This question was uploaded on 26/11/21 on social media accounts.
This question was asked at Cambridge university in 1801.
Solution:
Given series:
\[S = \frac1{1\times3} - \frac1{2\times4} + \frac1{3\times5} - \frac1{4\times6} -.....\]
All terms are in the form of:
\[\frac1{(n)(n+2)}\]
\[\rightarrow \frac1{(n)(n+2)} = \frac12\left( \frac1n - \frac1{n+2} \right)\]
Now our given series:
\[S = \frac12\left[\left( \frac11 - \frac13 \right) - \left( \frac12 - \frac14 \right) + \left( \frac13 - \frac15 \right) - \left( \frac14 - \frac16 \right) + ....\right]\]
\[S = \frac12\left[1 - \frac12\right] = \frac14\]
