Chapter 11.4, Problem 12OE

To determine

To tell:The series is same or not.

Answer to Problem 12OE

The terms of the series are not same, since the series is not same.

Explanation of Solution

Given information: ${\displaystyle \sum _{n=0}^4{\left(-1\right)}^n}\left(n+1\right),{\displaystyle \sum _{j=1}^5{\left(-1\right)}^j\left(j^2\right)}$

Take, ${\displaystyle \sum _{n=0}^4{\left(-1\right)}^n}\left(n+1\right)$

Put, $n=0,1,2,3,\text{ and 4}$ , we get the series.

  $\begin{array}{l}\Rightarrow {\displaystyle \sum _{n=0}^4{\left(-1\right)}^n}\left(n+1\right)={\left(-1\right)}^0\left(0+1\right)+{\left(-1\right)}^1\left(1+1\right)+{\left(-1\right)}^2\left(2+1\right)+{\left(-1\right)}^3\left(3+1\right)+{\left(-1\right)}^4\left(4+1\right)\\ \Rightarrow {\displaystyle \sum _{n=0}^4{\left(-1\right)}^n}\left(n+1\right)=1-2+3-4+5\\ \Rightarrow {\displaystyle \sum _{n=0}^4{\left(-1\right)}^n}\left(n+1\right)=9-6\\ \Rightarrow {\displaystyle \sum _{n=0}^4{\left(-1\right)}^n}\left(n+1\right)=3\end{array}$

Now, take ${\displaystyle \sum _{j=1}^5{\left(-1\right)}^j\left(j^2\right)}$

Put, $j=1,2,3,4,\text{ and}\text{5}$ , we get the series.

  $\begin{array}{l}\Rightarrow {\displaystyle \sum _{j=1}^5{\left(-1\right)}^j\left(j^2\right)}={\left(-1\right)}^1\left(1^2\right)+{\left(-1\right)}^2\left(2^2\right)+{\left(-1\right)}^3\left(3^2\right)+{\left(-1\right)}^4\left(4^2\right)+{\left(-1\right)}^5\left(5^2\right)\\ \Rightarrow {\displaystyle \sum _{j=1}^5{\left(-1\right)}^j\left(j^2\right)}=-1+4-9+16-25\\ \Rightarrow {\displaystyle \sum _{j=1}^5{\left(-1\right)}^j\left(j^2\right)}=20-35\\ \Rightarrow {\displaystyle \sum _{j=1}^5{\left(-1\right)}^j\left(j^2\right)}=-10\end{array}$

From the above series it is clear that the both series are not same.