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The sum of the series $ 2\left( {{1.7}^{-1}}+{{3}^{-1}}{{.7}^{-3}}+{{5}^{-1}}{{.7}^{-5}}+....... \right) $ .

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Last updated date: 25th Jun 2024
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Answer
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Hint: For this kind of problem, we need to simplify the given series into the know formulas. For this, we need to observe the terms in the given series and use suitable formulas for simplification. Here we will use the formula $ {{a}^{-n}}=\dfrac{1}{{{a}^{n}}} $ and modify all the series by using this formula. We have the formulas $ \log \left( 1+x \right) $ and $ \log \left( 1-x \right) $ to compare the obtained series and find the required result.

Complete step by step answer:
Given that,
 $ 2\left( {{1.7}^{-1}}+{{3}^{-1}}{{.7}^{-3}}+{{5}^{-1}}{{.7}^{-5}}+....... \right) $
Let $ S=2\left( {{1.7}^{-1}}+{{3}^{-1}}{{.7}^{-3}}+{{5}^{-1}}{{.7}^{-5}}+....... \right) $ .
Using the formula $ {{a}^{-n}}=\dfrac{1}{{{a}^{n}}} $ in the above equation, then we will get
 $ S=2\left( 1.\dfrac{1}{7}+\dfrac{1}{3}.\dfrac{1}{{{7}^{3}}}+\dfrac{1}{5}.\dfrac{1}{{{7}^{5}}}+....... \right) $
We know that $ {{1}^{n}}=1 $ , then we will get
 $ S=2\left( \dfrac{1}{7}+\dfrac{1}{3}.\dfrac{{{1}^{3}}}{{{7}^{3}}}+\dfrac{1}{5}.\dfrac{{{1}^{5}}}{{{7}^{5}}}+...... \right) $
We know that $ \dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}} $ , then we will have
 $ S=2\left( \dfrac{1}{7}+\dfrac{1}{3}.{{\left( \dfrac{1}{7} \right)}^{3}}+\dfrac{1}{5}.{{\left( \dfrac{1}{7} \right)}^{5}}+...... \right) $
Considering $ \dfrac{1}{7} $ as whole term in the above equation, then we will get
 $ S=2\left( \dfrac{{{\left( \dfrac{1}{7} \right)}^{1}}}{1}+\dfrac{{{\left( \dfrac{1}{7} \right)}^{3}}}{3}+\dfrac{{{\left( \dfrac{1}{7} \right)}^{5}}}{5}+...... \right)......\left( \text{i} \right) $
We have the formulas $ \log \left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+...... $ and $ \log \left( 1-x \right)=-x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-....... $
Now the value of \[\log \left( 1+x \right)-\log \left( 1-x \right)\] can be calculated as
\[\begin{align}
  & \log \left( 1+x \right)-\log \left( 1-x \right)=\left[ x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+.... \right]-\left[ -x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-.... \right] \\
 & \Rightarrow \log \left( 1+x \right)-\log \left( 1-x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+x+\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{4}}}{4}+..... \\
 & \Rightarrow \log \left( 1+x \right)-\log \left( 1-x \right)=2x+2.\dfrac{{{x}^{3}}}{3}+2.\dfrac{{{x}^{5}}}{5}+..... \\
\end{align}\]
Taking $ 2 $ as common from the above equation, then we will get
 $ \Rightarrow \log \left( 1+x \right)-\log \left( 1-x \right)=2\left( x+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}+..... \right) $
Let us substituting the value $ x=\dfrac{1}{7} $ in the above equation, then we will get
 $ \log \left( 1+\dfrac{1}{7} \right)-\log \left( 1-\dfrac{1}{7} \right)=2\left( \dfrac{1}{7}+\dfrac{{{\left( \dfrac{1}{7} \right)}^{3}}}{3}+\dfrac{{{\left( \dfrac{1}{7} \right)}^{5}}}{5}+.... \right) $
From equation $ \left( \text{i} \right) $ , we have the value of $ 2\left( \dfrac{1}{7}+\dfrac{{{\left( \dfrac{1}{7} \right)}^{3}}}{3}+\dfrac{{{\left( \dfrac{1}{7} \right)}^{5}}}{5}+.... \right)=S $ , then we will get
 $ S=\log \left( 1+\dfrac{1}{7} \right)-\log \left( 1-\dfrac{1}{7} \right) $
Simplifying the above equation by taking the LCM of $ 1+\dfrac{1}{7} $ , $ 1-\dfrac{1}{7} $ in the above equation, then we will get
 $ \begin{align}
  & \Rightarrow S=\log \left( \dfrac{1\times 7+1}{7} \right)-\log \left( \dfrac{1\times 7-1}{7} \right) \\
 & \Rightarrow S=\log \left( \dfrac{8}{7} \right)-\log \left( \dfrac{6}{7} \right) \\
\end{align} $
We have the logarithmic formula $ \log a-\log b=\log \left( \dfrac{a}{b} \right) $ , then we will get
 $ \begin{align}
  & S=\log \left( \dfrac{\dfrac{8}{7}}{\dfrac{6}{7}} \right) \\
 & \Rightarrow S=\log \left( \dfrac{8}{7}\times \dfrac{7}{6} \right) \\
\end{align} $
Cancelling the possible terms in the above equation, then we will get
 $ S=\log \left( \dfrac{4}{3} \right) $
Hence the value of the series $ 2\left( {{1.7}^{-1}}+{{3}^{-1}}{{.7}^{-3}}+{{5}^{-1}}{{.7}^{-5}}+....... \right) $ is $ \log \left( \dfrac{4}{3} \right) $ .

Note:
 The problems related to the series are very complicated i.e., we will get different answers for a small changes in the question. Here we have the series like $ 2\left( {{1.7}^{-1}}+{{3}^{-1}}{{.7}^{-3}}+{{5}^{-1}}{{.7}^{-5}}+....... \right) $ , if it is like $ 2\left( {{1.7}^{-1}}+{{3.7}^{-3}}+{{5.7}^{-5}}+....... \right) $ , then we need to use different formulas and different method and also, we will get different answer for this case.