# How do you use the properties of summation to evaluate the sum of Sigma i(i-1)^2 from i=1 to 15?

Nov 12, 2016

${\sum}_{i = 1}^{15} i {\left(i - 1\right)}^{2} = 12040$

#### Explanation:

We need to use these the standard results:
${\sum}_{r = 1}^{n} r = \frac{1}{2} n \left(n + 1\right)$
${\sum}_{r = 1}^{n} {r}^{2} = \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)$
${\sum}_{r = 1}^{n} {r}^{3} = \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2}$

We have;

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

So for the sum requested, using the standard results;

${\sum}_{i = 1}^{15} i {\left(i - 1\right)}^{2} = \frac{1}{2} \left(15\right) \left(15 + 1\right) - \left(2\right) \left(\frac{1}{6}\right) \left(15\right) \left(15 + 1\right) \left(30 + 1\right) + {\left[\frac{1}{2} \left(15\right) \left(15 + 1\right)\right]}^{2}$

$\therefore {\sum}_{i = 1}^{15} i {\left(i - 1\right)}^{2} = \frac{1}{2} \left(15\right) \left(16\right) - \left(\frac{1}{3}\right) \left(15\right) \left(16\right) \left(31\right) + {\left[\frac{1}{2} \left(15\right) \left(16\right)\right]}^{2}$
$\therefore {\sum}_{i = 1}^{15} i {\left(i - 1\right)}^{2} = 120 - 2480 + {120}^{2}$
$\therefore {\sum}_{i = 1}^{15} i {\left(i - 1\right)}^{2} = 120 - 2480 + 14400$
$\therefore {\sum}_{i = 1}^{15} i {\left(i - 1\right)}^{2} = 12040$

Note:
We could equally derive a single formula the sum of $n$ terms;

${\sum}_{i = 1}^{n} i {\left(i - 1\right)}^{2} = \frac{1}{2} n \left(n + 1\right) - 2 \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right) + \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2}$
$\therefore {\sum}_{i = 1}^{n} i {\left(i - 1\right)}^{2} = \frac{1}{2} n \left(n + 1\right) - \frac{1}{3} n \left(n + 1\right) \left(2 n + 1\right) + \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2}$
$\therefore {\sum}_{i = 1}^{n} i {\left(i - 1\right)}^{2} = \frac{1}{12} n \left(n + 1\right) \left\{6 - 4 \left(2 n + 1\right) + 3 n \left(n + 1\right)\right\}$

$\therefore {\sum}_{i = 1}^{n} i {\left(i - 1\right)}^{2} = \frac{1}{12} n \left(n + 1\right) \left(6 - 8 n - 4 - 3 {n}^{2} + 3 n\right)$
$\therefore {\sum}_{i = 1}^{n} i {\left(i - 1\right)}^{2} = \frac{1}{12} n \left(n + 1\right) \left(3 {n}^{2} - 5 n + 2\right)$

And with $n = 15$ we get:

${\sum}_{i = 1}^{15} i {\left(i - 1\right)}^{2} = \frac{1}{12} \left(15\right) \left(16\right) \left\{3 \left(15\right) \left(15\right) - 5 \left(15\right) + 2\right\}$
$\therefore {\sum}_{i = 1}^{15} i {\left(i - 1\right)}^{2} = \left(20\right) \left(675 - 75 + 2\right)$
$\therefore {\sum}_{i = 1}^{15} i {\left(i - 1\right)}^{2} = \left(20\right) \left(602\right)$
$\therefore {\sum}_{i = 1}^{15} i {\left(i - 1\right)}^{2} = 12040$, as before