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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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
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, $\binom{n}{3}=\binom{25}{3}=\dfrac{25\cdot24\cdot23}{6}=2300$.
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