Maths C Question Im having trouble with

[Digital Genius]

Homework Statement

Evaluate:

99
∑ 1 over √r+1 + √r
r=1

Homework Equations

this might be relevant..

the previous question was this, which i managed to solve but it still has the same fraction in it.

Show that: 1 over √r+1 + √r = √r+1 - √r , for r is greater or equal to 0

The Attempt at a Solution

i got as far as...

T1 =

99
∑ 1 over √r+1 + √r
r=1

=0.4142

T2= 2/√2+1 + √2 ...r=2
=0.6357

OR

T2=
99
∑ 1 over √r+1 + √r......ar^n-1
r=1

a=
1 over √r+1 + √r
.....ok nevermind i have no clue haha, i did get a bit more then that but just realized it doesn't make sense... thanks for the help in advance :)

 

[Saitama]

Hi Digital Genius! Welcome to PF! :)

I assume your question is:
$$\sum_{r=1}^{99} \frac{1}{\sqrt{r+1}+\sqrt{r}}$$

Hint: Multiply and divide by $\sqrt{r+1}-\sqrt{r}$.

 

[sankalpmittal]

If your question is same as stated by Pranav, then rationalize the denominator and convert it two sum of two terms. Then you see that the mid terms cancel.

If it is as you have shown in your attempt, then why, i consider it very difficult or at least very cumbersome.

 

[Digital Genius]

ok well according to my friends its this, but i don't see where they multiplied and divided...

=√99+1 - √r
=√99+1 -√1
=√100 - √1
=10-1
=9

any ideas??

 

[DrClaude]

Use your knowledge of the previous problem. If
$$
\frac{1}{\sqrt{r+1}+\sqrt{r}} = \sqrt{r+1}- \sqrt{r}
$$
how can you simplify
$$
\sum_{r=1}^{99} \frac{1}{\sqrt{r+1}+\sqrt{r}}
$$
?