
D.
8
1986
JEE Advanced
MCQ
IIT-JEE 1986
If ${C_r}$ stands for ${}^n{C_r},$ then the sum of the series
${{2\left( {{n \over 2}} \right){\mkern 1mu} !{\mkern 1mu} \left( {{n \over 2}} \right){\mkern 1mu} !} \over {n!}}\left[ {C_0^2 - 2C_1^2 + 3C_2^2 - } \right......... + {\left( { - 1} \right)^n}\left( {n + 1} \right)C_n^2\mathop ]\limits^ \sim \,,$
where $n$ is an even positive integer, is equal to
where $n$ is an even positive integer, is equal to
A.
0
B.
${\left( { - 1} \right)^{n/2}}\left( {n + 1} \right)$
C.
${\left( { - 1} \right)^{n/2}}\left( {n + 2} \right)$
D.
${\left( { - 1} \right)^n}n$
1983
JEE Advanced
MCQ
IIT-JEE 1983
Given positive integers $r > 1,\,n > 2$ and that the coefficient of $\left( {3r} \right)$th and $\left( {r + 2} \right)$th terms in the binomial expansion of ${\left( {1 + x} \right)^{2n}}$ are equal. Then
A.
$n = 2r$
B.
$n = 2r + 1$
C.
$n = 3r$
D.
none of these
1983
JEE Advanced
MCQ
IIT-JEE 1983
The coefficient of ${x^4}$ in ${\left( {{x \over 2} - {3 \over {{x^2}}}} \right)^{10}}$ is
A.
${{{405} \over {256}}}$
B.
${{{504} \over {259}}}$
C.
${{{450} \over {263}}}$
D.
none of these
2003
JEE Advanced
Numerical
IIT-JEE 2003
Prove that
${2^k}\left( {\matrix{ n \cr 0 \cr } } \right)\left( {\matrix{ n \cr k \cr } } \right) - {2^{^{k - 1}\left( {\matrix{ n \cr 2 \cr } } \right)}}\left( {\matrix{ n \cr 1 \cr } } \right)\left( {\matrix{ {n - 1} \cr {k - 1} \cr } } \right)$
$ + {2^{k - 2}}\left( {\matrix{ {n - 2} \cr {k - 2} \cr } } \right) - .....{\left( { - 1} \right)^k}\left( {\matrix{ n \cr k \cr } } \right)\left( {\matrix{ {n - k} \cr 0 \cr } } \right) = {\left( {\matrix{ n \cr k \cr } } \right)^ \cdot }$
${2^k}\left( {\matrix{ n \cr 0 \cr } } \right)\left( {\matrix{ n \cr k \cr } } \right) - {2^{^{k - 1}\left( {\matrix{ n \cr 2 \cr } } \right)}}\left( {\matrix{ n \cr 1 \cr } } \right)\left( {\matrix{ {n - 1} \cr {k - 1} \cr } } \right)$
$ + {2^{k - 2}}\left( {\matrix{ {n - 2} \cr {k - 2} \cr } } \right) - .....{\left( { - 1} \right)^k}\left( {\matrix{ n \cr k \cr } } \right)\left( {\matrix{ {n - k} \cr 0 \cr } } \right) = {\left( {\matrix{ n \cr k \cr } } \right)^ \cdot }$
Correct Answer: Solve it.
2002
JEE Advanced
Numerical
IIT-JEE 2002
Use mathematical induction to show that
${\left( {25} \right)^{n + 1}} - 24n + 5735$ is divisible by ${\left( {24} \right)^2}$ for all $ = n = 1,2,...$
${\left( {25} \right)^{n + 1}} - 24n + 5735$ is divisible by ${\left( {24} \right)^2}$ for all $ = n = 1,2,...$
Correct Answer: Solve it.
2000
JEE Advanced
Numerical
IIT-JEE 2000
A coin probability $p$ of showing head when tossed. It is tossed $n$ times. Let ${p_n}$ denote the probability that no two (or more) consecutive heads occur. Prove that ${p_1} = 1,\,\,{p_2} = 1 - {p^2}$ and ${p_n} = \left( {1 - p} \right).\,\,{p_{n - 1}} + p\left( {1 - p} \right){p_{n - 2}}$ for all $n \ge 3.$
Prove by induction on, that ${p_n} = A{\alpha ^n} + B{\beta ^n}$ for all $n \ge 1,$ where $\alpha $ and $\beta $ are the roots of quadratic equation ${x^2} - \left( {1 - p} \right)x - p\left( {1 - p} \right) = 0$ and $A = {{{p^2} + \beta - 1} \over {\alpha \beta - {\alpha ^2}}},B = {{{p^2} + \alpha - 1} \over {\alpha \beta - {\beta ^2}}}.$
Correct Answer: Solve it.
2000
JEE Advanced
Numerical
IIT-JEE 2000
For any positive integer $m$, $n$ (with $n \ge m$), let $\left( {\matrix{
n \cr
m \cr
} } \right) = {}^n{C_m}$
Prove that $\left( {\matrix{ n \cr m \cr } } \right) + \left( {\matrix{ {n - 1} \cr m \cr } } \right) + \left( {\matrix{ {n - 2} \cr m \cr } } \right) + ........ + \left( {\matrix{ m \cr m \cr } } \right) = \left( {\matrix{ {n + 1} \cr {m + 2} \cr } } \right)$
Prove that $\left( {\matrix{ n \cr m \cr } } \right) + \left( {\matrix{ {n - 1} \cr m \cr } } \right) + \left( {\matrix{ {n - 2} \cr m \cr } } \right) + ........ + \left( {\matrix{ m \cr m \cr } } \right) = \left( {\matrix{ {n + 1} \cr {m + 2} \cr } } \right)$
Hence or otherwise, prove that $\left( {\matrix{ n \cr m \cr } } \right) + 2\left( {\matrix{ {n - 1} \cr m \cr } } \right) + 3\left( {\matrix{ {n - 2} \cr m \cr } } \right) + ........ + \left( {n - m + 1} \right)\left( {\matrix{ m \cr m \cr } } \right) = \left( {\matrix{ {n + 2} \cr {m + 2} \cr } } \right).$.
Correct Answer: Solve it.
2000
JEE Advanced
Numerical
IIT-JEE 2000
Let $a,\,b,\,c$ be possitive real numbers such that ${b^2} - 4ac > 0$ and let ${\alpha _1} = c.$ Prove by induction that ${\alpha _{n + 1}} = {{a\alpha _n^2} \over {\left( {{b^2} - 2a\left( {{\alpha _1} + {\alpha _2} + ... + {\alpha _n}} \right)} \right)}}$ is well-defined and
${\alpha _{n + 1}} < {{{\alpha _n}} \over 2}$ for all $n = 1,2,....$ (Here, 'well-defined' means that the denominator in the expression for ${\alpha _{n + 1}}$ is not zero.)
${\alpha _{n + 1}} < {{{\alpha _n}} \over 2}$ for all $n = 1,2,....$ (Here, 'well-defined' means that the denominator in the expression for ${\alpha _{n + 1}}$ is not zero.)
Correct Answer: Solve it.
2000
JEE Advanced
Numerical
IIT-JEE 2000
For every possitive integer $n$, prove that
$\sqrt {\left( {4n + 1} \right)} < \sqrt n + \sqrt {n + 1} < \sqrt {4n + 2}.$
Hence or otherwise, prove that $\left[ {\sqrt n + \sqrt {\left( {n + 1} \right)} } \right] = \left[ {\sqrt {4n + 1} \,\,} \right],$
where $\left[ x \right]$ denotes the gratest integer not exceeding $x$.
$\sqrt {\left( {4n + 1} \right)} < \sqrt n + \sqrt {n + 1} < \sqrt {4n + 2}.$
Hence or otherwise, prove that $\left[ {\sqrt n + \sqrt {\left( {n + 1} \right)} } \right] = \left[ {\sqrt {4n + 1} \,\,} \right],$
where $\left[ x \right]$ denotes the gratest integer not exceeding $x$.
Correct Answer: Solve it.
1999
JEE Advanced
Numerical
IIT-JEE 1999
Let $n$ be any positive integer. Prove that
$$\sum\limits_{k = 0}^m {{{\left( {\matrix{
{2n - k} \cr
k \cr
} } \right)} \over {\left( {\matrix{
{2n - k} \cr
n \cr
} } \right)}}.{{\left( {2n - 4k + 1} \right)} \over {\left( {2n - 2k + 1} \right)}}{2^{n - 2k}} = {{\left( {\matrix{
n \cr
m \cr
} } \right)} \over {\left( {\matrix{
{2n - 2m} \cr
{n - m} \cr
} } \right)}}{2^{n - 2m}}} $$
for each non-be gatuve integer $m \le n.$ $\,\left( {Here\left( {\matrix{ p \cr q \cr } } \right) = {}^p{C_q}} \right).$
Correct Answer: Solve it.
1998
JEE Advanced
Numerical
IIT-JEE 1998
Let $p$ be a prime and $m$ a positive integer. By mathematical induction on $m$, or otherwise, prove that whenever $r$ is an integer such that $p$ does not divide $r$, $p$ divides ${}^{np}{C_r},$
[Hint: You may use the fact that ${\left( {1 + x} \right)^{\left( {m + 1} \right)p}} = {\left( {1 + x} \right)^p}{\left( {1 + x} \right)^{mp}}$]
Correct Answer: Solve it.
1997
JEE Advanced
Numerical
IIT-JEE 1997
Let $0 < {A_i} < n$ for $i = 1,\,2....,\,n.$ Use mathematical induction to prove that
$$\sin {A_1} + \sin {A_2}....... + \sin {A_n} \le n\,\sin \,\,\left( {{{{A_1} + {A_2} + ...... + {A_n}} \over n}} \right)$$
where $ \ge 1$ is a natural number. {You may use the fact that $p\sin x + \left( {1 - p} \right)\sin y \le \sin \left[ {px + \left( {1 - p} \right)y} \right],$ where $0 \le p \le 1$ and $0 \le x,y \le \pi .$}
Correct Answer: Solve it.
1996
JEE Advanced
Numerical
IIT-JEE 1996
Using mathematical induction prove that for every integer $n \ge 1,\,\,\left( {{3^{2n}} - 1} \right)$ is divisible by ${2^{n + 2}}$ but not by ${2^{n + 3}}$.
Correct Answer: Solve it.
1994
JEE Advanced
Numerical
IIT-JEE 1994
If $x$ is not an integral multiple of $2\pi $ use mathematical induction to prove that :
$$\cos x + \cos 2x + .......... + \cos nx = \cos {{n + 1} \over 2}x\sin {{nx} \over 2}\cos ec{x \over 2}$$
Correct Answer: Solve it.
1994
JEE Advanced
Numerical
IIT-JEE 1994
Let $n$ be a positive integer and ${\left( {1 + x + {x^2}} \right)^n} = {a_0} + {a_1}x + ............ + {a_{2n}}{x^{2n}}$
Show that $a_0^2 - a_1^2 + a_2^2...... + {a_{2n}}{}^2 = {a_n}$
Show that $a_0^2 - a_1^2 + a_2^2...... + {a_{2n}}{}^2 = {a_n}$
Correct Answer: Solve it.
1993
JEE Advanced
Numerical
IIT-JEE 1993
Using mathematical induction, prove that
${\tan ^{ - 1}}\left( {1/3} \right) + {\tan ^{ - 1}}\left( {1/7} \right) + ........{\tan ^{ - 1}}\left\{ {1/\left( {{n^2} + n + 1} \right)} \right\} = {\tan ^{ - 1}}\left\{ {n/\left( {n + 2} \right)} \right\}$
${\tan ^{ - 1}}\left( {1/3} \right) + {\tan ^{ - 1}}\left( {1/7} \right) + ........{\tan ^{ - 1}}\left\{ {1/\left( {{n^2} + n + 1} \right)} \right\} = {\tan ^{ - 1}}\left\{ {n/\left( {n + 2} \right)} \right\}$
Correct Answer: Solve it.
1993
JEE Advanced
Numerical
IIT-JEE 1993
Prove that $\sum\limits_{r = 1}^k {{{\left( { - 3} \right)}^{r - 1}}\,\,{}^{3n}{C_{2r - 1}} = 0,} $ where $k = \left( {3n} \right)/2$ and $n$ is an even positive integer.
Correct Answer: Solve it.
1992
JEE Advanced
Numerical
IIT-JEE 1992
If $\sum\limits_{r = 0}^{2n} {{a_r}{{\left( {x - 2} \right)}^r}\,\, = \sum\limits_{r = 0}^{2n} {{b_r}{{\left( {x - 3} \right)}^r}} } $ and ${a_k} = 1$ for all $k \ge n,$ then show that ${b_n} = {}^{2n + 1}{C_{n + 1}}$
Correct Answer: Solve it.
1992
JEE Advanced
Numerical
IIT-JEE 1992
Let $p \ge 3$ be an integer and $\alpha $, $\beta $ be the roots of ${x^2} - \left( {p + 1} \right)x + 1 = 0$ using mathematical induction show that ${\alpha ^n} + {\beta ^n}.$
(i) is an integer and (ii) is not divisible by $p$
(i) is an integer and (ii) is not divisible by $p$
Correct Answer: Solve it.
1991
JEE Advanced
Numerical
IIT-JEE 1991
Using induction or otherwise, prove that for any non-negative integers $m$, $n$, $r$ and $k$ ,
$\sum\limits_{m = 0}^k {\left( {n - m} \right)} {{\left( {r + m} \right)!} \over {m!}} = {{\left( {r + k + 1} \right)!} \over {k!}}\left[ {{n \over {r + 1}} - {k \over {r + 2}}} \right]$
$\sum\limits_{m = 0}^k {\left( {n - m} \right)} {{\left( {r + m} \right)!} \over {m!}} = {{\left( {r + k + 1} \right)!} \over {k!}}\left[ {{n \over {r + 1}} - {k \over {r + 2}}} \right]$
Correct Answer: Solve it.
1990
JEE Advanced
Numerical
IIT-JEE 1990
Prove that ${{{n^7}} \over 7} + {{{n^5}} \over 5} + {{2{n^3}} \over 3} - {n \over {105}}$ is an integer for every positive integer $n$
Correct Answer: Solve it.
1989
JEE Advanced
Numerical
IIT-JEE 1989
Prove that
${C_0} - {2^2}{C_1} + {3^2}{C_2}\,\, - \,..... + {\left( { - 1} \right)^n}{\left( {n + 1} \right)^2}{C_n} = 0,\,\,\,\,n > 2,\,\,$ where ${C_r} = {}^n{C_r}.$
${C_0} - {2^2}{C_1} + {3^2}{C_2}\,\, - \,..... + {\left( { - 1} \right)^n}{\left( {n + 1} \right)^2}{C_n} = 0,\,\,\,\,n > 2,\,\,$ where ${C_r} = {}^n{C_r}.$
Correct Answer: Solve it.
1989
JEE Advanced
Numerical
IIT-JEE 1989
Using mathematical induction, prove that ${}^m{C_0}{}^n{C_k} + {}^m{C_1}{}^n{C_{k - 1}}\,\,\, + .....{}^m{C_k}{}^n{C_0} = {}^{\left( {m + n} \right)}{C_k},$
where $m,\,n,\,k$ are positive integers, and ${}^p{C_q} = 0$ for $p < q.$
where $m,\,n,\,k$ are positive integers, and ${}^p{C_q} = 0$ for $p < q.$
Correct Answer: Solve it.
1988
JEE Advanced
Numerical
IIT-JEE 1988
Let $R$ $ = {\left( {5\sqrt 5 + 11} \right)^{2n + 1}}$ and $f = R - \left[ R \right],$ where [ ] denotes the greatest integer function. Prove that $Rf = {4^{2n + 4}}$
Correct Answer: Solve it.
1987
JEE Advanced
Numerical
IIT-JEE 1987
Prove by mathematical induction that $ - 5 - {{\left( {2n} \right)!} \over {{2^{2n}}{{\left( {n!} \right)}^2}}} \le {1 \over {{{\left( {3n + 1} \right)}^{1/2}}}}$ for all positive integers $n$.
Correct Answer: Solve it.
1985
JEE Advanced
Numerical
IIT-JEE 1985
Use method of mathematical induction ${2.7^n} + {3.5^n} - 5$ is divisible by $24$ for all $n > 0$
Correct Answer: Solve it.
1984
JEE Advanced
Numerical
IIT-JEE 1984
If $p$ be a natural number then prove that ${p^{n + 1}} + {\left( {p + 1} \right)^{2n - 1}}$ is divisible by ${p^2} + p + 1$ for every positive integer $n$.
Correct Answer: Solve it.
1984
JEE Advanced
Numerical
IIT-JEE 1984
Given ${s_n} = 1 + q + {q^2} + ...... + {q^2};$
${S_n} = 1 + {{q + 1} \over 2} + {\left( {{{q + 1} \over 2}} \right)^2} + ........ + {\left( {{{q + 1} \over 2}} \right)^n}\,\,\,,q \ne 1$
Prove that ${}^{n + 1}{C_1} + {}^{n + 1}{C_2}{s_1} + {}^{n + 1}{C_3}{s_2} + ..... + {}^{n + 1}{C_n}{s_n} = {2^n}{S_n}$
${S_n} = 1 + {{q + 1} \over 2} + {\left( {{{q + 1} \over 2}} \right)^2} + ........ + {\left( {{{q + 1} \over 2}} \right)^n}\,\,\,,q \ne 1$
Prove that ${}^{n + 1}{C_1} + {}^{n + 1}{C_2}{s_1} + {}^{n + 1}{C_3}{s_2} + ..... + {}^{n + 1}{C_n}{s_n} = {2^n}{S_n}$
Correct Answer: Solve it.
1983
JEE Advanced
Numerical
IIT-JEE 1983
If ${\left( {1 + x} \right)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ..... + {C_n}{x^n}$ then show that the sum of the products of the ${C_i}s$ taken two at a time, represented $\sum\limits_{0 \le i < j \le n} {\sum {{C_i}{C_j}} } $ is equal to ${2^{2n - 1}} - {{\left( {2n} \right)!} \over {2{{\left( {n!} \right)}^2}}}$
Correct Answer: Solve it.
1983
JEE Advanced
Numerical
IIT-JEE 1983
Use mathematical Induction to prove : If $n$ is any odd positive integer, then $n\left( {{n^2} - 1} \right)$ is divisible by 24.
Correct Answer: Solve it.
1982
JEE Advanced
Numerical
IIT-JEE 1982
Prove that ${7^{2n}} + \left( {{2^{3n - 3}}} \right)\left( {3n - 1} \right)$ is divisible by 25 for any natural number $n$.
Correct Answer: Solve it.
1979
JEE Advanced
Numerical
IIT-JEE 1979
Given that ${C_1} + 2{C_2}x + 3{C_3}{x^2} + ......... + 2n{C_{2n}}{x^{2n - 1}} = 2n{\left( {1 + x} \right)^{2n - 1}}$
where ${C_r} = {{\left( {2n} \right)\,!} \over {r!\left( {2n - r} \right)!}}\,\,\,\,\,r = 0,1,2,\,............,2n$
Prove that ${C_1}^2 - 2{C_2}^2 + 3{C_3}^2 - ............ - 2n{C_{2n}}^2 = {\left( { - 1} \right)^n}n{C_n}.$
where ${C_r} = {{\left( {2n} \right)\,!} \over {r!\left( {2n - r} \right)!}}\,\,\,\,\,r = 0,1,2,\,............,2n$
Prove that ${C_1}^2 - 2{C_2}^2 + 3{C_3}^2 - ............ - 2n{C_{2n}}^2 = {\left( { - 1} \right)^n}n{C_n}.$
Correct Answer: 5
1997
JEE Advanced
Numerical
IIT-JEE 1997
The sum of the rational terms in the expansion of ${\left( {\sqrt 2 + {3^{1/5}}} \right)^{10}}$ is ...............
Correct Answer: 41