Chapter 8.2, Problem 68E

To findThepartial sum of ${\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}$

To determine

The partial sum of ${\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}$ is $218625$ .

Given information:

The given sum is ${\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}$ .

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 ${\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}$ .

The above sum can be written as follows,

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}=\left(1000-1\right)+\left(1000-2\right)+\left(1000-3\right)+\left(1000-4\right)+\cdots \left(1000-250\right)\\ =999+998+997+996+\cdots +750\end{array}$

Compute the common difference as follows,

  $\begin{array}{c}{d}_1=a_2-a_1\\ =998-999\\ =-1\end{array}$

  $\begin{array}{c}{d}_2=a_3-a_2\\ =997-998\\ =-1\end{array}$

Therefore, the given sequence is an arithmetic sequence.

Compute the partial sum as follows,

  $\begin{array}{c}{S}_{250}=\frac{250\left(999+750\right)}{2}\\ =125\cdot 1749\\ =218625\end{array}$

Therefore, the sum of the partial arithmetic sequence is $218625$ .

Answer to Problem 68E

The partial sum of ${\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}$ is $218625$ .

Explanation of Solution

Given information:

The given sum is ${\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}$ .

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 ${\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}$ .

The above sum can be written as follows,

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}=\left(1000-1\right)+\left(1000-2\right)+\left(1000-3\right)+\left(1000-4\right)+\cdots \left(1000-250\right)\\ =999+998+997+996+\cdots +750\end{array}$

Compute the common difference as follows,

  $\begin{array}{c}{d}_1=a_2-a_1\\ =998-999\\ =-1\end{array}$

  $\begin{array}{c}{d}_2=a_3-a_2\\ =997-998\\ =-1\end{array}$

Therefore, the given sequence is an arithmetic sequence.

Compute the partial sum as follows,

  $\begin{array}{c}{S}_{250}=\frac{250\left(999+750\right)}{2}\\ =125\cdot 1749\\ =218625\end{array}$

Therefore, the sum of the partial arithmetic sequence is $218625$ .