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1 vote
12
∑ 20(0.5)^n-1
n=3

1 Answer

1 vote
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

User Caleb Shay
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