Question

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

Answer 1

Hint: In this question, we are to find the given expression. For this, we need to apply the logarithmic series formula. Since the given expression is in rational form, we need to calculate the sum of both series. By rewriting the series into the form of a logarithmic series, we can evaluate the sum of the given series.



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 series is
$\dfrac{\left( a-1 \right)-\dfrac{{{\left( a-1 \right)}^{2}}}{2}+\dfrac{{{\left( a-1 \right)}^{3}}}{3}-...\infty }{\left( b-1 \right)-\dfrac{{{\left( b-1 \right)}^{2}}}{2}+\dfrac{{{\left( b-1 \right)}^{3}}}{3}-...\infty }$
The series in the numerator is
$\left( a-1 \right)-\dfrac{{{\left( a-1 \right)}^{2}}}{2}+\dfrac{{{\left( a-1 \right)}^{3}}}{3}-...\infty \text{ }...(1)$
The series in the denominator is
$\left( b-1 \right)-\dfrac{{{\left( b-1 \right)}^{2}}}{2}+\dfrac{{{\left( b-1 \right)}^{3}}}{3}-...\infty \text{ }...(2)$
But we have a logarithmic series as
${{\log }_{e}}(1+x)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+...(3)$
So, by comparing (1) and (3), we get
$x=a-1$
Thus, on substituting this value into the logarithmic function in (3), we get
$\begin{align}
  & {{\log }_{e}}(1+x)={{\log }_{e}}(1+(a-1)) \\
 & \Rightarrow {{\log }_{e}}a \\
\end{align}$
Now, by comparing (2) and (3), we get
$x=b-1$
Thus, on substituting this value into the logarithmic function in (3), we get
$\begin{align}
  & {{\log }_{e}}(1+x)={{\log }_{e}}(1+(b-1)) \\
 & \Rightarrow {{\log }_{e}}b \\
\end{align}$
Thus, the given expression is
$\dfrac{\left( a-1 \right)-\dfrac{{{\left( a-1 \right)}^{2}}}{2}+\dfrac{{{\left( a-1 \right)}^{3}}}{3}-...\infty }{\left( b-1 \right)-\dfrac{{{\left( b-1 \right)}^{2}}}{2}+\dfrac{{{\left( b-1 \right)}^{3}}}{3}-...\infty }=\dfrac{{{\log }_{e}}a}{{{\log }_{e}}b}$
But we have the property
${{\log }_{n}}m=\dfrac{{{\log }_{e}}m}{{{\log }_{e}}n}$
So, we can write
$\dfrac{\left( a-1 \right)-\dfrac{{{\left( a-1 \right)}^{2}}}{2}+\dfrac{{{\left( a-1 \right)}^{3}}}{3}-...\infty }{\left( b-1 \right)-\dfrac{{{\left( b-1 \right)}^{2}}}{2}+\dfrac{{{\left( b-1 \right)}^{3}}}{3}-...\infty }=\dfrac{{{\log }_{e}}a}{{{\log }_{e}}b}={{\log }_{b}}a$



Option ‘A’ is correct



Note: In this question, the series is easy to compare, the only difference is the variable $x$. By using logarithmic functions and expansions, this type of sum would be evaluated. We have to apply logarithm properties if needed in the simplification. Here, on comparing we know that the series is the same type but the variable is different i.e., in the given series $x=a-1$ (numerator) and $x=b-1$ (denominator). So, by substituting this in the logarithmic function, the required sum will be obtained.