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Question: Find the value of \[\dfrac{1}{{3 \times 1}} + \dfrac{1}{{5 \times 7}} + \dfrac{1}{{11 \times 9}} + -...

Find the value of 13×1+15×7+111×9+−−−−\dfrac{1}{{3 \times 1}} + \dfrac{1}{{5 \times 7}} + \dfrac{1}{{11 \times 9}} + - - - - when 1−13+15−17+19−111=π41-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}=\dfrac{\pi }{4}

Explanation

Solution

Hint : The general term will be 1(2n−1)(2n+1)=12[(2n+1)−(2n−1)(2n−1)(2n+1)]=12[12n−1−12n+1]\dfrac{1}{(2n-1)(2n+1)}=\dfrac{1}{2}\left[ \dfrac{(2n+1)-(2n-1)}{(2n-1)(2n+1)} \right]=\dfrac{1}{2}\left[ \dfrac{1}{2n-1}-\dfrac{1}{2n+1} \right]. We will be solving the given question following this approach of the general term. First convert the equation given in the question to that of the general term and calculate the final answer.

Complete step-by-step answer :
We have,
1−13+15−17+19−111=π41-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}=\dfrac{\pi }{4}
can be written in general form in the following manner:
12n−1−12n+1\dfrac{1}{2n-1}-\dfrac{1}{2n+1}
Here, we have taken two terms, from the left side, at a time and the value of n starts with n=1.
Now, we need to evaluate
13×1+15×7+111×9+−−−−=\dfrac{1}{{3 \times 1}} + \dfrac{1}{{5 \times 7}} + \dfrac{1}{{11 \times 9}} + - - - - =
which can also be written as:
1(2n−1)(2n+1)\dfrac{1}{(2n-1)(2n+1)}.
So, we will rearrange this in the following manner:
1(2n−1)(2n+1)=12[(2n+1)−(2n−1)(2n−1)(2n+1)]=12[12n−1−12n+1] ⇒12[3−13×1+7−57×5+11−911×9+..........] ⇒12[1−13+15−17+19−111+........] ⇒12[π4] ⇒π8   \dfrac{1}{{(2n - 1)(2n + 1)}} = \dfrac{1}{2}\left[ {\dfrac{{(2n + 1) - (2n - 1)}}{{(2n - 1)(2n + 1)}}} \right] = \dfrac{1}{2}\left[ {\dfrac{1}{{2n - 1}} - \dfrac{1}{{2n + 1}}} \right] \\\ \Rightarrow \dfrac{1}{2}\left[ {\dfrac{{3 - 1}}{{3 \times 1}} + \dfrac{{7 - 5}}{{7 \times 5}} + \dfrac{{11 - 9}}{{11 \times 9}} + ..........} \right] \\\ \Rightarrow \dfrac{1}{2}\left[ {1 - \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} + \dfrac{1}{9} - \dfrac{1}{{11}} + ........} \right] \\\ \Rightarrow \dfrac{1}{2}\left[ {\dfrac{\pi }{4}} \right] \\\ \Rightarrow \dfrac{\pi }{8} \;
So, the correct answer is “π8\dfrac{\pi }{8}”.

Note : Whenever you have to find the sum of a series containing fractions where denominators can be expressed into two numbers, write the numerator as the difference of the numbers in the denominator and then separate one term into the difference of two in the following manner
a−ba×b\dfrac{{a - b}}{{a \times b}} = aa×b\dfrac{a}{a \times b} - ba×b\dfrac{b}{a \times b} = 1b−1a\dfrac{1}{b} - \dfrac{1}{a}