Question

evaluate 12sigma n=3 20(0.5)^n-1

Answer 1

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)