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Question: How do you find the sum of the finite geometric sequence of \(\sum {{2^{n - 1}}} \) from \(n = 1\;{\...

How do you find the sum of the finite geometric sequence of ∑2n−1\sum {{2^{n - 1}}} from n=1  to  9?n = 1\;{\text{to}}\;9?

Explanation

Solution

This is given that the given series is a geometric series and 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 ∑2n−1\sum {{2^{n - 1}}} 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=2(n+1)−12n−1=2\Rightarrow r = \dfrac{{{2^{(n + 1) - 1}}}}{{{2^{n - 1}}}} = 2
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)}}
To find value of its first term (a)(a) we will put n=1n = 1 in general term of the given geometric series
a=21−1=20=1a = {2^{1 - 1}} = {2^0} = 1
And according to the question n=9n = 9
∴S14=1(1−29)(1−2)=1−512−1=−511−1=511\therefore {S_{14}} = \dfrac{{1\left( {1 - {2^9}} \right)}}{{\left( {1 - 2} \right)}} = \dfrac{{1 - 512}}{{ - 1}} = \dfrac{{ - 511}}{{ - 1}} = 511

Therefore the required infinite sum of the given series is equals to 511511

Note: When this type of series is given to you, where series is indirectly given that is summation of its general term (as in this question) then to find common ratio or common difference just put n+1  and  nn + 1\;{\text{and}}\;n in the general term expression and find common ratio or common difference. Also put n=1n = 1 to find the first term.