Find the sum for each of the series

1. -10/5^n from n=0 to Infinity
2 (1/4^n)+(1/2)^(n-1)

1. -10/5^n from n=0 to Infinity SUM = -- 10 / [1 -- 1/5] = -- 50/4 = -- 12 1/2
2 (1/4^n)+(1/2)^(n-1) SUM = (1/4)[1 -- 1/4] + 1 / [ 1 -- 1/2] = 1/3 + 2 = 2 1/3

Use ∑[j=0,∞] r^j = 1/(1-r), for |r| < 1

1. ∑[n=0,∞] (-10/5^n) = -10∑[n=0,∞] (1/5)^n = -10/(1 - 1/5) = -25/2 = -12.5

2. You do also want to sum this one from 0 to ∞, right?
Note that (1/2)^(n-1) = 2/(2^n)
∑[n=0,∞] 1/(4^n) + 2∑[n=0,∞] 1/(2^n)
= 1/(1 - 1/4) + 2/(1 - 1/2) = 4/3 + 4 = 16/3

EDIT:
If the 2nd problem is to be summed from 1 to ∞, then EASY TO UNDERSTAND's answer is correct, because:
∑[n=0,∞] a[n] = 1/4^0 + (1/2)^-1 + ∑[n=1,∞] a[n]
= 1 + 2 + ∑[n=1,∞] a[n]
= 3 + ∑[n=1,∞] a[n]
= 3 + 2 &1/3 = 5 &1/3 = 16/3