# find the value for this expression?

• Aug 29th 2010, 04:00 AM
grgrsanjay
find the value for this expression?
find the value of $\sum_{k=1}^{99}\frac{1}{K\sqrt{K+1} + (K+1)\sqrt{K}})$
• Aug 29th 2010, 06:56 AM
Opalg
Quote:

Originally Posted by grgrsanjay
find the value of $\sum_{k=1}^{99}\frac{1}{K\sqrt{K+1} + (K+1)\sqrt{K}})$

Sketch solution: Rationalise the fraction by multiplying top and bottom by $(k+1)\sqrt k - k\sqrt{k+1}$, and you will find that you have a telescoping sum.

Spoiler:
• Aug 29th 2010, 05:37 PM
grgrsanjay
Quote:

Hello, grgrsanjay!

Opalg gave you an excellent game plan . . . Did you follow it?

Quote:

$\displaystyle \text{Find the value of: }\;\sum_{k=1}^{99}\,\frac{1}{(k\!+\!1)\sqrt{k} + k\sqrt{k\!+\!1}}$

Rationalize the denominator . . . multiply top and bottom by the conjugate:

. . $\displaystyle \frac{1}{(k\!+\!1)\sqrt{k} + k\sqrt{k\!+\!1}} \cdot\frac{(k\!+\!1)\sqrt{k} - k\sqrt{k\!+\!1}}{(k\!+\!1)\sqrt{k} - k\sqrt{k\!+\!1}} \;\;=\;\; \frac{(k\!+\!1)\sqrt{k} - k\sqrt{k\!+\!1}}{(k\!+\!1)^2k - k^2(k\!+\!1)} \;\;=\;\;\frac{(k\!+\!1)\sqrt{k} - k\sqrt{k\!+\!1}}{k(k\!+\!1)}$

. . . . $\displaystyle =\;\;\frac{(k\!+\!1)\sqrt{k}}{k(k\!+\!1)} - \frac{k\sqrt{k\!+\!1}}{k(k\!+\!1)} \;\;=\;\;\frac{\sqrt{k}}{k} - \frac{\sqrt{k\!+\!1}}{k\!+\!1} \;\;=\;\;\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k\!+\!1}}$

The sum becomes: . $\displaystyle \sum^{99}_{k=1}\left(\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k\!+\!1}}\right)$

. . . . . . . . . . . . $\displaystyle =\;\;\left(\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}\right) + \left(\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}}\right) + \hdots + \left(\frac{1}{\sqrt{99}} - \frac{1}{\sqrt{100}}\right)$

All terms cancel out . . . except the first and the last.

. . The sum is: . $\displaystyle \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{100}} \;\;=\;\;1 - \frac{1}{10} \;\;=\;\;\boxed{\frac{9}{10}}$

thanks you sorobon and thanks opalg