3 years ago

12 ∑ 20(0.5)^n-1 n=3

1 answer:

8 0

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

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