Question

Find $\dfrac{1}{2}{{x}^{2}}+\dfrac{2}{3}{{x}^{3}}+\dfrac{3}{4}{{x}^{4}}+......\infty =$
A. $\dfrac{x}{1+x}-{{\log }_{e}}(1-x)$
B. $\dfrac{x}{1+x}+{{\log }_{e}}(1-x)$
C. $\dfrac{x}{1-x}-{{\log }_{e}}(1-x)$
D. $\dfrac{x}{1-x}+{{\log }_{e}}(1-x)$

Answer 1

Hint: In this question, we are to find the sum of the given series. To solve this, the given series is to be rewritten in such a way that we can frame the series into a particular progression. So, by applying the appropriate formula, the required sum is to be calculated.



Formula Used:Logarithmic series:
If $x\in R$ and $\left| x \right|<1$, then the expansion is
${{\log }_{e}}(1+x)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+...$
If $x\in R$ and $\left| x \right|<1$, then the expansion is
${{\log }_{e}}(1-x)=-x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-...$
Some important formulae:
If $\left| x \right|<1$ then ${{\log }_{e}}\dfrac{1+x}{1-x}=2\left( x+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}+... \right)$
If $x>1$ then ${{\log }_{e}}\dfrac{x+1}{x-1}=2\left( \dfrac{1}{x}+\dfrac{1}{3{{x}^{3}}}+\dfrac{1}{5{{x}^{5}}}+... \right)$



Complete step by step solution:Given expansion is
$\dfrac{1}{2}{{x}^{2}}+\dfrac{2}{3}{{x}^{3}}+\dfrac{3}{4}{{x}^{4}}+......\infty $
The above sequence is rewritten as
$\begin{align}
  & \Rightarrow \left( 1-\dfrac{1}{2} \right){{x}^{2}}+\left( 1-\dfrac{1}{3} \right){{x}^{3}}+\left( 1-\dfrac{1}{4} \right){{x}^{4}}+...\infty \\
 & \Rightarrow \left( {{x}^{2}}+{{x}^{3}}+{{x}^{4}}+....\infty \right)-\left( \dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{4}}}{4}+...\infty \right) \\
\end{align}$
In the above expansion, the first sequence is in geometric progression. So, we can find its sum by ${{S}_{\infty }}=\dfrac{a}{1-r}$. I.e.,
$\left( {{x}^{2}}+{{x}^{3}}+{{x}^{4}}+....\infty \right)=\dfrac{{{x}^{2}}}{1-x}\text{ }...(1)$
The second sequence is a logarithmic series. So, we can write
$\left( \dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{4}}}{4}+... \right)=\{-{{\log }_{e}}(1-x)-x\}\text{ }...(2)$
Substituting (1) and (2) in the rewritten sequence, we get
$\begin{align}
  & \Rightarrow \left( {{x}^{2}}+{{x}^{3}}+{{x}^{4}}+....\infty \right)-\left( \dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{4}}}{4}+...\infty \right)=\dfrac{{{x}^{2}}}{1-x}-\left\{ -{{\log }_{e}}(1-x)-x \right\} \\
 & \Rightarrow \dfrac{{{x}^{2}}}{1-x}+x+{{\log }_{e}}(1-x) \\
 & \Rightarrow \dfrac{{{x}^{2}}+x(1-x)}{1-x}+{{\log }_{e}}(1-x) \\
 & \Rightarrow \dfrac{x}{1-x}+{{\log }_{e}}(1-x) \\
\end{align}$
Thus, the sum of the sequence is
$\dfrac{1}{2}{{x}^{2}}+\dfrac{2}{3}{{x}^{3}}+\dfrac{3}{4}{{x}^{4}}+......\infty =\dfrac{x}{1-x}+{{\log }_{e}}(1-x)$



Option ‘D’ is correct


Note: The given series is rewritten in order to get the appropriate progression in the series. Once the series represents a particular progression, we can find the sum of that series easily by applying the formula. Here the series is rewritten as the sum of the geometric series and the logarithmic series. So, by applying their formulae, we get the required sum. In such type of question, we need to observe the given series, so that we are able to find its progression. Then, the process of solving becomes easy for us.