Infinite series help

1. Compute (n=0 goes on the bottom of sigma, and ∞ goes on top) Σ(n=0,∞) 4/((-3)^n)-3/(3^n).

2. Compute Σ(n=0,∞)3/(2^n)+4/(5^n).

The answer to 1. is -3/2 and the answer to 2. is 11. I don't know how to do these types of problems and I need help setting them up and solving them.

These are the sum of geometric series.
Use a + ar + ar^2 + ... + ar^n + ... = a/(1 - r) when |r| < 1.

1) Σ(n=0 to ∞) [4/(-3)^n - 3/3^n]
= Σ(n=0 to ∞) 4(-1/3)^n - Σ(n=0 to ∞) (1/3)^(n-1)
= 4/(1 - (-1/3)) - 3/(1 - 1/3), via geometric series
= 3 - 9/2
= -3/2.

2) Σ(n=0 to ∞) [3/2^n + 4/5^n]
= Σ(n=0 to ∞) 3(1/2)^n + Σ(n=0 to ∞) 4(1/5)^n
= 3/(1 - 1/2) + 4/(1 - 1/5)
= 6 + 5
= 11.

I hope this helps!