Q- Find the sum of the series?
1/(2^1/2 + 1^1/2) + 1/(2^1/2 + 3^1/2) + ......... + 1/(120^1/2 + 121^1/2)
Any help would be greatly appreciated.
Thanks,
Ashish
Note that $\displaystyle \frac{1}{{\sqrt {k + 1} + \sqrt k }} = \sqrt {k + 1} - \sqrt k $.
You series is then $\displaystyle \sum\limits_{k = 1}^{120} {\frac{1}{{\sqrt {k + 1} + \sqrt k }}} = \sum\limits_{k = 1}^{120} {\left( {\sqrt {k + 1} - \sqrt k } \right)} $
Write out several terms to see what is going to happen.