Evaluate 12sigma n=3 20 (0.5) ^n-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)