91Ó°ÊÓ

$\underset{n=1}{\overset{∞}{∑}}\frac{(-1{)}^n{n}^2}{n!}$

The first term of the series $\underset{n=1}{\overset{∞}{∑}}\frac{(-1{)}^n{n}^2}{n!}$is obtained by substituting $n=1$in $\frac{(-1{)}^n{n}^2}{n!}$. Therefore, the value at $n=1$is:

$\frac{(-1{)}^n{n}^2}{n!}=\frac{(-1{)}^1\left(1^2\right)}{1!}$(Substituting)

$=\frac{-1}{1!}=-1$

Therefore, first term of the series $\underset{n=1}{\overset{∞}{∑}}\frac{(-1{)}^n{n}^2}{n!}$is -1.

The second term of the series $\underset{n=1}{\overset{∞}{∑}}\frac{(-1{)}^n{n}^2}{n!}$is obtained by substituting $n=2$in $\frac{(-1{)}^n{n}^2}{n!}$. Therefore, the value at $n=2$is:

$\frac{(-1{)}^n{n}^2}{n!}=\frac{(-1{)}^2(2{)}^2}{2!}$( Substituting)

$=\frac{4}{2!}$

The second term of the series $\underset{n=1}{\overset{∞}{∑}}\frac{(-1{)}^n{n}^2}{n!}$is $\frac{4}{2!}$.

The third term of the series $\underset{n=1}{\overset{∞}{∑}}\frac{(-1{)}^n{n}^2}{n!}$is obtained by substituting$n=3$in $\frac{(-1{)}^n{n}^2}{n!}$. Therefore, the value at $n=3$ is:

$\frac{(-1{)}^n{n}^2}{n!}=\frac{(-1{)}^3(3{)}^2}{3!}$ (Substituting)

$=\frac{-9}{3!}$

The third term of the series $\underset{n=1}{\overset{∞}{∑}}\frac{(-1{)}^n{n}^2}{n!}$is $\frac{-9}{3!}$.

The fourth term of the series $\underset{n=1}{\overset{∞}{∑}}\frac{(-1{)}^n{n}^2}{n!}$is obtained by substituting $n=4$in $\frac{(-1{)}^n{n}^2}{n!}$. Therefore, the value at $n=4$is:

$\frac{(-1{)}^n{n}^2}{n!}=\frac{(-1{)}^4(4{)}^2}{4!}$( Substituting)

$=\frac{16}{4!}$

The fourth term of the series $\underset{n=1}{\overset{∞}{∑}}\frac{(-1{)}^n{n}^2}{n!}$is$\frac{16}{4!}$.

The fifth term of the series $\underset{n=1}{\overset{∞}{∑}}\frac{(-1{)}^n{n}^2}{n!}$is obtained by substituting $n=5$in $\frac{(-1{)}^n{n}^2}{n!}$. Therefore, the value at $n=5$is:

$\frac{(-1{)}^n{n}^2}{n!}=\frac{(-1{)}^5(5{)}^2}{5!}$(Substituting)

$=-\frac{25}{5!}$

The fifth term of the series $\underset{n=1}{\overset{∞}{∑}}\frac{(-1{)}^n{n}^2}{n!}$is $-\frac{25}{5!}$.