Chapter 12, Problem 41RE

To determine

Calculate the sum of the ${\displaystyle \sum _{k=1}^{10}{\left(k-1\right)}^2}$ without using sigma notation and do not evaluate it.

Answer to Problem 41RE

The final summation of ${\displaystyle \sum _{k=1}^{10}{\left(k-1\right)}^2}$ is ${\left(0\right)}^2+{\left(1\right)}^2+{\left(2\right)}^2+\mathrm{.........}+{\left(9\right)}^2$ .

Explanation of Solution

Given information:

It is given in the question that, ${\displaystyle \sum _{k=1}^{10}{\left(k-1\right)}^2}$

By evaluating the given summation ${\displaystyle \sum _{k=1}^{10}{\left(k-1\right)}^2}$ ,

by using the method of expanding

  $\begin{array}{l}{\left(k-1\right)}^2+{\left(k-1\right)}^2+{\left(k-1\right)}^2+\mathrm{.........}+{\left(k-1\right)}^2\\ {\left(1-1\right)}^2+{\left(2-1\right)}^2+{\left(3-1\right)}^2+\mathrm{.........}+{\left(10-1\right)}^2\\ {\left(0\right)}^2+{\left(1\right)}^2+{\left(2\right)}^2+\mathrm{.........}+{\left(9\right)}^2\end{array}$

The final summation of ${\displaystyle \sum _{k=1}^{10}{\left(k-1\right)}^2}$ is ${\left(0\right)}^2+{\left(1\right)}^2+{\left(2\right)}^2+\mathrm{.........}+{\left(9\right)}^2$ .