algebra problem

**the undertaker** · June 2nd 2009, 10:09 PM

1/sqrt(n) - 1/sqrt(n+1)

is there a quick way to find out the sum of the series where n=1 + n=2 + n=3, all the way up to...99?

**yeongil** · June 2nd 2009, 10:38 PM

Couldn't you have replied to your own thread at http://www.mathhelpforum.com/math-he...-real-one.html ?

$\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$
for n = 1, 2, 3,... 99

See what happens when you plug in the numbers:

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

Everything cancels except the first and last terms!

$\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}$

$\frac{1}{1}-\frac{1}{10}$

$\frac{9}{10}$

01

**Amer** · June 2nd 2009, 10:46 PM

you mean

$\sum_{n=1}^{99} \frac {1}{\sqrt n} -\frac {1}{\sqrt{n+1}}$

lats expand some

$\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt4}....\frac{1}{\sqrt{n+1}}$

as you can see the terms delete each other finally you get just

$1-\frac{1}{\sqrt{n+1}}$

now substitute 99 instead of n see what you will get ...

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