Evaluate 20(0.5)^(n-1) when n= 3 to 12. Use sigma notation .

Answer 1

what you have is a geometric series, with
a = 20
r = 1/2

S2 = 20(1-r^2)/(1-r)
S12 = 20(1-r^12)/1-r

sigma(3,12) = S12-S2
= 20[(1-r^12)/(1-r) - (1-r^2)/(1-r)]
= 20/(1-r) ((1-r^12)-(1-r^2))
= 5115/512