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Algebra Level 3

$\frac { 2 }{ 15 } +\frac { 2 }{ 35 } +\frac { 2 }{ 63 } +.......+\frac { 2 }{ 9999 }$

2 solutions

Sujoy Roy
Nov 24, 2014

Given expression $=\sum_{n=1}^{49}\frac{2}{(2n+1)(2n+3)}$

$=\sum_{n=1}^{49}[\frac{(2n+3)-(2n+1)}{(2n+1)(2n+3)}]$

$=\sum_{n=1}^{49}[\frac{1}{2n+1}-\frac{1}{2n+3}]$

$=(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+\ldots+(\frac{1}{99}-\frac{1}{101})$

$=(\frac{1}{3}-\frac{1}{101})=\boxed{0.323}$

But can't it be $\dfrac{2}{n^2-1}$ ? @sujoy roy

Abdur Rehman Zahid - 6 years, 5 months ago

Yes , the answer comes by this also good observation

$\displaystyle \sum_{n=4}^{100} \dfrac{2}{(n-1)(n+1)}$

$\displaystyle \sum_{n=4}^{100} \left(\dfrac{1}{n-1} - \dfrac{1}{n + 1}\right)$

$\dfrac{1}{3} - \dfrac{1}{101}$

U Z - 6 years, 4 months ago

Abhiram Rao
Apr 25, 2016

This one came in NSEJS some time back.