# Root

Algebra

### Root

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

### Re: Root

?? 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 $$\frac{1}{\sqrt{1}+ \sqrt{2}}+\frac{1}{\sqrt{2}+ \sqrt{3}}+\frac{1}{\sqrt{3}+ \sqrt{4}}+ \cdot\cdot\cdot+\frac{1}{\sqrt{2016}+\sqrt{2017}}$$.

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

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

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

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

### Re: Root

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