Chapter 12.1, Problem 53AYU

In Problems 51-60, write out each sum.

${\displaystyle \underset{k=1}{\overset{n}{\sum }}\frac{k^2}{2}}$

To determine

To find: Each sum: ${\displaystyle \sum }_{k=1}^n{\frac{k}{2}}^2$.

Answer to Problem 53AYU

${\displaystyle \sum }_{k=1}^n{\frac{k}{2}}^2=\frac{1}{2}+\frac{4}{2}+\frac{9}{2}+\dots \dots .+{\frac{n}{2}}^2$

Explanation of Solution

Given:

${\displaystyle \sum }_{k=1}^n{\frac{k}{2}}^2$

Calculation:

${\displaystyle \sum }_{k=1}^n{\frac{k}{2}}^2={\frac{1}{2}}^2+{\frac{2}{2}}^2+{\frac{3}{2}}^2+\dots \dots .+{\frac{n}{2}}^2$

${\displaystyle \sum }_{k=1}^n{\frac{k}{2}}^2=\frac{1}{2}+\frac{4}{2}+\frac{9}{2}+\dots \dots .+{\frac{n}{2}}^2$