Question

Sum of first $n$ terms of the sequence $5,7,11,17,25,\dots$ is equal to

• A. $\frac{n}{6}(2n^2+15)$
• B. $\frac{n}{3}(n^2+14)$
• C. $n(2n+3)$
• D. $\frac{n}{6}(n^2+14)$

Answer 1

The correct option is B $\frac{n}{3}(n^2+14)$

$S_n=5+7+11+17+\cdots +T_n$
$S_n=5+7+11+\cdots +T_{n-1}+T_n$
Subtracting the above equations,
$0=5+2+4+6+\cdots -T_n$
$\therefore T_n=5+2(1+2+3+\cdots +(n-1))=5+2\left(\frac{n-1}{2}\right)n$
$\Rightarrow T_n=5+n(n-1)$

$\therefore S_n=\sum _{n=1}^n(n^2-n+5)=\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+5n=\frac{n}{3}(n^2+14)$