Suppose A and B are the coefficients of 30^ th and 12^ th terms respectively in the binomi…

Question

Suppose $A$ and $B$ are the coefficients of $30^{	ext {th }}$ and $12^{	ext {th }}$ 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:

Accepted Answer

$$\begin{aligned}
& A={ }^{2 n-1} C_{29} \quad B={ }^{2 n-1} C_{11} \
& 2{ }^{2 n-1} C_{29}=5{ }^{2 n-1} C_{11} \
& 2 \frac{(2 n-1)!}{29!(2 n-30)!}=5 \frac{(2 n-1)!}{(2 n-12)!11!} \
& \frac{1}{29 \ldots 12 \cdot 5}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 2} \
& \frac{1}{30 \cdot 29 \ldots 12}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 12} \
& 2 n-12=30 \
& n=21
\end{aligned}$$

Solutions · 1