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Question: How do you find the sum of the infinite geometric series \(\dfrac{{{2^n}}}{{{5^{2n + 1}}}}?\)...

How do you find the sum of the infinite geometric series 2n52n+1?\dfrac{{{2^n}}}{{{5^{2n + 1}}}}?

Explanation

Solution

To find the sum of infinite geometric series, firstly, find the common ratio between two consecutive terms, and get consecutive terms by putting n=a  and  a+1n = a\;{\text{and}}\;a + 1 in the given geometric expression.

Formula used:

  1. Common ratio of a G.P.: r=un+1unr = \dfrac{{{u_{n + 1}}}}{{{u_n}}}
  2. Infinite sum of G.P.: S∞=a1−r{S_\infty } = \dfrac{a}{{1 - r}}, where S∞{S_\infty }, aa and rr are the sum of infinite geometric series, first term and the common ratio of the series respectively.

Complete step by step solution:
In order to find the sum of the infinite geometric series 2n52n+1\dfrac{{{2^n}}}{{{5^{2n + 1}}}} we will first find the common ratio of the series by 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
r=2n+152(n+1)+12n52n+1=225r = \dfrac{{\dfrac{{{2^{n + 1}}}}{{{5^{2(n + 1) + 1}}}}}}{{\dfrac{{{2^n}}}{{{5^{2n + 1}}}}}} = \dfrac{2}{{25}}
Now, we will use the formula for sum of infinite terms of geometric series which is given as follows
S∞=a1−r,  where  S∞,  a  and  r{S_\infty } = \dfrac{a}{{1 - r}},\;{\text{where}}\;{S_\infty },\;a\;{\text{and}}\;r are sum of infinite geometric series, first term of the series and common ratio of the series respectively
In the series 2n52n+1\dfrac{{{2^n}}}{{{5^{2n + 1}}}} the first term can be find out by substituting the value of n=1n = 1 in the given geometric expression
⇒a=2152×1+1=253=2125\Rightarrow a = \dfrac{{{2^1}}}{{{5^{2 \times 1 + 1}}}} = \dfrac{2}{{{5^3}}} = \dfrac{2}{{125}}
Putting a=2125  and r=225a = \dfrac{2}{{125}}\;{\text{and}}\,r = \dfrac{2}{{25}} in the above formula, we will get
S∞=21251−225=2125×2525−2=23×23=269{S_\infty } = \dfrac{{\dfrac{2}{{125}}}}{{1 - \dfrac{2}{{25}}}} = \dfrac{{\dfrac{2}{{125}} \times 25}}{{25 - 2}} = \dfrac{2}{{3 \times 23}} = \dfrac{2}{{69}}

Therefore the required infinite sum of the given series is equal to 269\dfrac{2}{{69}}.

Note:
The value of the common ratio in a geometric should never be equal to one, since the progression in terms will not occur when it equals one and thus there will be no geometric sequence. In an arithmetic series, the common ratio equal to one in the geometric series is identical to the common difference equal to zero.