# Thread: Problem related to sequence and series

1. ## Problem related to sequence and series

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

2. Write
1 = sqrt(2)^2 - sqrt(1)^2

1 = sqrt(3)^2 - sqrt(2)^2

1 = sqrt(4)^2 - sqrt(3)^2

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Remember that

sqrt(a)^2 - sqrt(b)^2 ={sqrt(a) - sqrt(b)} x {sqrt(a) + sqrt(b)}

Use this formula for each term many things will be cancelled .. ..

3. Originally Posted by a69356
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.