Question

If $n\ge 2$ is a positive integer, then the sum of the series ${}^{n+1}{C}_2+2({}^2{C}_2+{}^3{C}_2+{}^4{C}_2+\cdots +{}^n{C}_2)$ is

• A. $\frac{n(n+1{)}^2(n+2)}{12}$
• B. $\frac{n(n-1)(2n+1)}{6}$
• C. $\frac{n(n+1)(2n+1)}{6}$
• D. $\frac{n(2n+1)(3n+1)}{6}$

Answer 1

The correct option is C $\frac{n(n+1)(2n+1)}{6}$

We know ${}^2{C}_2={}^3{C}_3$
Let $S={}^3{C}_3+{}^3{C}_2+\cdots +{}^n{C}_2={}^{n+1}{C}_3(\because {}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r)$
$\therefore {}^{n+1}{C}_2+{}^{n+1}{C}_3+{}^{n+1}{C}_3$
$={}^{n+2}{C}_3+{}^{n+1}{C}_3$
$=\frac{(n+2)!}{3!(n-1)!}+\frac{(n+1)!}{3!(n-2)!}$
$=\frac{(n+2)(n+1)n}{6}+\frac{(n+1)\left(n\right)(n-1)}{6}$
$=\frac{n(n+1)(2n+1)}{6}$

$TRICK:$ Put $n=2$ and verify the options.