This question was uploaded on 09/10/21 on social media accounts.
Question taken from YouTube uploaded on "letsthinkcritically".
This question is a good example of the telescopic sum of infinite series. The numerator is added by 2 in each term and in the denominator an extra term is multiplied which is greater than the previous term by 4.
Solution:
Given series:
\[S = \frac{1}{3}+\frac{3}{3\cdot7}+\frac{5}{3\cdot7\cdot11}+\frac{7}{3\cdot7\cdot11\cdot15}+\cdot\cdot\cdot\]
\[\Rightarrow 2S = \frac{2}{3}+\frac{6}{3\cdot7}+\frac{10}{3\cdot7\cdot11}+\frac{14}{3\cdot7\cdot11\cdot15}+\cdot\cdot\cdot\]
\[\Rightarrow 2S = \frac{2}{3}+\left(\frac{1}{3}-\frac{1}{3\cdot7}\right)+\left(\frac{1}{3\cdot7}-\frac{1}{3\cdot7\cdot11}\right)+\cdot\cdot\cdot\]
This is a telescopic sum and also this series is a reducing series.
\[\Rightarrow 2S = \frac{2}{3}+\frac{1}{3}\]
\[\Rightarrow 2S = 1\]
\[\Rightarrow S = \frac12\]
Question related to Algebra, Summation.
