Question

How do you find the sum of finite geometric series?

Answer 1

Hint: Here in this question, we have to find the sum of finite geometric series. The geometric series is defined as the series with a constant ratio between the two successive terms. Then by considering the geometric series we have found the sum of the series.

Complete step-by-step answer:
In the mathematics we have three types of series namely, arithmetic series, geometric series and harmonic series.
The geometric series is defined as the series with a constant ratio between the two successive terms. The finite geometric series is generally represented as $ a,ar,a{r^2},...,a{r^n} $ , where a is first term and r is a common ratio.
Now we have to find the sum of finite geometric series, the sum for finite geometric series is defined by $ {S_n} $
The sum of finite geometric series is
 $ {S_n} = a + ar + a{r^2} + ... + a{r^n} $ ------(1)
Now we multiply the equation (1) by r so we get
 $ \Rightarrow r{S_n} = r\left( {a + ar + a{r^2} + ... + a{r^n}} \right) $
On multiplying we have
 $ \Rightarrow r{S_n} = ar + a{r^2} + a{r^3} + ... + a{r^{n + 1}} $ ------(2)
So, the equation (1) can be written as $ \Rightarrow {S_n} - a = ar + a{r^2} + ... + a{r^n} $ -------(3)
The equation (2) written as $ \Rightarrow r{S_n} = ar + a{r^2} + a{r^3} + ... + a{r^n} + a{r^{n + 1}} $ -----------(4)
Substitute the equation (3) in equation (4) we get
 $ \Rightarrow r{S_n} = {S_n} - a + a{r^{n + 1}} $ ----------(5)
Take $ {S_n} $ to the LHS we get
 $ \Rightarrow r{S_n} - {S_n} = a{r^{n + 1}} - a $ -------(6)
Take $ {S_n} $ common in LHS and a common in RHS we get
 $ \Rightarrow {S_n}(r - 1) = a({r^{n + 1}} - 1) $
The above equation is written as
 $ \Rightarrow {S_n} = \dfrac{{a({r^{n + 1}} - 1)}}{{(r - 1)}} $
So, the correct answer is “ $ \dfrac{{a({r^{n + 1}} - 1)}}{{(r - 1)}} $ ”.

The above formula or sum formula is used for the geometric series which is having the common ratio greater than 1.
The sum of finite geometric series for the common ratio less than 1, we have
 $ {S_n} = \dfrac{{a(1 - {r^{n + 1}})}}{{(1 - r)}} $
So, the correct answer is “ $ {S_n} = \dfrac{{a(1 - {r^{n + 1}})}}{{(1 - r)}} $ ”.

Note: Three different forms of series are arithmetic series, geometric series and harmonic series. For the arithmetic series is the series with common differences. The geometric series is the series with a common ratio. The sum is known as the total value of the given series.