$\sum _{k\=1}^{n}1$ = $n$ 

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

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

Table 4.1.3(a)   Closed-form expressions for sums of powers of the index $k$ 

Initialize and define the function $f$ 

Loading Student:-Calculus1

$f\left(x\right)\=6\+x-x^2$$\stackrel{\text{assign as function}}{\to }$$f$

Obtain the right Riemann sum

$\mathrm{RiemannSum}\left(f\left(x\right)\,x\=-2..3\,\mathrm{partition}\=n\,\mathrm{method}\=\mathrm{right}\,\mathrm{output}\=\mathrm{sum}\right)$

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

$\=$

$\frac{5\left(\frac{25{\left(n+1\right)}^2}{2n}-\frac{25\left(n+1\right)}{2n}-\frac{25{\left(n+1\right)}^3}{3n^2}+\frac{25{\left(n+1\right)}^2}{2n^2}-\frac{25\left(n+1\right)}{6n^2}\right)}{n}$

$\stackrel{\text{simplify}}{\=}$

$\frac{125\left(n^2-1\right)}{6n^2}$

Table 4.1.3(b)   Maple's calculation of the right Riemann sum for $f\left(x\right)$ on the interval $\left[-2\,3\right]$ 

$-\frac{125}{n^3}\sum _{i\=1}^n\left(i^2-i n\right)\=$

$-\frac{125}{n^3}\left(\sum _{i\=1}^n{i}^2-n\sum _{i\=1}^{n}i\right)$

$\=$

$-\frac{125}{n^3}\left(\left(\frac{1}{6}n\left(n\+1\right)\left(2n\+1\right)\right)-n\left(\frac{n}{2}\left(n\+1\right)\right)\right)$

$\=$

$-\frac{125}{n^3}\left(\frac{1}{6}\left(2n^3\+3n^2\+n\right)-\frac{1}{2}\left(n^3\+n^2\right)\right)$

$\=$

$-\frac{125}{n^3}\left(-\frac{n^3}{6}\+\frac{n}{6}\right)$

$\=$

$\frac{125}{6} \frac{n^2-1}{n^2}$