Question

The sum of the following series,$\frac{1}{3}{}^n{c}_1+\frac{(1^2+2^2)}{5}{}^n{c}_2+\frac{(1^2+2^2+3^2)}{7}{}^n{c}_3$ + ...... upto n terms is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

General term ($T_r$) = $\frac{(1^2+2^2+3^2+......+r^2)}{(2r+1)}{}^n{C}_r$

= $\frac{r(r+1)(2r+1)}{6(2r+1)}{}^n{c}_r$

= $\frac{1}{6}(r^2{}^n{c}_r+r.{}^n{c}_r)$

S = $\sum _{r=1}^{n}Tr=\frac{1}{6}\sum _{r=1}^n{r}^2.{}^n{c}_r+\frac{1}{6}\sum _{r=1}^{n}r.{}^n{c}_r$

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

= $\frac{2^{n-2}n}{6}(n+1+2)=\frac{2^{n-2}.n(n+3)}{6}$