Geometric series

​​In a nutshell

An geometric series finds the sum of a given number of terms in an geometric sequence, a sequence whose terms are separated by a common ratio. 

Calculating the sum of a sequence

To find the sum of a given number of terms of an geometric sequence, use a formula rather than manually adding up the terms. The sum $$S_n$$ is given using the first term, $$a$$, the common ratio, $$r$$, and the $$n^{th}$$ term.​

​$$\boxed {S_n = \dfrac {a(1-r^n)}{1-r} \quad or \quad S_n = \dfrac {a(r^n-1)}{r-1}}$$​​

Note: Either formula can be used to find the sum but the formula on the left is easier when $$r<1$$ and the right when $$r>1$$.

Deriving the formula

The terms in a finite geometric sequence are as follows:

​$$a, ar^1 ,ar^2,... , ar^{n-2}, ar^{n-1}$$​​

This means the sum of a finite geometric sequence is equal to: 

​$$S_n=a+ar^1 +ar^2+...+ ar^{n-2}+ ar^{n-1}$$​​

Multiplying the series by $$r$$ will give: 

$$\begin{aligned} rS_n &=&r &[a+ar^1 +...+ ar^{n-2}+ ar^{n-1}] \\ rS_n &=& &[ar+ar^2 +...+ ar^{n-1}+ ar^{n}] \end{aligned}$$​​

​

Subtract this from $$S_n$$:

$$S_n - rS_n = [a+ar^1 +...+ ar^{n-2}+ ar^{n-1}]- [ar+ar^2 +...+ ar^{n-1}+ ar^{n}]$$​​

​$$S_n(1-r ) = a+[ar^1-ar^1] +...+ [ar^{n-2}-ar^{n-2}]+ [ar^{n-1}-ar^{n-1}]- ar^{n}$$​​

$$S_n(1-r ) = a- ar^{n}$$

​$$\boxed{S_n= \dfrac{a(1-r^{n})}{1-r}}$$

​Multiplying the top and bottom of the fraction by $$-1$$ which will maintain equivalence to $$S_n$$:

$$S_n= \dfrac{a(1-r^{n})}{1-r} \times \dfrac{-1}{-1}$$

$$\boxed{S_n= \dfrac{a(r^{n}-1)}{r-1}}$$​​

Example 1

Find the sum of the first $$12$$ terms of the geometric sequence with a first term of, $$7$$, and a common ratio of $$0.9$$. 

Identify the formula to use:

$$S_n = \dfrac{a(1-r^n)}{1-r}$$​​

​

Substitute values to find the sum: 

$$S_{12} = \dfrac{7(1-0.9^{12})}{1-0.9}$$​​

$$\underline{S_{12} = 50.23\ (2 \ d.p.)}$$​​

Example 2

How many terms are needed for the sum of a geometric series in the form $$2+4+8+16 +...$$ to exceed $$1000$$?

Identify the common ratio and first term:

$$r=\dfrac 42 = 2 \\ a = 2$$​​

Substitute the given values and solve:​

$$1000 = \dfrac{2(2^n-1)}{2- 1}$$​​

​$$500= {2^n-1}$$​​

​$$501 = 2^n$$

​$$\ln (501) = n \ln(2)$$​​

​$$n = \dfrac{\ln(501)}{\ln(2)} = 8.968666$$

Round up since it is the sum which exceeds $$1000$$:

​$$\underline{n = 9}$$​​