Chapter 8.2, Problem 74E

To determine

To find:Thepartial sum ${\displaystyle \sum _{n=51}^{100}n}-{\displaystyle \sum _{n=1}^{50}n}$ .

Answer to Problem 74E

  ${\displaystyle \sum _{n=51}^{100}n}-{\displaystyle \sum _{n=1}^{50}n}=2500$

Explanation of Solution

Given information:

A sequence is given as ${\displaystyle \sum _{n=51}^{100}n}-{\displaystyle \sum _{n=1}^{50}n}$ .

Concept used:

An arithmetic sequence of n terms, has the form $a_1,a_2,a_3,\mathrm{...},a_n$ and should have a common difference between each two consecutive terms.

That is

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}$

Common difference can be defined by d .

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}=d$

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.

Sum of an arithmetic finite sequence has the form

  $S_n=\frac{n}{2}\left(a_1+a_n\right)$

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

Calculation:

Consider the given partial sum.

  ${\displaystyle \sum _{n=51}^{100}n}-{\displaystyle \sum _{n=1}^{50}n}$

The above sum can be written as

  ${\displaystyle \sum _{n=51}^{100}n}-{\displaystyle \sum _{n=1}^{50}n}=\left(51+52+53+\mathrm{...}+100\right)-\left(1+2+3+\mathrm{...}+50\right)$

The expression ${\displaystyle \sum _{n=51}^{100}n}-{\displaystyle \sum _{n=1}^{50}n}$ is difference of two arithmetic sequence.

Now, first term of the sequence ${\displaystyle \sum _{n=51}^{100}n}$ is $a_1=50$ .

The difference is

  $\begin{array}{c}{d}_1=a_2-a_1\\ =51-50\\ =1\end{array}$

Last term is $a_{50}=100$

Number of term for the required sumis indicated as 50.

So, partial sum ${\displaystyle \sum _{n=51}^{100}n}$ is calculated as

  $\begin{array}{c}{\displaystyle \sum _{n=51}^{100}n}=S_n=\frac{n}{2}\left(a_1+a_n\right)\\ S_{50}=\frac{50}{2}\left(51+100\right)\\ =25\times 151\\ =3775\end{array}$

Now, first term of the sequence ${\displaystyle \sum _{n=1}^{50}n}$ is $a_1=1$ .

The difference is

  $\begin{array}{c}{d}_1=a_2-a_1\\ =2-1\\ =1\end{array}$

Last term is $a_{50}=50$

Number of term for the required sum is indicated as 50.

So, partial sum ${\displaystyle \sum _{n=1}^{50}n}$ is calculated as

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{50}n}=S_n=\frac{n}{2}\left(a_1+a_n\right)\\ S_{50}=\frac{50}{2}\left(1+50\right)\\ =25\times 51\\ =1275\end{array}$

Now, subtract ${\displaystyle \sum _{n=1}^{50}n}$ from ${\displaystyle \sum _{n=51}^{100}n}$ .

  $\begin{array}{c}{\displaystyle \sum _{n=51}^{100}n}-{\displaystyle \sum _{n=1}^{50}n}=3775-1275\\ =2500\end{array}$

Thus, ${\displaystyle \sum _{n=51}^{100}n}-{\displaystyle \sum _{n=1}^{50}n}=2500$ .