Chapter 11, Problem 38CLT

To determine

To calculate:

The sum of the following summation:

  ${\displaystyle \sum _{i=1}^{50}\left(1-i^2\right)}$

Answer to Problem 38CLT

The value of the ${\displaystyle \sum _{i=1}^{50}\left(1-i^2\right)}$ is $-42875$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{i=1}^{50}\left(1-i^2\right)}$

Formula Used:

The property of the summation is given by the following expression:

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

And

  $\begin{array}{l}{\displaystyle \sum _{k=1}^n{i}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ {\displaystyle \sum _{k=1}^{n}c}=cn\end{array}$

Calculation:

The given summation is:

  ${\displaystyle \sum _{i=1}^{50}\left(1-i^2\right)}$

Use the property of the summation:

  ${\displaystyle \sum _{i=1}^{50}\left(1-i^2\right)}={\displaystyle \sum _{i=1}^{50}1}+{\displaystyle \sum _{i=1}^{50}{i}^2}$

So, the value of the summation is calculated as:

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{50}\left(1-i^2\right)}={\displaystyle \sum _{i=1}^{50}1}+{\displaystyle \sum _{i=1}^{50}{i}^2}\\ {\displaystyle \sum _{i=1}^{50}\left(1-i^2\right)}=cn+\frac{n\left(n+1\right)\left(2x+1\right)}{6}\\ {\displaystyle \sum _{i=1}^{50}\left(1-i^2\right)}=\left(1\times 50\right)-\frac{50\left(50+1\right)\left(2\left(50\right)+1\right)}{6}\\ {\displaystyle \sum _{i=1}^{50}\left(1-i^2\right)}=50-42925\\ {\displaystyle \sum _{i=1}^{50}\left(1-i^2\right)}=-42875\end{array}$