Calculate sum[i=4,15][1/(sqrt[i+1]+sqrt[i])]

[laurence000]

Hi, I'm struggling with sums a lot. The following is what I need to calculate. I could do it "the long way", but how should I approach the problem in order not to calculate it the long way?

Thank you so much!
Laurence

 

[soroban]

Hello, Laurence!

Someone already explained this . . . I'll give you the details.

\(\displaystyle \L\sum^{15}_{i=4}\frac{1}{\sqrt{i\,+\,1}\,+\,\sqrt{i}}\)

Multiply top and bottom by the conjugate:

. . \(\displaystyle \L\frac{1}{\sqrt{i\,+\,1}\,+\,\sqrt{i}}\cdot\frac{\sqrt{1\,+\,i}\,-\,\sqrt{i} }{\sqrt{1\,+\,1}\,-\,\sqrt{i}} \;= \;\frac{\sqrt{1\,+\,i}\,-\,\sqrt{i}}{(1\,+\,i)\,-\,i} \;=\;\sqrt{1\,+\,i}\,-\,\sqrt{i}\)

So we have: \(\displaystyle \L\:\sum^{15}_{i=4}\left(\sqrt{1\,+\,i}\,-\,\sqrt{i}\right)\)

. . \(\displaystyle \L=\;(\not{\sqrt{5}}\,-\,\sqrt{4})\,+\,(\not{\sqrt{6}}\,-\,\not{\sqrt{5}})\,+\,(\not{\sqrt{7}}\,-\,\not{\sqrt{6}})\,+\,(\not{\sqrt{8}}\,-\,\not{\sqrt{7}})\,+\,\cdots\,+\,(\not{\sqrt{15}}\,-\,\not{\sqrt{14}})\,+\,(\sqrt{16}\,-\,\not{\sqrt{15}})\)

. . \(\displaystyle \L= \;-\sqrt{4}\,+\,\sqrt{16}\;=\;-2\,+\,4\;=\;\fbox{2}\)

 

[laurence000]

thanks so much, I didn't think about using identity!!! it helps a lot!!!