Find the value of \[1{\text{ }} + {\text{ }}2/1.2.3{\text{ }} + {\text{ }}2/3.4.5{\text{ }} + {\text{ }}2/5.6.7{\text{ }} + ....\infty {\text{ }}\]
A. \[{\text{2 log 2}}\]
B. \[{\text{2 log 4}}\]
C. \[{\text{2 log 3}}\]
D. None of these

Hint: In this question, we need to find the value of the given expression of \[\log \left( {1 + x} \right)\]\[1{\text{ }} + {\text{ }}2/1.2.3{\text{ }} + {\text{ }}2/3.4.5{\text{ }} + {\text{ }}2/5.6.7{\text{ }} + ....\infty {\text{ }}\]. For this, we have to consider the expansion series of \[\log \left( {1 + x} \right)\]. After that by putting \[x = 1\] and by simplifying it further, we get the final result.

Formula used: We know that the expansion of \[\log \left( {1 + x} \right)\] is \[\log \left( {1 + x} \right) = x{\text{ }} - {\text{ }}{x^2}/2{\text{ }} + {\text{ }}{x^3}/3{\text{ }} - {\text{ }}{x^4}/4{\text{ }} + {\text{ }}......\infty \]

Complete step-by-step answer:
We have to consider \[\log \left( {1 + x} \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4}.....\]
Now, put \[x = 1\] in the above equation.
Thus, we get
\[\log \left( {1 + 1} \right) = 1 - \dfrac{{{{\left( 1 \right)}^2}}}{2} + \dfrac{{{{\left( 1 \right)}^3}}}{3} - \dfrac{{{{\left( 1 \right)}^4}}}{4} + \dfrac{{{{\left( 1 \right)}^5}}}{5} - \dfrac{{{{\left( 1 \right)}^6}}}{6}.....\]
\[\log \left( 2 \right) = \underbrace {1 - \dfrac{1}{2}}_{consider} + \underbrace {\dfrac{1}{3} - \dfrac{1}{4}}_{consider} + \underbrace {\dfrac{1}{5} - \dfrac{1}{6}}_{consider}.....\]
By simplifying, we get
\[\log \left( 2 \right) = \dfrac{{2 - 1}}{2} + \dfrac{{4 - 3}}{{3 \times 4}} - \dfrac{{6 - 5}}{{5 \times 6}} + .....\]
\[\log \left( 2 \right) = \dfrac{1}{{1 \times 2}} + \dfrac{1}{{1 \times 3 \times 4}} - \dfrac{1}{{1 \times 5 \times 6}} + .....\] (1)
Now consider \[\log \left( 2 \right) = 1 - \underbrace {\dfrac{1}{2} + \dfrac{1}{3}}_{consider} - \underbrace {\dfrac{1}{4} + \dfrac{1}{5}}_{consider} - \dfrac{1}{6}.....\] again for simplification.
Thus, we get
\[\log \left( 2 \right) = 1 + \dfrac{{ - 3 + 2}}{{1 \times 2 \times 3}} + \dfrac{{ - 5 + 4}}{{1 \times 4 \times 5}} - .....\]
\[\log \left( 2 \right) = 1 - \dfrac{1}{{1 \times 2 \times 3}} - \dfrac{1}{{1 \times 4 \times 5}} - .....\] (2)
Let us add (1) and (2)
Thus, we get
\[\log \left( 2 \right) + \log \left( 2 \right) = \left[ {\dfrac{1}{{1 \times 2}} + \dfrac{1}{{1 \times 3 \times 4}} - \dfrac{1}{{1 \times 5 \times 6}} + .....} \right] + \left[ {1 - \dfrac{1}{{1 \times 2 \times 3}} - \dfrac{1}{{1 \times 4 \times 5}} - .....} \right]\]
\[2\log \left( 2 \right) = 1 + \left( {\dfrac{1}{{1 \times 2}} - \dfrac{1}{{1 \times 2 \times 3}}} \right) + \left( {\dfrac{1}{{1 \times 3 \times 4}} - \dfrac{1}{{1 \times 4 \times 5}}} \right) + ....\]
\[2\log \left( 2 \right) = 1 + \left( {\dfrac{{6 - 2}}{{12}}} \right) + \left( {\dfrac{{20 - 12}}{{240}}} \right) + ....\]
By simplifying further, we get
\[2\log \left( 2 \right) = 1 + \left( {\dfrac{4}{{12}}} \right) + \left( {\dfrac{8}{{240}}} \right) + ....\]
Let us rearrange the terms here.
\[2\log \left( 2 \right) = 1 + \left( {\dfrac{{2 \times 2}}{{2 \times 2 \times 3}}} \right) + \left( {\dfrac{{2 \times 2 \times 2}}{{4 \times 60}}} \right) + ....\]
\[2\log \left( 2 \right) = 1 + \left( {\dfrac{2}{{1 \times 2 \times 3}}} \right) + \left( {\dfrac{2}{{1 \times 3 \times 4 \times 5}}} \right) + ....\]
So, we can write it as
\[{\text{2 log 2}} = {\text{1 }} + {\text{ }}2/1.2.3{\text{ }} + {\text{ }}2/3.4.5{\text{ }} + {\text{ }}2/5.6.7{\text{ }} + ....\infty {\text{ }}\]
Hence, we can say that the value of \[1{\text{ }} + {\text{ }}2/1.2.3{\text{ }} + {\text{ }}2/3.4.5{\text{ }} + {\text{ }}2/5.6.7{\text{ }} + ....\infty {\text{ }}\]is \[{\text{2 log 2}}\].
Therefore, the correct option is (A).

Additional information: A series expansion is defined as an illustration of a specific function as the sum of powers in one of the variables or as the sum of powers of the other (typically fundamental) function \[f\left( x \right)\]. We know that there are several types of series expansions such as the Taylor series, Fourier series, Laurent series, Dirichlet series, etc.

Note: Here, students generally make mistakes in writing the expansion of \[\log \left( {1 + x} \right)\]. They may forget to take alternate signs. So, ultimately the final result may get wrong. Also, it is necessary to rearrange the terms while simplifying the expansion. Here, the main trick is to get two series by doing simplifications and finally add them to get the desired result.