Chapter 12.1, Problem 78AYU

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

${\displaystyle \underset{k=0}{\overset{14}{\sum }}\left(k^2-4\right)}$

To determine

To find: The sum of given sequence: ${\displaystyle \sum }_{k=\text{ 0}}^{14}\left(k^2-4\right)$.

Answer to Problem 78AYU

${\displaystyle \sum }_{k=\text{ 0}}^{14}\left(k^2-4\right)=\text{ 959}$

Explanation of Solution

Given:

${\displaystyle \sum }_{k=\text{ 0}}^{14}\left(k^2-4\right)$

Calculation:

Using ${\displaystyle \sum }_{\text{k }=1}^{\text{n}}{k}^2=1^2+2^2+3^2+{\text{ 4}}^2+\dots +{\text{ n}}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ and ${\displaystyle \sum }_{\text{k }=1}^{\text{n}}\text{c }=\text{ cn}$ we have

${\displaystyle \sum }_{k=\text{ 0}}^{14}\left(k^2-4\right)={\displaystyle \sum }_{k=\text{ 0}}^{14}{k}^2-{\displaystyle \sum }_{k=\text{ 0}}^{14}\text{4 }=\frac{14\left(14+1\right)\left(2*14+1\right)}{6}-\text{ 14 }*\text{ 4 }=1015-\text{ 56 }=\text{ 959}$