Question

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

Answer 1

`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)`