Chapter 7.1, Problem 62E

Solution

To state: Whether the statement is true o false. If the statement is true, prove it. It the statement is false, give a counterexample.

  ${\displaystyle \sum _{i=1}^n{\left(a_i\right)}^k}={\left({\displaystyle \sum _{i=1}^n{a}_i}\right)}^k$ .

The statement is false.

Given information:

The given statement is ${\displaystyle \sum _{i=1}^n{\left(a_i\right)}^k}={\left({\displaystyle \sum _{i=1}^n{a}_i}\right)}^k$ .

Explanation:

Now consider the given statement,

  ${\displaystyle \sum _{i=1}^n{\left(a_i\right)}^k}={\left({\displaystyle \sum _{i=1}^n{a}_i}\right)}^k$

Let’s assume $a_i=i,n=3,\text{and}k=2$ .

Substitute the values in the given summation notation,

  ${\displaystyle \sum _{i=1}^3{\left(i\right)}^2}={\left({\displaystyle \sum _{i=1}^{3}i}\right)}^2$

Evaluate ${\displaystyle \sum _{i=1}^3{\left(i\right)}^2}$ ,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^3{\left(i\right)}^2}=1^2+2^2+3^2\\ =1+4+9\\ =14\end{array}$

Similarly, evaluate ${\left({\displaystyle \sum _{i=1}^{3}i}\right)}^2$ .

  $\begin{array}{c}{\left({\displaystyle \sum _{i=1}^{3}i}\right)}^2={\left(1+2+3\right)}^2\\ ={\left(6\right)}^2\\ =36\end{array}$

It can be seen that from both sides the resultant answers are different.

Therefore, the given statement ${\displaystyle \sum _{i=1}^n{\left(a_i\right)}^k}={\left({\displaystyle \sum _{i=1}^n{a}_i}\right)}^k$ is false.