Sum of the First $n$ Terms of a Series

Example 1:

Find the sum of the first $20$ terms of the arithmetic series if $a_1=5$ and $a_{20}=62$ .

$\begin{array}{l}{S}_{20}=\frac{20(5+62)}{2}\\ S_{20}=670\end{array}$

Example 2:

Find the sum of the first $40$ terms of the arithmetic sequence
$2,5,8,11,14,\cdots$

First find the $40$ ^th term:

$\begin{array}{l}{a}_{40}=a_1+(n-1)d\\ =2+39\left(3\right)=119\end{array}$

Then find the sum:

$\begin{array}{l}{S}_n=\frac{n(a_1+a_n)}{2}\\ S_{40}=\frac{40(2+119)}{2}=2420\end{array}$

Example 3:

Find the sum:

${\displaystyle \underset{k=1}{\overset{50}{\sum }}(3k+2)}$

First find $a_1$ and $a_{50}$ :

$\begin{array}{l}{a}_1=3\left(1\right)+2=5\\ a_{20}=3\left(50\right)+2=152\end{array}$

Then find the sum:

$\begin{array}{l}{S}_k=\frac{k(a_1+a_k)}{2}\\ S_{50}=\frac{50(5+152)}{2}=3925\end{array}$

Example 4:

Find the sum of the first $8$ terms of the geometric series if $a_1=1$ and $r=2$ .

$S_8=\frac{1(1-2^8)}{1-2}=255$

Example 5:

Find $S_{10}$ of the geometric series $24+12+6+\cdots$ .

First, find $r$ . 

$r=\frac{r_2}{r_1}=\frac{12}{24}=\frac{1}{2}$

Now, find the sum:

$S_{10}=\frac{24\left(1-{\left(\frac{1}{2}\right)}^{10}\right)}{1-\frac{1}{2}}=\frac{3069}{64}$

Example 6:

Evaluate.

${\displaystyle \underset{n=1}{\overset{10}{\sum }}3{(-2)}^{n-1}}$

(You are finding $S_{10}$ for the series $3-6+12-24+\cdots$ , whose common ratio is $-2$ .)

$\begin{array}{l}{S}_n=\frac{a_1(1-r^n)}{1-r}\\ S_{10}=\frac{3\left[1-{(-2)}^{10}\right]}{1-(-2)}=\frac{3(1-1024)}{3}=-1023\end{array}$