# How do you use the properties of summation to evaluate the sum of Sigma (i^2-1) from i=1 to 10?

Oct 31, 2016

${\sum}_{i = 1}^{10} \left({i}^{2} - 1\right) = 375$

#### Explanation:

We could just add up the 10 terms but the numbers get a bit horrific, so it is actually easier in this case to derive a general formula standard formula for $\sum {i}^{2}$ we have:

${\sum}_{i = 1}^{n} \left({i}^{2} - 1\right) = {\sum}_{i = 1}^{n} {i}^{2} - {\sum}_{i = 1}^{n} 1$
$\therefore {\sum}_{i = 1}^{n} \left({i}^{2} - 1\right) = \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right) - n$
$\therefore {\sum}_{i = 1}^{n} \left({i}^{2} - 1\right) = \frac{1}{6} n \left\{\left(n + 1\right) \left(2 n + 1\right) - 6\right\}$
$\therefore {\sum}_{i = 1}^{n} \left({i}^{2} - 1\right) = \frac{1}{6} n \left\{2 {n}^{2} + 3 n + 1 - 6\right\}$
$\therefore {\sum}_{i = 1}^{n} \left({i}^{2} - 1\right) = \frac{1}{6} n \left\{2 {n}^{2} + 3 n - 5\right\}$
$\therefore {\sum}_{i = 1}^{n} \left({i}^{2} - 1\right) = \frac{1}{6} n \left(n - 1\right) \left(2 n + 5\right)$

So with n=10 we have:
${\sum}_{i = 1}^{10} \left({i}^{2} - 1\right) = \frac{1}{6} \left(10\right) \left(9\right) \left(25\right)$
${\sum}_{i = 1}^{10} \left({i}^{2} - 1\right) = 375$