How do you find the sum of the finite geometric sequence of Sigma (5/2)^(n-1) from n=1 to 10?

1 Answer
May 24, 2018

sum_(n=1)^10 (5/2)^(n-1)=3254867/512=6357.162109375

Explanation:

Solution

sum_(n=1)^10 (5/2)^(n-1)
=sum_(m=0)^9 (5/2)^(m) (you can still use n as the variable, but in effect you are letting m=n-1)
=((5/2)^10-1)/((5/2)-1) (geometric series formula)
=3254867/512=6357.162109375

Geometric Series Formula

From the series

sum_(n=0)^x k^n = 1 + k + k^2 + ... + k^x

We multiply throughout by k

ksum_(n=0)^x k^n = k + k^2 + k^3 + ... + k^(x+1)

Now subtract the original series from that sum

ksum_(n=0)^x k^n - sum_(n=0)^x k^n = (k-1)sum_(n=0)^x k^n

= cancel(k) + cancel(k^2) + cancel(k^3) + ... + cancel(k^x) + k^(x+1) - (1 + cancel(k) + cancel(k^2) + ... + cancel(k^x))

= k^(x+1) - 1

So, dividing by (k-1), we have
sum_(n=0)^x k^n = (k^(x+1) - 1)/(k-1)