Question

Find the sum to n terms of the series 3+6+10+16+...:

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

Answer 1

The correct option is B $n(n+1)+2^n-1$

Go through options
Let n = 2, then $S_n=3+6=9$
$\therefore S_n=2\left(3\right)+2^2-1=9$
at n = 3, $S_n=19$
$\therefore S_n=3\times 4+2^3-1=19$
Hence choice (b) is correct.

Alternatively: 3+6+10+16+...
$=(2+4+6+8+\cdots )+(1+2+4+8+\cdots )$
$=n(n+1)+(2^n-1)$