Question

If $n\ge 2$is a positive integer, then the sum of the series ${}^{n+1}C_2+2\left({}^{2}c_2+{}^{3}c_2+{}^{4}c_2+..+{}^{n}c_2\right)$ is

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

Answer 1

The correct option is C

$\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

Explanation for correct option

Given that $n\ge 2$is a positive integer

$\therefore {}^{n+1}C_2+2\left({}^{2}c_2+{}^{3}c_2+{}^{4}c_2+..+{}^{n}c_2\right)$

$$\begin{gathered} ={}^{n+1}C_2+2\left({}^{3}c_3+{}^{3}c_2+{}^{4}c_2+..+{}^{n}c_2\right)\left[\because {}^{n}c_n={}^{m}c_m\right] \\ ={}^{n+1}C_2+2\left({}^{4}c_3+{}^{4}c_2+..+{}^{n}c_2\right)\left[\because {}^{n}c_m+{}^{n}c_{m-1}={}^{n+1}c_m\right] \\ ={}^{n+1}C_2+2\left({}^{5}c_3+..+{}^{n}c_2\right)\left[\because {}^{n}c_m+{}^{n}c_{m-1}={}^{n+1}c_m\right] \end{gathered}$$

Proceeding in this way we get

$$\begin{gathered} ={}^{n+1}C_2+2\left({}^{n}c_3+{}^{n}c_2\right) \\ ={}^{n+1}C_2+2.{}^{n+1}c_3 \\ =\frac{\left(n+1\right)!}{2!\left(n+1-2\right)!}+\frac{2·\left(n+1\right)!}{3!\left(n+1-3\right)!}\left[\because {}^n{C}_m=\frac{n!}{m!\left(n-m\right)!}\right] \\ =\frac{\left(n+1\right)!}{2!\left(n-1\right)!}+\frac{2·\left(n+1\right)!}{3!\left(n-2\right)!} \\ =\frac{n\left(n+1\right)}{2}+\frac{\left(n-1\right)\left(n\right)\left(n+1\right)}{3} \\ =n\left(n+1\right)\left[\frac{1}{2}+\frac{n-1}{3}\right] \\ =n\left(n+1\right)\left[\frac{3+2n-2}{6}\right] \\ =n\left(n+1\right)\left[\frac{2n+1}{6}\right] \\ =\frac{n\left(n+1\right)\left(2n+1\right)}{6} \end{gathered}$$

Hence, option(C) i.e. $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ is correct