Question

If $b=a-\dfrac{{{a}^{2}}}{2}+\dfrac{{{a}^{3}}}{3}-\dfrac{{{a}^{4}}}{4}+...$ then $b+\dfrac{{{b}^{2}}}{2!}+\dfrac{{{b}^{3}}}{3!}+\dfrac{{{b}^{4}}}{4!}+...\infty $
A. ${{\log }_{e}}a$
B. ${{\log }_{e}}b$
C. $a$
D. ${{e}^{a}}$

Answer 1

Hint: In this question, we are to find the sum of the given series. For this, we need to apply the logarithmic series formula. By rewriting the given series into the form of required series, we can evaluate its value.



Formula Used:Exponential series:
${{e}^{x}}=1+\dfrac{x}{1!}+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+...$
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
$b=a-\dfrac{{{a}^{2}}}{2}+\dfrac{{{a}^{3}}}{3}-\dfrac{{{a}^{4}}}{4}+...(1)$
But we have a logarithmic series as
${{\log }_{e}}(1+x)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+...(2)$
On comparing (1) and (2), we get
$x=a$
Thus, on substituting this value into the logarithmic function, we get
$\begin{align}
  & b=a-\dfrac{{{a}^{2}}}{2}+\dfrac{{{a}^{3}}}{3}-\dfrac{{{a}^{4}}}{4}+...\infty ={{\log }_{e}}(1+a) \\
 & \Rightarrow b={{\log }_{e}}(1+a) \\
\end{align}$
Since we know that \[{{n}^{x}}=m\Leftrightarrow {{\log }_{n}}m=x\], then
$\begin{align}
  & b={{\log }_{e}}(1+a) \\
 & \Rightarrow 1+a={{e}^{b}} \\
\end{align}$
We have the exponential series as,
${{e}^{x}}=1+\dfrac{x}{1!}+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+...$
So, we can write
$1+a={{e}^{b}}=1+\dfrac{b}{1!}+\dfrac{{{b}^{2}}}{2!}+\dfrac{{{b}^{3}}}{3!}+....\infty $
On simplifying,
$\begin{align}
  & 1+a=1+\dfrac{b}{1!}+\dfrac{{{b}^{2}}}{2!}+\dfrac{{{b}^{3}}}{3!}+....\infty \\
 & \Rightarrow a=b+\dfrac{{{b}^{2}}}{2!}+\dfrac{{{b}^{3}}}{3!}+....\infty \\
\end{align}$
Thus, the value of the series is $a$.



Option ‘C’ 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 same type but the variable is different i.e., in the given series $x=b$. So, by substituting this in the logarithmic function, the required sum will be obtained.