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