Q) Find the sum of the series:
a).
1/(2^1/2 + 1^1/2) + 1/(2^1/2 + 3^1/2) + .......+ 1/(120^1/2 +121^1/2)
b).
1/(1 x 5) + 1/(5 x 9) + 1/(9 x 13) + .......+ 1/(221 x 225).
Any help would be appreciated.
Thanks
$\displaystyle \frac1{(4n-3)(4n+1)} = \frac{1/4}{4n-3} - \frac{1/4}{4n+1}$. So
$\displaystyle \begin{aligned}\frac1{1\times5} + \frac1{5\times9}\, + &\,\frac1{9\times13} + \ldots + \frac1{221\times225}\\ &= \frac14\biggl(\Bigl(\frac11-\frac15\Bigr) + \Bigl(\frac15-\frac19\Bigr) + \Bigl(\frac19-\frac1{13}\Bigr) + \ldots + \Bigl(\frac1{221}-\frac1{225}\Bigr)\biggr).\end{aligned}$