Question

12
∑ 20(0.5)^n-1
n=3

Answer 1

Assuming n starts from 1.
∑$20 (0.5)^{n-1}$; n = 1 to 3
$=20 (0.5)^{1-1}+20 (0.5)^{2-1}+20 (0.5)^{3-1} \\ =20 (0.5)^{0}+20 (0.5)^{1}+20 (0.5)^{2} \\ =20(1)+20(0.5)+20(0.25) \\ =20+10+5 \\ =35$