Q .24 - Calculus

Question

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

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

Step-by-Step Solution

Answer

Ans: The five terms of the series are $7,9,11,13,15$

Step 1. Given information:

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

Step 2. Finding the first term of the series:

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:

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

=6+1 

=7

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

Step 3. Finding the second term of the series:

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

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

$$\begin{gathered} =8+1 \\ =9 \end{gathered}$$

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

Step 4. Finding the third term of the series:

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

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

=10+1

=11

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

Step 5. Finding the fourth term of the series:

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

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

=12+1

=13

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

Step 6. Finding the fifth term of the series:

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

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

=14+1 

=15

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