Chapter 9, Problem 74RE

To determine

To find: the sum ${\displaystyle \sum _{n=1}^6\left(n^5-n^2\right)}$ using the formulas for the sums of powers of integers..

Answer to Problem 74RE

  $12,110$

Explanation of Solution

Concept Used:

  $1^5+2^5+\cdots +n^5=\frac{n^2{\left(n+1\right)}^2\left(2n^2+2n-1\right)}{12}$

  $1^2+2^2+\cdots +n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

Consider,

  $\begin{array}{c}{\displaystyle \sum _{n=1}^6\left(n^5-n^2\right)}={\displaystyle \sum _{n=1}^6{n}^5}-{\displaystyle \sum _{n=1}^6{n}^2}\\ =\frac{6^2{\left(6+1\right)}^2\left(2\times 6^2+2\times 6-1\right)}{12}-\frac{6\left(6+1\right)\left(2\times 6+1\right)}{6}\\ =12,201-91\\ =12,110\end{array}$