Binomial Theorem JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Binomial Theorem, free to read — no sign-in needed. The full chapter has 10 questions; sign in to attempt the remaining 5 in the exam simulator.

1. Q1JEE Advanced Adv 2025 (Paper 2)
   Let $a_0, a_1, \ldots, a_{23}$ be real numbers such that $\left(1+\frac{2}{5} x\right)^{23}=\sum_{i=0}^{23} a_i x^i$ for every real number $x$. let $a_r$ be the largest among the numbers $a_j$ for $0 \leq j \leq 23$. The the value of $r$ is ______
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   Answer: 6

   For $\mathrm{x}=1$ $\left(1+\frac{2}{5}\right)^{23}=a_0+a_1+a_2+\ldots+a_{23}$ for numerically greatest term $\frac{n+1}{1+\left|\frac{a}{b}\right|}=\frac{23+1}{1+\frac{5}{2}}=\frac{48}{7}$ $\Rightarrow\left[\frac{48}{7}\right]=6=\mathrm{m}$ (where [.] greatest integer function) so, $\mathrm{T}_7$ is numerical greatest term Hence $r=6$
2. Q2JEE Advanced Adv 2023 (Paper 1)
   Let $a$ and $b$ be two nonzero real numbers. If the coefficient of ${x}^{5}$ in the expansion of ${\left(a{x}^{2}+\dfrac{70}{27bx}\right)}^{4}$ is equal to the coefficient of ${x}^{-5}$ in the expansion of ${\left(ax-\dfrac{1}{b{x}^{2}}\right)}^{7}$, then the value of $2b$ is
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   Answer: 3

   Given, The coefficient of ${x}^{5}$ in ${\left(a{x}^{2}+\dfrac{70}{27bx}\right)}^{4}$ is equal to coefficient of ${x}^{-5}$ in ${\left(ax-\dfrac{1}{b{x}^{2}}\right)}^{7}$, Now finding the coefficient of ${x}^{5}$ in ${\left(a{x}^{2}+\dfrac{70}{27bx}\right)}^{4}$ we get, ${T}_{r+1}=Cr4{\left(a{x}^{2}\right)}^{4-r}\cdot {\left(\dfrac{70}{27bx}\right)}^{r}$ $\Rightarrow {T}_{r+1}=Cr4{\left(a\right)}^{4-r}\cdot {\left(\dfrac{70}{27b}\right)}^{r}\cdot {x}^{8-2r-r}$ So, $8-2r-r=5\Rightarrow r=1$ So, ${T}_{2}=C14{a}^{3}\cdot \dfrac{70}{27b}\cdot {x}^{5}$ Now finding the coefficient of ${x}^{-5}$ in ${\left(ax-\dfrac{1}{b{x}^{2}}\right)}^{7}$ we get, ${T}_{r+1}=Cr7{\left(ax\right)}^{7-r}\cdot {\left(\dfrac{1}{b{x}^{2}}\right)}^{r}$ $\Rightarrow {T}_{r+1}=Cr7{\left(a\right)}^{7-r}\cdot {\left(\dfrac{1}{b}\right)}^{r}\cdot {x}^{7-r-2r}$ So, $7-r-2r=-5\Rightarrow r=4$ So, ${T}_{5}=C47{\left(a\right)}^{3}\cdot {\left(\dfrac{1}{b}\right)}^{4}\cdot {x}^{-5}$ Now equating the coefficient of ${x}^{5}\&{x}^{-5}$ we get, $C47{\left(a\right)}^{3}\cdot {\left(\dfrac{1}{b}\right)}^{4}=C14{a}^{3}\cdot \dfrac{70}{27b}$ $\Rightarrow 35{a}^{3}\cdot {\left(\dfrac{1}{b}\right)}^{4}=4{a}^{3}\cdot \dfrac{70}{27b}$ $\Rightarrow {\left(\dfrac{1}{b}\right)}^{3}=4\times \dfrac{2}{27}$ $\Rightarrow b=\dfrac{3}{2}\Rightarrow 2b=3$
3. Q3JEE Advanced Adv 2020 (Paper 2)
   For nonnegative integers $s$ and $r$, let $\left(\begin{matrix}s \\ r\end{matrix}\right)=\left\{\begin{matrix}\dfrac{s!}{r!\left(s-r\right)!} & \text{if}r\leq s \\ 0 & \text{if}r\gt s\end{matrix}\right.$. For positive integers $m$ and $n,$ let $g\left(m,n\right)=\sum _{p=0}^{m+n}\dfrac{f\left(m,n,p\right)}{\left(\begin{matrix}n+p \\ p\end{matrix}\right)}$, where for any nonnegative integer $p$, $f\left(m,n,p\right)=\sum _{i=0}^{p}\left(\begin{matrix}m \\ i\end{matrix}\right)\left(\begin{matrix}n+i \\ p\end{matrix}\right)\left(\begin{matrix}p+n \\ p-i\end{matrix}\right)$. Then which of the following statements is/are TRUE?
   1. A.$g\left(m,n\right)=g\left(n,m\right)$ for all positive integers $m,n$
   2. B.$g\left(m,n+1\right)=g\left(m+1,n\right)$ for all positive integers $m,n$
   3. C.$g\left(2m,2n\right)=2g\left(m,n\right)$ for all positive integers $m,n$
   4. D.$g\left(2m,2n\right)={\left(g\left(m,n\right)\right)}^{2}$ for all positive intergers $m,n$
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   Answer: A,B,D

   $f\left(m,n,p\right)=\sum _{i=0}^{p}\left(\begin{matrix}m \\ i\end{matrix}\right)\left(\begin{matrix}n+i \\ p\end{matrix}\right)\left(\begin{matrix}p+n \\ p-i\end{matrix}\right)$ $=\sum _{i=0}^{p}Cim\cdot \dfrac{\left(n+i\right)!}{p!\left(n+i-p\right)!}\cdot \dfrac{(p+n)!}{\left(p-i\right)!\left(n+i\right)!}$ $=\sum _{i=0}^{p}Cim\cdot \dfrac{(p+n)!}{p!n!}\cdot \dfrac{n!}{\left(n+i-p\right)!\left(p-i\right)!}$ $=\sum _{i=0}^{p}Cim.Cpp+n\cdot Cp-in$ $Cpp+n\cdot \sum _{i=0}^{p}Cim\cdot Cp-in$ $\left(∵\sum _{i=0}^{p}Cim\cdot Cp-in=Cpm+n\right)$ $∴f\left(m,n,p\right)=Cpp+n.Cpm+n$ $\Rightarrow \dfrac{f\left(m,n,p\right)}{Cpp+n}=Cpm+n$. $g\left(m,n\right)=\sum _{p=0}^{m+n}Cpm+n={2}^{m+n}$ (A) $g(2m,2n)={2}^{2m+2n}$ $2g\left(m,n\right)=2.{2}^{m+n}$ (B) $g\left(m,n+1\right)={2}^{m+n+1}=g\left(m+1,n\right)$ (C) $g\left(2m,2n\right)={2}^{2m+2n}={\left({2}^{m+n}\right)}^{2}={\left(g\left(m,n\right)\right)}^{2}$ (D) $g\left(m,n\right)=g\left(n,m\right)$.
4. Q4JEE Advanced Adv 2019 (Paper 2)
   Suppose $\det ⁡\left[\begin{matrix}\sum _{k=0}^{n}k & \sum _{k=0}^{n}{C}_{k}n{k}^{2} \\ \sum _{k=0}^{n}{C}_{k}nk & \sum _{k=0}^{n}{C}_{k}n{3}^{k}\end{matrix}\right]=0,$ holds for some positive integer $n.$ Then $\sum _{k=0}^{n}\dfrac{{C}_{k}n}{k+1}$ equals
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   Answer: 6.2

   As given $\left|\begin{matrix}\sum _{k=0}^{n}k & \sum _{k=0}^{n}Cn{}_{k}{k}^{2} \\ \sum _{k=0}^{n}Cn{}_{k}k & \sum _{k=0}^{n}Cn{}_{k}{3}^{k}\end{matrix}\right|=0$ $\left(a\right)\sum _{k=0}^{n}k=\dfrac{n\left(n+1\right)}{2}$ $\left(b\right)\sum _{k=0}^{n}{C}_{k}n\left({k}^{2}\right)=\sum _{k=0}^{n}{\left({k}^{2}-k+k\right)}^{n}{C}_{k}$ $=\sum _{k=0}^{n}\left({k}^{2}-k\right){C}_{k}n+\sum _{k=0}^{n}k{C}_{k}n$ $=\sum _{k=0}^{n}k\left(k-1\right)\dfrac{n}{k}.\dfrac{n-1}{k-1}{C}_{k-2}n-2+\sum _{k=0}^{n}k\dfrac{n}{k}{C}_{k-1}n-1$ $=n\left(n-1\right)\sum _{k=0}^{n}{C}_{k-2}n-2+n\sum _{k=0}^{n}{C}_{k-1}n-1$ $=n\left(n-1\right){2}^{n-2}+n{2}^{n-1}=n{2}^{n-2}\left(n-1+2\right)=n\left(n+1\right){2}^{n-2}$ $\left(\begin{matrix}∴r{C}_{r}n=n{C}_{r-1}n-1 \\ ∴\sum _{r=0}^{n}{C}_{r}n={2}^{n}\end{matrix}\right)$ $\left(c\right)\sum _{k=0}^{n}{C}_{k}nk=\sum _{k=0}^{n}k{C}_{k}n=\sum _{k=0}^{n}k\dfrac{n}{k}{C}_{k-1}n-1$ $=n\sum _{k=0}^{n}{C}_{k-1}n-1=n{2}^{n-1}$ $\left(d\right)\sum _{k=0}^{n}{C}_{k}n{3}^{k}={C}_{0}n+{C}_{1}3+{C}_{2}n{3}^{2}+\ldots +{C}_{n}n{3}^{n}n$ $={\left(1+3\right)}^{n}$ $={4}^{n}$ now $\left|\begin{matrix}\dfrac{n\left(n+1\right)}{2} & n\left(n+1\right){2}^{n-2} \\ n{2}^{n-1} & {4}^{n}\end{matrix}\right|=0$ $n\left(n+1\right){2}^{2n-1}-{n}^{2}\left(n+1\right){2}^{2n-3}=0$ ${2}^{2n-1}-n{2}^{2n-3}=0$ $n=4$ Now $\sum _{k=0}^{4}\dfrac{{C}_{k}4}{k+1}=\dfrac{1}{5}\sum _{k=0}^{4}{C}_{k+1}5$ $\begin{matrix}=\dfrac{1}{5}\left({2}^{5}-1\right) & \left(∴\dfrac{{C}_{r}n}{r+1}=\dfrac{{C}_{r+1}n+1}{n+1}\right)\end{matrix}$ $\begin{matrix}=\dfrac{31}{5}=6.20 & \left(∴\sum _{r=1}^{n}{C}_{r}={2}^{r}-1n\right)\end{matrix}$
5. Q5JEE Advanced Adv 2018 (Paper 2)
   Let $X={\left({}^{10}{C}_{1}\right)}^{2}+2{\left({}^{10}{C}_{2}\right)}^{2}+3{\left({}^{10}{C}_{3}\right)}^{2}+\ldots +10{\left({}^{10}{C}_{10}\right)}^{2}$ , where ${}^{10}{C}_{r},r\in \left\{1,2,\ldots .,10\right\}$ denote binomial coefficients. Then, the value of $\dfrac{1}{1430}X$ is ______ .
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   Answer: 646

   $X=\sum _{r=0}^{n}r.{\left({}^{n}{C}_{r}\right)}^{2};n=10$ $X=n.\sum _{r=0}^{n}{}^{n}{C}_{r}.{}^{n-1}{C}_{r-1}$ $X=n.\sum _{r=1}^{n}{}^{n}{C}_{n-r}.{}^{n-1}{C}_{r-1}$ $X=n.{}^{2n-1}{C}_{n-1};n=10$ $X=10.{}^{19}{C}_{9}$ $\dfrac{X}{1430}=\dfrac{1}{143}.{}^{19}{C}_{9}$ $=646$