Chapter 9.4, Problem 52E

To determine

To find the sum ${\displaystyle \sum _{n=1}^{20}\left(n^3-n\right)}$ using the formulas for the sums of powers of integers.

Answer to Problem 52E

  ${\displaystyle \sum _{n=1}^{20}\left(n^3-n\right)}=43890$

Explanation of Solution

Given:

Let $S={\displaystyle \sum _{n=1}^{20}\left(n^3-n\right)}$

Concept used:

  $1+2+3+4+\cdots +n=\frac{n\left(n+1\right)}{2}$

  $1^3+2^3+3^3+4^3+\cdots +n^3=\frac{n^2{\left(n+1\right)}^2}{4}$

Calculation:

Apply the formula for the sum and obtain

  $\begin{array}{c}S={\displaystyle \sum _{n=1}^{20}\left(n^3-n\right)}\\ ={\displaystyle \sum _{n=1}^{20}{n}^3}-{\displaystyle \sum _{n=1}^{20}n}\\ =\left(1^3+2^3+\cdots +{20}^2\right)-\left(1+2+\cdots +20\right)\\ =\frac{{20}^2{\left(20+1\right)}^2}{4}-\frac{20\left(20+1\right)}{2}\\ =5\times 20\times {21}^2-10\times 21\\ =44100-210\\ =43890\end{array}$