Question

$C_0^2+3.C_1^2+5.C_2^2+\dots \dots +(2n+1).C_n^2=$

• A. $(n+1)2^n$
• B. $(2n+1{)}^{2n}{C}_n$
• C. $(n+1){.}^{2n}{C}_n$
• D. $(2n-1{)}^{2n}{C}_n$

Answer 1

Answer: (C)

$(n+1){.}^{2n}{C}_n$

$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+\cdots +C_n{x}^n$ ------(1)
$(1+\frac{1}{x}{)}^n=C_0+C_1\frac{1}{x}+C_2\frac{1}{x^2}+\cdots +C_n\frac{1}{x^n}$ -----(2)
Multiplying both sides, we get
$\frac{(1+x{)}^{2n}}{(x{)}^n}=\sum C_0^2+x\sum C_0{C}_1+x^2\sum C_0{C}_2+....+x^r\sum C_0{C}_r+...$
The various sums are the coefficients of $x^0,x,x^2,....x^r$ in $\frac{(1+x{)}^{2n}}{(x{)}^n}$ or coefficients of $x^n,x^{n+1},x^{n+2},....,x^{n+r}$ in the expansion of $(1+x{)}^{2n}$ which occur in $T_{n+1},T_{n+2}....$
Clearly the coefficent of $x^n$ in $(1+x{)}^{2n}$ is ${}^{2n}{C}_n$
$\therefore C_0^2+C_1^2+C_2^2+\dots \dots +C_n^2{=}^{2n}{C}_n$
Let $S=C_0^2+3.C_1^2+5.C_2^2+\dots \dots +(2n+1).C_n^2$ ----(3)
$S=(2n+1).C_n^2+(2n-1).C_{n-1}^2+\dots \dots +C_0^2$ -----(4)
Adding (3) and (4) gives
$2S=(2n+2)[C_0^2+C_1^2+C_2^2+\dots \dots +C_n^2]$
$\therefore S=(n+1){.}^{2n}{C}_n$
Hence, option C.