Chapter 8.2, Problem 65E

To find The partial sum of ${\displaystyle \sum _{n=1}^{500}\left(n+8\right)}$

To determine

The partial sum of ${\displaystyle \sum _{n=1}^{500}\left(n+8\right)}$ is $129250$ .

Given information:

The given sum is ${\displaystyle \sum _{n=1}^{500}\left(n+8\right)}$ .

Definition used:

The nth term of the arithmetic sequence has the form $a_n=a_1+\left(n-1\right)d$ ,where $a_1$ is the first term of the sequence, and d is the common difference.

The sum of finite arithmetic sequence is given by $S_n=\frac{n\left(a_1+a_n\right)}{2}$ .

Here n is the number of terms, $a_1$ is the first term of the sequence, and $a_n$ is the last terms of sequence.

Calculation:

It is given that ${\displaystyle \sum _{n=1}^{500}\left(n+8\right)}$ .

The above sum can be written as follows,

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{500}\left(n+8\right)}=\left(1+8\right)+\left(2+8\right)+\left(3+8\right)+\left(4+8\right)+\cdots \left(500+8\right)\\ =9+10+11+12+\cdots +508\end{array}$

Compute the common difference as follows,

  $\begin{array}{c}{d}_1=a_2-a_1\\ =10-9\\ =1\end{array}$

  $\begin{array}{c}{d}_2=a_3-a_2\\ =12-11\\ =1\end{array}$

Therefore, the given sequence is an arithmetic sequence.

Compute the partial sum as follows,

  $\begin{array}{c}{S}_{500}=\frac{500\left(9+508\right)}{2}\\ =250\cdot 517\\ =129250\end{array}$

Therefore, the sum of the partial arithmetic sequence is $129250$ .

Answer to Problem 65E

The partial sum of ${\displaystyle \sum _{n=1}^{500}\left(n+8\right)}$ is $129250$ .

Explanation of Solution

Given information:

The given sum is ${\displaystyle \sum _{n=1}^{500}\left(n+8\right)}$ .

Definition used:

The nth term of the arithmetic sequence has the form $a_n=a_1+\left(n-1\right)d$ ,where $a_1$ is the first term of the sequence, and d is the common difference.

The sum of finite arithmetic sequence is given by $S_n=\frac{n\left(a_1+a_n\right)}{2}$ .

Here n is the number of terms, $a_1$ is the first term of the sequence, and $a_n$ is the last terms of sequence.

Calculation:

It is given that ${\displaystyle \sum _{n=1}^{500}\left(n+8\right)}$ .

The above sum can be written as follows,

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{500}\left(n+8\right)}=\left(1+8\right)+\left(2+8\right)+\left(3+8\right)+\left(4+8\right)+\cdots \left(500+8\right)\\ =9+10+11+12+\cdots +508\end{array}$

Compute the common difference as follows,

  $\begin{array}{c}{d}_1=a_2-a_1\\ =10-9\\ =1\end{array}$

  $\begin{array}{c}{d}_2=a_3-a_2\\ =12-11\\ =1\end{array}$

Therefore, the given sequence is an arithmetic sequence.

Compute the partial sum as follows,

  $\begin{array}{c}{S}_{500}=\frac{500\left(9+508\right)}{2}\\ =250\cdot 517\\ =129250\end{array}$

Therefore, the sum of the partial arithmetic sequence is $129250$ .