Chapter 12.1, Problem 50AYU

In Problems 51-60, write out each sum.

${\displaystyle \underset{k=1}{\overset{n}{\sum }}\left(2k+1\right)}$

To determine

To find: Each sum: ${\displaystyle \sum }_{k=1}^n\left(2k+1\right)$.

Answer to Problem 50AYU

${\displaystyle \sum }_{k=1}^n\left(2k+1\right)=3+5+\dots \dots \dots \dots .+\left(2\left(n\right)+1\right)$

Explanation of Solution

Given:

${\displaystyle \sum }_{k=1}^n\left(2k+1\right)$

Calculation:

${\displaystyle \sum }_{k=1}^n\left(2k+1\right)=\left(2\left(1\right)+1\right)+\left(2\left(2\right)+1\right)+\dots \dots \dots \dots .+\left(2\left(n\right)+1\right)$

${\displaystyle \sum }_{k=1}^n\left(2k+1\right)=3+5+\dots \dots \dots \dots .+\left(2\left(n\right)+1\right)$