1. ## Sum of Series

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

2. Originally Posted by a69356
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).
These are both telescoping series. For a), the n'th term is $\displaystyle \frac1{\sqrt{n+1}+\sqrt n} = \sqrt{n+1}-\sqrt n$. For b), the n'th term is $\displaystyle \frac1{(4n-3)(4n+1)}$, which you can split into partial fractions.

3. Thanks Opalg,

I found the sum of the first series it comes out to be 10.

But for the second one I found that there are 56 terms in the series and the series is neither A.P nor G.P.

How do we proceed further to find the sum of it.

Thanks,
Ashish

4. $\displaystyle \frac1{(4n-3)(4n+1)} = \frac{1/4}{4n-3} - \frac{1/4}{4n+1}$. 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}