# How do you use the summation formulas to rewrite the expression Sigma (6k(k-1))/n^3 as k=1 to n without the summation notation and then use the result to find the sum for n=10, 100, 1000, and 10000?

Nov 15, 2016

${\sum}_{k = 1}^{n} \frac{1}{n} ^ 3 6 k \left(k - 1\right) = \frac{2}{n} ^ 2 \left({n}^{2} - 1\right)$

#### Explanation:

Let ${S}_{n} = {\sum}_{k = 1}^{n} \frac{1}{n} ^ 3 6 k \left(k - 1\right)$
$\therefore {S}_{n} = \frac{6}{n} ^ 3 {\sum}_{k = 1}^{n} \left({k}^{2} - k\right)$
$\therefore {S}_{n} = \frac{6}{n} ^ 3 \left\{{\sum}_{k = 1}^{n} \left({k}^{2}\right) - {\sum}_{j = 1}^{n} \left(k\right)\right\}$

And using the standard results:
${\sum}_{r = 1}^{n} r = \frac{1}{2} n \left(n + 1\right)$

We have;

${S}_{n} = \frac{6}{n} ^ 3 \left\{\frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right) - \frac{1}{2} n \left(n + 1\right)\right\}$
${S}_{n} = \frac{1}{n} ^ 3 \left\{n \left(n + 1\right) \left(2 n + 1\right) - 3 n \left(n + 1\right)\right\}$
${S}_{n} = \frac{1}{n} ^ 3 \left\{n \left(n + 1\right) \left(2 n + 1 - 3\right)\right\}$
${S}_{n} = \frac{1}{n} ^ 3 n \left(n + 1\right) \left(2 n - 2\right)$
${S}_{n} = \frac{2}{n} ^ 2 \left(n + 1\right) \left(n - 1\right)$
${S}_{n} = \frac{2}{n} ^ 2 \left({n}^{2} - 1\right)$

And this has been calculated using Excel for $n = 10 , 100 , 1000 , 10000$

What happens as $n \rightarrow \infty$?

[ NB As an additional task we could possibly conclude that as $n \rightarrow \infty$ then ${S}_{n} \rightarrow 2$; This is probably the conclusion of this question]

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

$\therefore {S}_{n} = 2 {n}^{2} / {n}^{2} - \frac{2}{n} ^ 2$
$\therefore {S}_{n} = 2 - \frac{2}{n} ^ 2$

And so,

${\lim}_{n \rightarrow \infty} {S}_{n} = {\lim}_{n \rightarrow \infty} \left(2 - \frac{2}{n} ^ 2\right)$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = {\lim}_{n \rightarrow \infty} \left(2\right) - {\lim}_{n \rightarrow \infty} \left(\frac{2}{n} ^ 2\right)$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = 2 + 2 {\lim}_{n \rightarrow \infty} \left(\frac{1}{n} ^ 2\right)$
$\therefore {\lim}_{n \rightarrow \infty} {S}_{n} = 2$, $\text{ as } \left({\lim}_{n \rightarrow \infty} \left(\frac{1}{n} ^ 2\right) = 0\right)$
Which confirms our assumption!