Find the sum to n terms of GP whose nth term is 3(2)^n-1

Answer 1

For a GP, term(n) = a r^(n-1)
comparing this with 3(2)^(n-1)
we have a = 3, r = 2

sum(n) = a(r^n - 1)/(r-1)
= 3(2^n) -1)/(1)
= 3(2)^n - 3

check:
the terms would be:
3, 6, 12, 24, ...
just adding them up, sum(4) = 45
using my formula, sum(4) = 3(2^4) - 3 = 48-3 = 45
looks good