Chapter 12.1, Problem 82AYU

In Problems 71-82, find the sum of each sequence.

${\displaystyle \underset{k=4}{\overset{24}{\sum }}{k}^3}$

To determine

To find: The sum of given sequence: ${\displaystyle \sum }_{k=4}^{24}{k}^3$.

Answer to Problem 82AYU

${\displaystyle \sum }_{k=4}^{24}{k}^3=89964$

Explanation of Solution

Given:

${\displaystyle \sum }_{k=4}^{24}{k}^3$

Calculation:

Using ${\displaystyle \sum }_{k=1}^n{k}^3=1^3+2^3+3^3+4^3+\dots +n^3={\left[\frac{n\left(n+1\right)}{2}\right]}^2$ we have,

${\displaystyle \sum }_{k=4}^{24}{k}^3={\displaystyle \sum }_{k=1}^{24}{k}^3-{\displaystyle \sum }_{k=1}^3{k}^3={\left[\frac{24\left(24+1\right)}{2}\right]}^2-{\left[\frac{3\left(3+1\right)}{2}\right]}^2=90000-36=89964$