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

May 24, 2018

${\sum}_{n = 1}^{10} {\left(\frac{5}{2}\right)}^{n - 1} = \frac{3254867}{512} = 6357.162109375$

### Solution

${\sum}_{n = 1}^{10} {\left(\frac{5}{2}\right)}^{n - 1}$
$= {\sum}_{m = 0}^{9} {\left(\frac{5}{2}\right)}^{m}$ (you can still use $n$ as the variable, but in effect you are letting $m = n - 1$)
$= \frac{{\left(\frac{5}{2}\right)}^{10} - 1}{\left(\frac{5}{2}\right) - 1}$ (geometric series formula)
$= \frac{3254867}{512} = 6357.162109375$

### Geometric Series Formula

From the series

${\sum}_{n = 0}^{x} {k}^{n} = 1 + k + {k}^{2} + \ldots + {k}^{x}$

We multiply throughout by k

$k {\sum}_{n = 0}^{x} {k}^{n} = k + {k}^{2} + {k}^{3} + \ldots + {k}^{x + 1}$

Now subtract the original series from that sum

$k {\sum}_{n = 0}^{x} {k}^{n} - {\sum}_{n = 0}^{x} {k}^{n} = \left(k - 1\right) {\sum}_{n = 0}^{x} {k}^{n}$

$= \cancel{k} + \cancel{{k}^{2}} + \cancel{{k}^{3}} + \ldots + \cancel{{k}^{x}} + {k}^{x + 1} - \left(1 + \cancel{k} + \cancel{{k}^{2}} + \ldots + \cancel{{k}^{x}}\right)$

$= {k}^{x + 1} - 1$

So, dividing by (k-1), we have
${\sum}_{n = 0}^{x} {k}^{n} = \frac{{k}^{x + 1} - 1}{k - 1}$