Mathematics
Qiqe
2017-08-13 07:40:00
evaluate 12sigma n=3 20(0.5)^n-1
ANSWERS
Laurenlove9777
2017-08-13 08:58:35

For the same reasons as the last problem... s(n)=20(1-(1/2)^n)/(1-1/2) s(n)=40(1-(1/2)^n) But this time we are not starting at one so we must subtract the sum of s(2) from the sum of s(12) so: 40((1-(1/2)^12-(1-(1/2)^2) 40(0.249755859375) ≈9.99  (to nearest one-thousandth)

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