Suppose $A$ and $B$ are the coefficients of $30^{\text {th }}$ and $12^{\text {th }}$ terms respectively in the binomial expansion of $(1+x)^{2 \mathrm{n}-1}$. If $2 \mathrm{~A}=5 \mathrm{~B}$, then n is equal to:
Solution
<p>$$\begin{aligned}
& A={ }^{2 n-1} C_{29} \quad B={ }^{2 n-1} C_{11} \\
& 2{ }^{2 n-1} C_{29}=5{ }^{2 n-1} C_{11} \\
& 2 \frac{(2 n-1)!}{29!(2 n-30)!}=5 \frac{(2 n-1)!}{(2 n-12)!11!} \\
& \frac{1}{29 \ldots 12 \cdot 5}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 2} \\
& \frac{1}{30 \cdot 29 \ldots 12}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 12} \\
& 2 n-12=30 \\
& n=21
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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