Root

Algebra

Root

Postby Guest » Sun Mar 29, 2020 12:24 am

Can you solve this problem?
https://youtu.be/qygYkoPWjOs
Guest
 

Re: Root

Postby Guest » Mon Mar 30, 2020 6:06 pm

?? The website you link to gives a complete solution! Did you not understand it?

I will try to make it a little clearer.

The sum is [tex]\frac{1}{\sqrt{1}+ \sqrt{2}}+\frac{1}{\sqrt{2}+ \sqrt{3}}+\frac{1}{\sqrt{3}+ \sqrt{4}}+ \cdot\cdot\cdot+\frac{1}{\sqrt{2016}+\sqrt{2017}}[/tex].

What they do is simplify each fraction by "completing the square". That uses the fact that, for any numbers, a and b, [tex](a+ b)(a- b)= a^2- b^2[/tex]. In particular, for any n, [tex](\sqrt{n}+\sqrt{n+1})(\sqrt{n}-\sqrt{n+1})= (\sqrt{n})^2- (\sqrt{n+1})^2= n- (n+1)= -1[/tex].

We can "complete the square" by multiplying both numerator and denominator by [tex]\sqrt{n}- \sqrt{n+1}[/tex]:
[tex]\frac{1}{\sqrt{n}+\sqrt{n+1}}\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}-\sqrt{n+1}}=[/tex][tex]-(\sqrt{n}-\sqrt{n+1})= -(\sqrt{n}- \sqrt{n+1})[/tex].

Now, look what happens when we replace the fractions in the original sum with these equal sums!
[tex]-(\sqrt{1}- \sqrt{2})-(\sqrt{2}- \sqrt{3})- \cdot\cdot\cdot- (\sqrt{2016}- \sqrt{2017})[/tex]
[tex]= (\sqrt{2}- \sqrt{1})+ (\sqrt{3}- \sqrt{2})+ (\sqrt{4}- \sqrt{3})+ \cdot\cdot\cdot+ (\sqrt{2017}- \sqrt{2016})[/tex].

Now this is what is called a "telescoping" sum because all the interior terms cancel! It reduces to [tex]\sqrt{2017}- \sqrt{1}= \sqrt{2017}- 1[/tex].
Guest
 

Re: Root

Postby Guest » Mon Apr 27, 2020 9:00 pm

That is NOT "completing the square", it is "rationalizing the denominator".
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