If for positive integers r>1, n>2, the coefficients of the (3 r)^ th and (r+2)^ th powers of x in the expansion of (1+x)^2 n are equal, then n is…

Expansion of $(1+x)^{2 n}$(1+x)2n is $1+{ }^{2 n} \mathrm{C}_1 x+{ }^{2 n} \mathrm{C}_2 x^2$1+2nC1​x+2nC2​x2 $+\ldots \ldots .+{ }^{2 n} \mathrm{C}_r x^n+{ }^{2 n} \mathrm{C}_{r+1} x^{n+1}+\ldots \ldots+{ }^{2 n} \mathrm{C}_{2 n} x^{2 n}$+…….+2nCr​xn+2nCr+1​xn+1+……+2nC2n​x2n As given ${ }^{2 n} \mathrm{C}_{r+2}={ }^{2 n} \mathrm{C}_{3 r}$2nCr+2​=2nC3r​ $$\begin{aligned} &\Rightarrow \frac{(2 n) !}{(r+2) !(2 n-r-2) !}=\frac{(2 n) !}{(3 r) !(2 n-3 r) !} \\ &\Rightarrow(3 r) !(2 n-3 r) !=(r+2) !(2 n-r-2) ! \end{aligned}$$​⇒(r+2)!(2n−r−2)!(2n)!​=(3r)!(2n−3r)!(2n)!​⇒(3r)!(2n−3r)!=(r+2)!(2n−r−2)!​ Now, put value of $n$n from the given choices. Choice (a) put $n=2 r+1$n=2r+1 in (1) LHS : $(3 r) !(4 r+2-3 r) !=(3 r) !(r+2) !$(3r)!(4r+2−3r)!=(3r)!(r+2)! RHS : $(r+2) !(3 r) !$(r+2)!(3r)! $$\Rightarrow \text { LHS }=\text { RHS }$$⇒ LHS = RHS