JEE MAIN Binomial Theorem Previous Year Questions (PYQs) – Page 1 of 7
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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
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 $1^{2}\cdot{^{15}C_{1}}+2^{2}\cdot{^{15}C_{2}}+3^{2}\cdot{^{15}C_{3}}+\cdots+15^{2}\cdot{^{15}C_{15}}=2^{m}\cdot3^{n}\cdot5^{k}$, 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
If the constant term in the binomial expansionof ${\left( {\sqrt x - {k \over {{x^2}}}} \right)^{10}}$ is 405, then |k| equals :
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Solution
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