Chapter 9.5, Problem 31E

Solution

To calculate : The sum in terms of $n$ for the series ${\displaystyle \sum _{k=1}^n\left(k^2-3k+4\right)}$ .

The sum of series ${\displaystyle \sum _{k=1}^n\left(k^2-3k+4\right)}$ is $\frac{n\left(n^2-3n+8\right)}{3}$ .

Given information :

The series ${\displaystyle \sum _{k=1}^n\left(k^2-3k+4\right)}$ .

Formula used :

The formula ${\displaystyle \sum _{k=1}^n{k}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ and ${\displaystyle \sum _{k=1}^{n}k}=\frac{n\left(n+1\right)}{2}$ .

Calculation :

Consider the series ${\displaystyle \sum _{k=1}^n\left(k^2-3k+4\right)}$ ,

Rewrite the sum.

  $\begin{array}{c}S={\displaystyle \sum _{k=1}^n\left(k^2-3k+4\right)}\\ ={\displaystyle \sum _{k=1}^n{k}^2}-{\displaystyle \sum _{k=1}^{n}3k}+{\displaystyle \sum _{k=1}^{n}4}\end{array}$

Substitute $n$ for $k$ into the formula of ${\displaystyle \sum _{k=1}^n{k}^2}$ and ${\displaystyle \sum _{k=1}^{n}k}=\frac{n\left(n+1\right)}{2}$ .

  $\begin{array}{c}S=\frac{n\left(n+1\right)\left(2n+1\right)}{6}-\frac{3n\left(n+1\right)}{2}+4n\\ =\frac{n\left[\left(n+1\right)\left(2n+1\right)-9\left(n+1\right)+24\right]}{6}\\ =\frac{n\left(2n^2+n+2n+1-9n-9+24\right)}{6}\\ =\frac{n\left(2n^2-6n+16\right)}{6}\end{array}$

  $S=\frac{n\left(n^2-3n+8\right)}{3}$

Thus, the sum of series ${\displaystyle \sum _{k=1}^n\left(k^2-3k+4\right)}$ is $\frac{n\left(n^2-3n+8\right)}{3}$ .