Question

Use the summation formulas to rewrite the expression without the summation notation. Use the result to find the sums for $n=10,100,1000\text{and}10,000$.

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

Answer 1

Step-1 : Summation of series :

Consider the expression $\sum _{k=1}^n\frac{6k\left(k-1\right)}{n^3}$.

Since the summation runs over $k$, take the constant $\frac{6}{n^3}$ out of summation:

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

Simplify the expression:

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

Now, use the standard result $\begin{array}{rcl}\sum _{k=1}^n{k}^2& =& \frac{n\left(n+1\right)\left(2n+1\right)}{6}\text{and}\sum _{k=1}^{n}k=\frac{n\left(n+1\right)}{2}\end{array}$in the above expression:

$\begin{array}{rcl}\frac{6}{n^3}\left(\sum _{k=1}^n{k}^2-\sum _{k=1}^{n}k\right)& =& \frac{6}{n^3}\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}-\frac{n\left(n+1\right)}{2}\right]\end{array}$

Further simplify the expression:

$\begin{array}{rcl}\frac{6}{n^3}\left(\sum _{k=1}^n{k}^2-\sum _{k=1}^{n}k\right)& =& \frac{6}{n^3}\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}-\frac{n\left(n+1\right)}{2}\right]\\ & =& \frac{6}{n^2}\left[\frac{\left(n+1\right)\left(2n+1\right)}{6}-\frac{\left(n+1\right)}{2}\right]\\ & =& \frac{\left(n+1\right)\left(2n+1\right)}{n^2}-\frac{3\left(n+1\right)}{n^2}\end{array}$

Hence, the rewritten form of the expression $\sum _{k=1}^n\frac{6k\left(k-1\right)}{n^3}$ without the summation notation is $\begin{array}{rcl}& & \frac{\left(n+1\right)\left(2n+1\right)}{n^2}-\frac{3\left(n+1\right)}{n^2}\end{array}$.

Now find the sum for $n=10,100,1000\text{and}10,000$ by substituting the respective values in the rewritten expression $\frac{\left(n+1\right)\left(2n+1\right)}{n^2}-\frac{3\left(n+1\right)}{n^2}$:

Step-2 : Summation for $n=10$,

$\begin{array}{rcl}\frac{\left(n+1\right)\left(2n+1\right)}{n^2}-\frac{3\left(n+1\right)}{n^2}& =& \frac{\left(10+1\right)\left(2·10+1\right)}{{\left(10\right)}^2}-\frac{3\left(10+1\right)}{{\left(10\right)}^2}\\ & =& \frac{11·21}{100}-\frac{3·11}{100}\\ & =& 2.31-0.33\\ & =& 1.98\end{array}$

Step-3 : Summation for $n=100$:

$\begin{array}{rcl}\frac{\left(n+1\right)\left(2n+1\right)}{n^2}-\frac{3\left(n+1\right)}{n^2}& =& \frac{\left(100+1\right)\left(2·100+1\right)}{{\left(100\right)}^2}-\frac{3\left(100+1\right)}{{\left(100\right)}^2}\\ & =& 2.0301-0.0303\\ & =& 1.9998\end{array}$

Step-4 : Summation for $n=1000$ :

$\begin{array}{rcl}\frac{\left(n+1\right)\left(2n+1\right)}{n^2}-\frac{3\left(n+1\right)}{n^2}& =& \frac{\left(1000+1\right)\left(2·1000+1\right)}{{\left(1000\right)}^2}-\frac{3\left(1000+1\right)}{{\left(1000\right)}^2}\\ & =& 2.003001-0.003003\\ & =& 1.999998\end{array}$

Step-5 : Summation for $n=10,000$:

$\begin{array}{rcl}\frac{\left(n+1\right)\left(2n+1\right)}{n^2}-\frac{3\left(n+1\right)}{n^2}& =& \frac{\left(10,000+1\right)\left(2·10,000+1\right)}{{\left(10,000\right)}^2}-\frac{3\left(10,000+1\right)}{{\left(10,000\right)}^2}\\ & =& 2.00030001-0.00030003\\ & =& 1.99999998\end{array}$

Hence the sum for $\mathbf{n}\mathbf{=}\mathbf{10}\mathbf{,}\mathbf{}\mathbf{100}\mathbf{,}\mathbf{}\mathbf{1000}\mathbf{}\mathbf{\text{and}}\mathbf{}\mathbf{10}\mathbf{,}\mathbf{000}$i is $\mathbf{1}\mathbf{.}\mathbf{98}\mathbf{,}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{9998}\mathbf{,}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{999998}\mathbf{}\mathbf{\text{and}}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{99999998}$ respectively.