Question

Sum of series $\frac{1}{1.2.3}+\frac{5}{3.4.5}+\frac{9}{5.6.7}+\dots$ is equal to

• A. $\frac{3}{2}-3{\log }_e\left(2\right)$
• B. $\frac{5}{2}-3{\log }_e\left(2\right)$
• C. $1-4{\log }_e\left(2\right)$
• D. None of these

Answer 1

The correct option is B

$\frac{5}{2}-3{\log }_e\left(2\right)$

Explanation for the correct answer:

Finding the sum of series:

$\mathrm{Let}\mathrm{S}=\frac{1}{1.2.3}+\frac{5}{3.4.5}+\frac{9}{5.6.7}+...$

The nth term is:

$$\begin{gathered} {\mathrm{T}}_{\mathrm{n}}=\frac{1+\left(\mathrm{n}-1\right)4}{\left(2\mathrm{n}-1\right)\left(2\mathrm{n}\right)\left(2\mathrm{n}+1\right)} \\ =\frac{\left(4\mathrm{n}-3\right)}{\left(2\mathrm{n}-1\right)\left(2\mathrm{n}\right)\left(2\mathrm{n}+1\right)} \end{gathered}$$

Now, write ${\mathrm{T}}_{\mathrm{n}}$ as partial fraction:

$$\begin{gathered} {\mathrm{T}}_{\mathrm{n}}=\frac{\mathrm{A}}{(2\mathrm{n}-1)}+\frac{\mathrm{B}}{2\mathrm{n}}+\frac{\mathrm{C}}{(2\mathrm{n}+1)} \\ \text{Considering the denominators} \\ 2\mathrm{n}-1=0 \\ \Rightarrow n=\frac{1}{2}\text{at}A \\ 2n=0 \\ \Rightarrow n=0\text{at}B \\ 2\mathrm{n}+1=0 \\ \Rightarrow n=-\frac{1}{2}\text{at}C \end{gathered}$$

Now substituting respective values we get,

$\begin{array}{rcl}\text{For}A\text{at}n& =& \frac{1}{2}\\ \frac{\left(4\left({\displaystyle \frac{1}{2}}\right)-3\right)}{\left(2\left(\frac{1}{2}\right)\right)\left(2\left(\frac{1}{2}\right)+1\right)}& =& A\\ \frac{\left(2-3\right)}{\left(1\right)\left(1+1\right)}& =& A\\ A& =& -\frac{1}{2}\\ \text{For}B\text{at}n& =& 0\\ \frac{\left(4\left({\displaystyle 0}\right)-3\right)}{\left(2\left(0\right)-1\right)\left(2\left(0\right)+1\right)}& =& B\\ \frac{\left(-3\right)}{\left(-1\right)\left(1\right)}& =& B\\ B& =& 3\\ \text{For}C\text{at}n& =& \frac{-1}{2}\\ \frac{\left(4\left({\displaystyle -\frac{1}{2}}\right)-3\right)}{\left(2\left(-\frac{1}{2}\right)-1\right)\left(2\left(-\frac{1}{2}\right)\right)}& =& C\\ \frac{\left(-2-3\right)}{\left(-1-1\right)\left(-1\right)}& =& C\\ C& =& \frac{-5}{2}\end{array}$

Hence,

$$\begin{gathered} \mathrm{A}=\frac{-1}{2},\mathrm{B}=3\mathrm{and}\mathrm{C}=\frac{-5}{2} \\ \mathrm{So}{\mathrm{T}}_{\mathrm{n}}=\frac{{\displaystyle -1}}{2\left(2\mathrm{n}-1\right)}+\frac{3}{2\mathrm{n}}-\frac{5}{2\left(2\mathrm{n}+1\right)} \end{gathered}$$

In the third bracket, add and subtract

$$\begin{gathered} \mathrm{S}=\frac{-1}{2}\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+.........\right)+3\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.........\right)-\frac{5}{2}\left(1-1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+.........\right) \\ =\frac{-1}{2}\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+.........\right)+3\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.........\right)-\frac{5}{2}\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+.........\right)+\frac{5}{2} \\ =-3\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+.........\right)+3\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.........\right)+\frac{5}{2} \\ =-3\left(1-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+........\right)+\frac{5}{2} \\ =\frac{5}{2}-3{\log }_{\mathrm{e}}\left(2\right) \end{gathered}$$

Hence, option(B) is the correct answer