In Exercises 21–28 provide the first five terms of the series.

$\sum _{n=3}^{\infty }(2n+1)$

$\sum _{n=3}^{\infty }(2n+1)$

The first term of the series $\sum _{n=3}^{\infty }(2n+1)$is obtained by substituting $n=3$in $2 n+1$. Therefore, the value at $n=3$is:

$2n+1=2\left(3\right)+1$(Substituting)

=6+1

=7

The first term of the series $\sum _{n=3}^{\infty }(2n+1)$ is 7 .

The second term of the series $\sum _{n=3}^{\infty }(2n+1)$is obtained by substituting $n=4$in $2n+1$. Therefore, the value at $n=4$is:

$2n+1=2\left(4\right)+1$(Substituting)

The second term of the series $\sum _{n=3}^{\infty }(2n+1)$is 9 .

The third term of the series $$\sum _{n=3}^{\infty }(2n+1)$$ is obtained by substituting $n=5$in $2n+1$. Therefore, the value at $n=5$is:

$2n+1=2\left(5\right)+1$(Substituting)

=10+1

=11

The third term of the series $\sum _{n=3}^{\infty }(2n+1)$is 11 .

The fourth term of the series $\sum _{n=3}^{\infty }(2n+1)$is obtained by substituting $n=6$in $2n+1$. Therefore, the value at $n=6$is:

$2n+1=2\left(6\right)+1$(Substituting)

=12+1

=13

The fourth term of the series $\sum _{n=3}^{\infty }(2n+1)$is 13 .

The fifth term of the series $\sum _{n=3}^{\infty }(2n+1)$is obtained by substituting $n=7$in $2n+1$. Therefore, the value at $n=7$is:

$2n+1=2\left(7\right)+1$(Substituting)

=14+1

=15

The fifth term of the series $\sum _{n=3}^{\infty }(2n+1)$is 15 .