Find the sum of the series SUMMATION of n=0 to infinity of (3^n+5)/(4^n)

Note that we can split the sum up as follows:
∑ (3^n + 5)/4^n (from n=0 to infinity)
= ∑ (3^n/4^n + 5/4^n) (from n=0 to infinity), by splitting up the fraction
= ∑ [(3/4)^n + 5(1/4)^n] (from n=0 to infinity), by the laws of exponents
= ∑ (3/4)^n (from n=0 to infinity) + 5 ∑ (1/4)^n (from n=0 to infinity).

Both of these infinite series are geometric. The first series has a first term of (3/4)^0 = 1 and a common ratio of 3/4. The second series has a first term of 1 and a common ratio of 1/4. Therefore:
∑ (3/4)^n (from n=0 to infinity) + 5 ∑ (1/4)^n (from n=0 to infinity)
= 1/(1 - 3/4) + 5/(1 - 1/4)
= 32/3.

I hope this helps!

sum = 1/(1 - 3/4) + 5/(1 - 1/4) = 32/3

(3^n+5)/(4^n) = (3/4)^n + 5* (1/4^n)
Two infinite sums of GPs each using "S = a/(1 - r)"

SUMMATION of n=0 to infinity of (3^n+5)/(4^n)
= 1/[1 - (3/4) ] + 5/[1 - (4/4) ]

SUMMATION = 32/3

Regards - Ian