Solution to Summation of Series Homework Statement

[sooyong94]

Homework Statement

Someone please check my work... :D

If $f(x)=\sqrt{x}+\sqrt{x+1}$ , find the value of
$\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+...+\frac{1}{f(24)}$

Homework Equations

Summation of series, rationalizing the denominator.

The Attempt at a Solution

$f(x)=\sqrt{x}+\sqrt{x+1}$
$f(x)=\sqrt{x}+\sqrt{x+1}(\frac{\sqrt{x}-\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}})$
$f(x)=\frac{x-(x+1)}{\sqrt{x}-\sqrt{x+1}}$
$\frac{1}{f(x)}=\sqrt{x+1}-\sqrt{x}$
$\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+...+\frac{1}{f(24)}$
$=\sum_{r=1}^{24} \sqrt{x+1}-\sqrt{x}$
$=(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+...+(\sqrt{24}-\sqrt{23})+(\sqrt{25}-\sqrt{24})$
$=\sqrt{25}-\sqrt{1}$
$=4$

 

[vanhees71]

I would have done it a bit shorter
$$\frac{1}{f(x)}=\frac{1}{\sqrt{x+1}+\sqrt{x}}=\frac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})}=\sqrt{x+1}-\sqrt{x}.$$
Your result is nevertheless correct.

 

[sooyong94]

Thanks a lot! :D

 

[Borek]

Your parentheses were slightly off.

 

[sooyong94]

Edited.