How would you solve for the sum of this series? Summation from n=3 to infinity (3^(n+3))/(5^(n-3))

I don't understand how to go about solving this. Any help would be greatly appreciated!!!

Thank you!

By the laws of exponents, you can write:
(a) 3^(n + 3) = 3^n * 3^3 = 27(3^n)
(b) 5^(n - 3) = 5^n * 5^(-3) = (1/125)(5^n).

So, we have:
sum(n=3 to infinity) 3^(n + 3)/5^(n - 3)
= sum(n=3 to infinity) [27(3^n)]/[(1/125)(5^n)]
= 3375*sum(n=3 to infinity) (3/5)^n.

Then, sum(n=3 to infinity) (3/5)^n is an infinite geometric series with a first term of (3/5)^3 = 27/125 and a common ratio of 3/5. Thus:
sum(n=3 to infinity) (3/5)^n = (27/125)/(1 - 3/5) = 27/50.

Therefore, the series sums to 3375(27/50) = 3645/2.

I hope this helps!