If n≥2 is a positive integer, then the sum of the series C2 n+1+2(C2 2+C2 3+C2 4+.....+ nC2) is

C2 n+1+2(C2 2+C2 3+C2 4+.....+C2 n)=C2 n+1+2 ∑r=2n C2 r=C2 n+1+∑ (2) r!(r-2 ! 2!)=C2 n+1+∑r=2n r(r-1)=C2 n+1+∑r=2n r2-∑r=2nr=(n+1) n2+n (n+1) (2n+1)6-1-n(n+1)2-1=(n+1) n2+n(n+1) (2n+1)6-n(n+1)2=n(n+1) (2n+1)6

If n≥2 is a positive integer, then the sum of the series C2 n+1+2(C2 2+C2 3+C2 4+.....+ nC2) is

C2 n+1+2(C2 2+C2 3+C2 4+.....+C2 n)=C2 n+1+2 ∑r=2n C2 r=C2 n+1+∑ (2) r!(r-2 ! 2!)=C2 n+1+∑r=2n r(r-1)=C2 n+1+∑r=2n r2-∑r=2nr=(n+1) n2+n (n+1) (2n+1)6-1-n(n+1)2-1=(n+1) n2+n(n+1) (2n+1)6-n(n+1)2=n(n+1) (2n+1)6

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If n $\geq$ 2 is a positive integer, then the sum of the series ${}^{n + 1}C_{2} + 2 ({}^{2}C_{2} + {}^{3}C_{2} + {}^{4}C_{2} + . . . . . +^{n} C_{2}) i s$

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

$\frac{n (2 n + 1) (3 n + 1)}{6}$

$\frac{n {(n + 1)}^{2} (n + 2)}{12}$

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

answer is A.

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Detailed Solution

$\begin{array}{l} {}^{n + 1}C_{2} + 2 ({}^{2}C_{2} + {}^{3}C_{2} + {}^{4}C_{2} + . . . . . + {}^{n}C_{2}) \\ = {}^{n + 1}C_{2} + 2 \sum_{r = 2}^{n} {}^{r}C_{2} \\ = {}^{n + 1}C_{2} + \sum (2) \frac{r!}{(r - 2! 2!)} \\ = {}^{n + 1}C_{2} + \sum_{r = 2}^{n} r (r - 1) \\ = {}^{n + 1}C_{2} + \sum_{r = 2}^{n} r^{2} - \sum_{r = 2}^{n} r \\ = \frac{(n + 1) n}{2} + \frac{n (n + 1) (2 n + 1)}{6} - 1 - [\frac{n (n + 1)}{2} - 1] \\ = \frac{(n + 1) n}{2} + \frac{n (n + 1) (2 n + 1)}{6} - \frac{n (n + 1)}{2} \\ = \frac{n (n + 1) (2 n + 1)}{6} \end{array}$