Answer
$S(n)=2-\displaystyle \frac{2}{n^{2}}$
$S(10)=1.98$
$S(100)=1.9998$
$S(1000)=1.999998$
$S(10,000)=1.99999998$
Work Step by Step
$\displaystyle \sum_{k=1}^{n}\frac{6k(k-1)}{n^{3}}=$
... the $\displaystyle \frac{6}{n^{3}}$ is constant,...$\displaystyle \sum_{i=1}^{n}ka_{i}=k\sum_{i=1}^{n}a_{i}$
$=\displaystyle \frac{6}{n^{3}}\sum_{k=1}^{n}(k^{2}-k)$
... property: $ \displaystyle \sum_{i=1}^{n}(a_{i}\pm b_{i})=\sum_{i=1}^{n}a_{i}\pm\sum_{i=1}^{n}b_{i}$
$=\displaystyle \frac{6}{n^{3}}[\sum_{k=1}^{n}k^{2}-\sum_{k=1}^{n}k]$
$... $T4.2.3. $\displaystyle \sum_{i=1}^{n}i^{2}=\frac{n(n+1)(2n+1)}{6}$ ,
$... $Th4.2.2. $\displaystyle \sum_{i=1}^{n}i=\frac{n(n+1)}{2}$
$=\displaystyle \frac{6}{n^{3}}[\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}]$
... factor out n and reduce ...
$=\displaystyle \frac{6}{n^{2}}[\frac{(2n^{2}+3n+1)}{6}-\frac{(n+1)}{2}]$
$=\displaystyle \frac{6}{n^{2}}[\frac{2n^{2}+3n+1-3n-3}{6}]$
... reduce 6 , simplify numerator....
$=\displaystyle \frac{1}{n^{2}}[2n^{2}-2]$
$=2-\displaystyle \frac{2}{n^{2}}$
$S(n)=2-\displaystyle \frac{2}{n^{2}}$
$S(10)=2-\displaystyle \frac{2}{10^{2}}=1.98$
$S(100)=2-\displaystyle \frac{2}{100^{2}}=1.9998$
$S(1000)=2-\displaystyle \frac{2}{1000^{2}}=1.999998$
$S(10,000)=2-\displaystyle \frac{2}{10,000^{2}}=1.99999998$
