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

Oct 31, 2016

${\sum}_{i = 1}^{10} i \left({i}^{2} + 1\right) = 3080$

#### 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 $\sum i$ and $\sum {i}^{3}$ we have:

${\sum}_{i = 1}^{n} i \left({i}^{2} + 1\right) = {\sum}_{i = 1}^{n} \left({i}^{3} + i\right)$
$\therefore {\sum}_{i = 1}^{n} i \left({i}^{2} + 1\right) = {\sum}_{i = 1}^{n} {i}^{3} + {\sum}_{i = 1}^{n} i$
$\therefore {\sum}_{i = 1}^{n} i \left({i}^{2} + 1\right) = \frac{{n}^{2} {\left(n + 1\right)}^{2}}{4} + \frac{n \left(n + 1\right)}{2}$
$\therefore {\sum}_{i = 1}^{n} i \left({i}^{2} + 1\right) = \frac{n \left(n + 1\right)}{4} \left\{n \left(n + 1\right) + 2\right\}$
$\therefore {\sum}_{i = 1}^{n} i \left({i}^{2} + 1\right) = \frac{n \left(n + 1\right)}{4} \left({n}^{2} + n + 2\right)$

So with n=10 we have:
$\therefore {\sum}_{i = 1}^{10} i \left({i}^{2} + 1\right) = \frac{10 \left(11\right)}{4} \left(100 + 10 + 2\right)$
$\therefore {\sum}_{i = 1}^{10} i \left({i}^{2} + 1\right) = \frac{10 \left(11\right) \left(112\right)}{4}$
$\therefore {\sum}_{i = 1}^{10} i \left({i}^{2} + 1\right) = 3080$