Chapter 9, Problem 6CT

To determine

To calculate: The value of the expression ${\displaystyle \sum _{i=1}^{50}\left(2i^2+5\right)}$ .

Answer to Problem 6CT

The value of the expression ${\displaystyle \sum _{i=1}^{50}\left(2i^2+5\right)}$ is $86,100$ .

Explanation of Solution

Given information:

The expression ${\displaystyle \sum _{i=1}^{50}\left(2i^2+5\right)}$ .

Formula used:

Summation property, ${\displaystyle \sum \left(a+b\right)={\displaystyle \sum a}}+{\displaystyle \sum b}$ and ${\displaystyle \sum _{n=1}^{m}an}=a{\displaystyle \sum _{n=1}^{m}n}$ .

The sum of square of first n terms of sequence is given by $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ .

Calculation:

Consider expression ${\displaystyle \sum _{i=1}^{50}\left(2i^2+5\right)}$ . Apply the summation property, ${\displaystyle \sum \left(a+b\right)={\displaystyle \sum a}}+{\displaystyle \sum b}$ and ${\displaystyle \sum _{n=1}^{m}an}=a{\displaystyle \sum _{n=1}^{m}n}$ to obtain,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^{50}\left(2i^2+5\right)}={\displaystyle \sum _{i=1}^{50}2i^2}+{\displaystyle \sum _{i=1}^{50}5}\\ =2{\displaystyle \sum _{i=1}^{50}{i}^2}+{\displaystyle \sum _{i=1}^{50}5}\end{array}$

Expand the expression as,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^{50}\left(2i^2+5\right)}={\displaystyle \sum _{i=1}^{50}2i^2}+{\displaystyle \sum _{i=1}^{50}5}\\ =2{\displaystyle \sum _{i=1}^{50}{i}^2}+{\displaystyle \sum _{i=1}^{50}5}\\ =2\left(1^2+2^2+3^2+\cdots +{50}^2\right)+\left(5+5+\cdots 5\right)\left[\text{50times}\right]\end{array}$

Recall that the sum of square of first n terms of sequence is given by $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ .

So, sum of square of first 50 terms is $\frac{50\left(50+1\right)\left(2\left(50\right)+1\right)}{6}$ . Apply it,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^{50}\left(2i^2+5\right)}=2\frac{50\left(50+1\right)\left(2\left(50\right)+1\right)}{6}+250\\ =\frac{\left(50\right)\left(51\right)\left(101\right)}{3}+250\\ =86100\end{array}$

Thus, the value of the expression ${\displaystyle \sum _{i=1}^{50}\left(2i^2+5\right)}$ is $86,100$ .