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Question: How do you find the sum of the first \(14\) terms of the series \(1 + 4 + 16 + 64 + ...?\)...

How do you find the sum of the first 1414 terms of the series 1+4+16+64+...?1 + 4 + 16 + 64 + ...?

Explanation

Solution

First of all find the nature of the given series then proceed accordingly. The series is an infinite geometric series and as we know that the common ratio of consecutive terms in a geometric series is fixed. To find the sum of infinite geometric series, firstly, find the common ratio between terms, by dividing a term with its progressive term. And then use the following formula:
Sn=a(1−rn)(1−r),  where  Sn,  n,  a  and  r{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}},\;{\text{where}}\;{S_n},\;n,\;a\;{\text{and}}\;r are sum of first “n” terms of geometric series, number of terms, first term of the series and common ratio of the series respectively.

Formula used:
Common ratio of a G.P.: r=un+1unr = \dfrac{{{u_{n + 1}}}}{{{u_n}}}
Sum of nn terms of G.P.: Sn=a(1−rn)(1−r){S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}

Complete step by step solution:
In order to find the sum of the infinite geometric series 1+4+16+64+...1 + 4 + 16 + 64 + ... we will first find the common ratio of the series as following
r=un+1un,  where  un+1  and  unr = \dfrac{{{u_{n + 1}}}}{{{u_n}}},\;{\text{where}}\;{u_{n + 1}}\;{\text{and}}\;{u_n} are “n+1th” and “nth” term of the geometric series respectively
We will take second and first term, to find the common ratio,
⇒r=41=4\Rightarrow r = \dfrac{4}{1} = 4
Now, we will use the formula for sum of first “n” terms of geometric series which is given as follows
Sn=a(1−rn)(1−r){S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}
In the series 1+4+16+64+...1 + 4 + 16 + 64 + ... the first term is a=1a = 1
And according to the question n=14n = 14
∴S14=1(1−414)(1−4)=1−268435456−3=−268435455−3=89478485\therefore {S_{14}} = \dfrac{{1\left( {1 - {4^{14}}} \right)}}{{\left( {1 - 4} \right)}} = \dfrac{{1 - 268435456}}{{ - 3}} = \dfrac{{ - 268435455}}{{ - 3}} = 89478485

Therefore the required infinite sum of the given series is equals to 8947848589478485

Note: The common ratio of geometric series or progression often tells us about the essence of the series, which either increases or decreases depending on the value of the common ratio being greater than one or less than one. And we're going to get an alternate geometric sequence if the common ratio is negative.