Question

Evaluate $\displaystyle\sum_{k=1}^{11}$ (2 + 3k).

Answer 1

$\displaystyle\sum_{k=1}^{11} (2+3^{k})$

$\displaystyle\sum_{k=1}^{11} (2)+ \displaystyle\sum_{k=1}^{11} (3^{k})= 2(11)+\displaystyle\sum_{k=1}^{11} (3^{k})=22+\displaystyle\sum_{k=1}^{11} (3^{k}) ....(1) \displaystyle\sum_{k=1}^{11} 3^{k}= 3^1+3^2+3^3+...3^{11}$

The terms of this sequence 3, $3^2,3^3$ , … forms a G.P.

$s_{n} = \frac{a(r^n-1)}{r-1}$

⇒ $s_{11}$ = $\frac{3[(3)^{11}-1]}{3-1}$

⇒ $s_{11}=\frac{3}{2}3^{11}-1$

∴$\displaystyle\sum_{k=1}^{11} 3^{k}=\frac{3}{2}(3^{11}-1)$

Substituting this value in equation (1), we obtain

$\displaystyle\sum_{k=1}^{11} (2+3^{k}) = 22+\frac{3}{2}(3^{11}-1)$