JEE MAIN Binomial Theorem Previous Year Questions (PYQs) – Page 1 of 8
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In the expansion of $\left(\sqrt[3]{2}+\dfrac{1}{\sqrt[3]{3}}\right)^{n},\ n\in\mathbb{N}$, if the ratio of $15^{\text{th}}$ term from the beginning to the $15^{\text{th}}$ term from the end is $\dfrac{1}{6}$, then the value of ${}^nC_3$ is
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Solution
$T_r=\binom{n}{r-1}a^{,n-r+1}b^{,r-1}$ for $(a+b)^n$.
$15^{\text{th}}$ from the beginning:
$T_{15}^{(beg)}=\binom{n}{14}a^{,n-14}b^{14}$.
$15^{\text{th}}$ from the end (swap $a,b$):
$T_{15}^{(end)}=\binom{n}{14}b^{,n-14}a^{14}$.
Given $\dfrac{T_{15}^{(beg)}}{T_{15}^{(end)}}=\dfrac{1}{6}$,
coefficients cancel:
$\left(\dfrac{a}{b}\right)^{n-28}=\dfrac{1}{6}$.
Here $a=2^{1/3},\ b=3^{-1/3}$
$\ \Rightarrow\ \dfrac{a}{b}=2^{1/3}\cdot 3^{1/3}=6^{1/3}$.
So $(6^{1/3})^{,n-28}=6^{-1}$
$\ \Rightarrow\ n-28=-3\ \Rightarrow\ n=25$.
Therefore, $\binom{n}{3}=\binom{25}{3}=\dfrac{25\cdot24\cdot23}{6}=2300$.
The coefficient of $x^{18}$ in the product
$(1+x)(1-x)^{10}(1+x+x^{2})^{9}$ is:
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Solution
$(1+x)(1-x)^{10} = (1-x^2)(1-x)^9$
$(1-x)^9(1+x+x^2)^9 = [(1-x)(1+x+x^2)]^9 = (1-x^3)^9$
$\Rightarrow (1+x)(1-x)^{10}(1+x+x^2)^9 = (1-x^2)(1-x^3)^9$
$(1-x^3)^9 = \sum_{k=0}^{9} \binom{9}{k}(-1)^k x^{3k}$
For $x^{18}$:
$3k = 18 \Rightarrow k=6$
Coefficient $= \binom{9}{6}(-1)^6 = \binom{9}{6} = 84$
If the coefficients of $x^4$, $x^5$, and $x^6$ in the expansion of $(1+x)^n$ are in arithmetic progression,
then the maximum value of $n$ is:
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Solution
The positive value of $\lambda$ for which the coefficient of $x^2$ in the expression
$x^2 \left( \sqrt{x} + \dfrac{\lambda}{x^2} \right)^{10}$ is $720$, is –
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Solution
The term independent of $x$ in the expansion of
$\left(\dfrac{1}{60} - \dfrac{x^{8}}{81}\right)\left(2x^{2} - \dfrac{3}{x^{2}}\right)^{6}$
is equal to:
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Solution
If the coefficients of x7 in ${\left( {{x^2} + {1 \over {bx}}} \right)^{11}}$ and x$-$7 in ${\left( {{x} - {1 \over {bx^2}}} \right)^{11}}$, b $\ne$ 0, are equal, then the value of b is equal to :
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Solution
If $\displaystyle \sum_{r=0}^{25} \left\{ {^{50}C_{r}} \cdot {^{\,50-r}C_{\,25-r}} \right\} = K \binom{50}{25}$, then $K$ is equal to :
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Solution
If
$\displaystyle \left(\frac{1}{^{15}C_0} + \frac{1}{^{15}C_1}\right)\left(\frac{1}{^{15}C_1} + \frac{1}{^{15}C_2}\right)\cdots\left(\frac{1}{^{15}C_{12}} + \frac{1}{^{15}C_{13}}\right) = \frac{\alpha^{13}}{^{14}C_0\cdot ^{14}C_1\cdots ^{14}C_{12}}$
then $30\alpha$ is equal to ______.
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Solution
$\prod_{r=0}^{12} \left(\frac{1}{^{15}C_r} + \frac{1}{^{15}C_{r+1}}\right)$
$= \prod_{r=0}^{12} \frac{16}{(r+1)\cdot \frac{15!}{r!(15-r)!}}$
$= \prod_{r=0}^{12} \frac{16}{^{15}C_r \cdot \frac{r+1}{15}}$
$= \prod_{r=0}^{12} \frac{16}{^{14}C_r}$
$= \frac{\left(\frac{16}{15}\right)^{13}}{^{14}C_0\cdot ^{14}C_1\cdots ^{14}C_{12}}$
$\Rightarrow \alpha = \frac{16}{15}$
$\Rightarrow 30\alpha = 32$
If $1^{2}\cdot{^{15}C_{1}}+2^{2}\cdot{^{15}C_{2}}+3^{2}\cdot{^{15}C_{3}}+\cdots+15^{2}\cdot{^{15}C_{15}}=2^{m}\cdot3^{n}\cdot5^{k}$, where $m,n,k\in\mathbb{N}$, then $m+n+k$ is equal to:
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Solution
If in the expansion of $(1+x)^p(1-x)^q$, the coefficients of $x$ and $x^2$ are $1$ and $-2$, respectively, then $p^2+q^2$ is equal to:
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Solution
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