If for positive integers $r>1, n>2$, the coefficients of the $(3 r)^{\text {th }}$ and $(r+2)^{\text {th }}$…
If for positive integers $r>1, n>2$, the coefficients of the $(3 r)^{\text {th }}$ and $(r+2)^{\text {th }}$ powers of $x$ in the expansion of $(1+x)^{2 n}$ are equal, then $n$ is equal to:
$2 r+1$
$2 r-1$
$3 r$
$r+1$
Solution
Expansion of $(1+x)^{2 n}$ is $1+{ }^{2 n} \mathrm{C}_1 x+{ }^{2 n} \mathrm{C}_2 x^2$ $+\ldots \ldots .+{ }^{2 n} \mathrm{C}_r x^n+{ }^{2 n} \mathrm{C}_{r+1} x^{n+1}+\ldots \ldots+{ }^{2 n} \mathrm{C}_{2 n} x^{2 n}$ As given ${ }^{2 n} \mathrm{C}_{r+2}={ }^{2 n} \mathrm{C}_{3 r}$
$
\begin{aligned}
&\Rightarrow \frac{(2 n) !}{(r+2) !(2 n-r-2) !}=\frac{(2 n) !}{(3 r) !(2 n-3 r) !} \\
&\Rightarrow(3 r) !(2 n-3 r) !=(r+2) !(2 n-r-2) !
\end{aligned}
$
Now, put value of $n$ from the given choices.
Choice (a) put $n=2 r+1$ in (1)
LHS : $(3 r) !(4 r+2-3 r) !=(3 r) !(r+2) !$
RHS : $(r+2) !(3 r) !$
$
\Rightarrow \text { LHS }=\text { RHS }
$