Question

$2+4+7+11+16+...........$ to n terms.

• A. $\frac{1}{6}(n^2+3n+8)$
• B. $\frac{n}{6}(n^2+3n+8)$
• C. $\frac{1}{6}(n^2-3n+8)$
• D. $\frac{n}{6}(n^2-3n+8)$

Answer 1

The correct option is B $\frac{n}{6}(n^2+3n+8)$

Let $S_n=2+4+7+11+16+...........n$ terms
$\therefore S_{n-1}=2+4+7+11+16+...........n-1$ terms
Now, $T_n=S_n-S_{n-1}=2+2+3+4+5+6+..................+n=1+\frac{n(n+1)}{2}=1+\frac{1}{2}(n^2+n)$
Hence required summation is
$=\sum T_n=\sum 1+\frac{1}{2}(\sum n^2+\sum n)$
$=n+\frac{1}{2}(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2})=\frac{n}{6}(n^2+3n+8)$
Hence, option 'B' is correct.