# How to evaluate this sum?

## Evaluate ${\sum}_{k = 1}^{99} k \left({k}^{2} - 1\right)$ Given that: ${\sum}_{k = 1}^{n} k = \frac{n \left(n + 1\right)}{2}$ and ${\sum}_{k = 1}^{n} {k}^{3} = {\left[{\sum}_{k = 1}^{n} k\right]}^{2}$

Mar 12, 2018

${S}_{99} = 24497550$

#### Explanation:

We want to evaluate

${S}_{99} = {\sum}_{k = 1}^{99} k \left({k}^{2} - 1\right)$

Let's take the general example

${S}_{n} = {\sum}_{k = 1}^{n} k \left({k}^{2} - 1\right) = {\sum}_{k = 1}^{n} {k}^{3} - k = {\sum}_{k = 1}^{n} {k}^{3} - {\sum}_{k = 1}^{n} k$

Using

• ${\sum}_{k = 1}^{n} k = \frac{n \left(n + 1\right)}{2}$
• ${\sum}_{k = 1}^{n} {k}^{3} = {\left({\sum}_{k = 1}^{n} k\right)}^{2} = {\left(\frac{n \left(n + 1\right)}{2}\right)}^{2}$

Thus

${S}_{n} = {\left(\frac{n \left(n + 1\right)}{2}\right)}^{2} - \frac{n \left(n + 1\right)}{2}$

For $n = 99$

${S}_{99} = {\left(\frac{99 \left(99 + 1\right)}{2}\right)}^{2} - \frac{99 \left(99 + 1\right)}{2}$

${S}_{99} = {\left(\frac{9900}{2}\right)}^{2} - \frac{9900}{2}$

${S}_{99} = \frac{{9900}^{2} - 19800}{4} = 24497550$

Mar 12, 2018

$24497550$

#### Explanation:

Given that:

${\sum}_{k = 1}^{n} k = \frac{n \left(n + 1\right)}{2}$

and

${\sum}_{k = 1}^{n} {k}^{3} = {\left[{\sum}_{k = 1}^{n} k\right]}^{2}$

follows

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

hence

${\sum}_{k = 1}^{n} k \left({k}^{2} - 1\right) = {\left(\frac{n \left(n + 1\right)}{2}\right)}^{2} - \frac{n \left(n + 1\right)}{2} =$

$= \frac{1}{4} \left({n}^{2} + n - 2\right) n \left(n + 1\right)$

and for $n = 99$ gives

${\sum}_{k = 1}^{n} k \left({k}^{2} - 1\right) = 24497550$