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Leibniz Series Regrouped: Sum 1/(1·3)+1/(5·7)+... | JEE

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Question
If 1−13+15−17+19−111+⋯=π41-\frac13+\frac15-\frac17+\frac19-\frac{1}{11}+\cdots=\frac{\pi}{4}, then the value of 11⋅3+15⋅7+19⋅11+⋯\frac{1}{1\cdot3}+\frac{1}{5\cdot7}+\frac{1}{9\cdot11}+\cdots is
Aπ8\frac{\pi}{8}correct
Bπ6\frac{\pi}{6}
Cπ4\frac{\pi}{4}
Dπ36\frac{\pi}{36}
Solution
Step 1: Pair the given series:
π4=(1−13)+(15−17)+(19−111)+⋯ .\frac{\pi}{4}=\left(1-\frac13\right)+\left(\frac15-\frac17\right)+\left(\frac19-\frac{1}{11}\right)+\cdots.
 Step 2: Each pair is twice a required term, since 1−13=21⋅31-\frac13=\frac{2}{1\cdot3}, 15−17=25⋅7\frac15-\frac17=\frac{2}{5\cdot7}, …\ldots:
π4=2(11⋅3+15⋅7+19⋅11+⋯ ).\frac{\pi}{4}=2\left(\frac{1}{1\cdot3}+\frac{1}{5\cdot7}+\frac{1}{9\cdot11}+\cdots\right).
 Step 3: So the required sum is π8.\dfrac{\pi}{8}. Correct answer: (1)
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