Question

If for positive integers $r>1,n>2,$ the coefficients of the ${\left(3r\right)}^{th}$ and ${\left(r+2\right)}^{th}$ powers of x in the expansion of ${\left(1+x\right)}^{2n}$ are equal, then $n$ is equal to:

• A. 2r + 1
• B. 2r - 1
• C. 3 r
• D. r + 1

Answer 1

Answer: (A)

Expansion of ${\left(1+x\right)}^{2n}is1{+}^{2n}{C}_{1}x{+}^{2n}{C}_2{x}^2$
$+......{+}^{2n}{C}_r{x}^r{+}^{2n}{C}_{r+1}{x}^{r+1}+......{+}^{2n}{C}_{2n}{x}^{2n}$
As given ${}^{2n}{C}_{r+2}{=}^{2n}{C}_{3r}$
$\Rightarrow \frac{\left(2n\right)!}{\left(r+2\right)!\left(2n-r-2\right)!}=\frac{\left(2n\right)!}{\left(3r\right)!\left(2n-3r\right)!}$
$\Rightarrow \left(3r\right)!\left(2n-3r\right)!=\left(r+2\right)!\left(2n-r-2\right)!\dots \left(1\right)$
Now, put value of n from the given choices.
Choice $(a)$ put $n=2r+1in(1)$
LHS : $(3r)!(4r+2-3r)!=(3r)!(r+2)!$
$RHS:(r+2)!(3r)!$
$\Rightarrow LHS=RHS$