# Evaluating $\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{16}+\sqrt{20}+\sqrt{25}}=?$

It's a question (not hw) I bumped into few years back. Couldn't make any real progress with. Maybe you can help?

$$\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{16}+\sqrt{20}+\sqrt{25}}=?$$

Thanks.

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Your expression cannot be simplified further, except for taking out perfect-square-part out of each square root. – Sangchul Lee Jun 22 '12 at 12:33
Can you explain a bit more? It's supposed to be an Olympiad question or something like that. – Amihai Zivan Jun 22 '12 at 12:35
In denominator general term is $k$ + $sqrt(k(k+1))$ + (k+1) where k starts from 2. – Aang Jun 22 '12 at 12:40
I suspect that you are asking $$\frac{1}{\sqrt[3]{4}+\sqrt[3]{6}+\sqrt[3]{9}}+\frac{1}{\sqrt[3]{9}+\sqrt[3]{12‌​}+\sqrt[3]{16}}+\frac{1}{\sqrt[3]{16}+\sqrt[3]{20}+\sqrt[3]{25}} = \sqrt[3]{5}-\sqrt[3]{2}.$$ – Sangchul Lee Jun 22 '12 at 12:40
To see how sos440's equality can be obtained, invoke $$\rm \frac{1}{a^2+ab+b^2}=\frac{b~-~a~}{b^3-a^3}$$ telescopically by setting $\rm a=\sqrt[3]{k}$ and $\rm b=\sqrt[3]{k+1}$ for $\rm k=2,3,4$. – anon Jun 22 '12 at 12:58

The case of cube roots is probably more interesting than square roots; namely, simplifying

$$\frac{1}{\sqrt[3]{4}+\sqrt[3]{6}+\sqrt[3]{9}}+\frac{1}{\sqrt[3]{9}+\sqrt[3]{12‌​}+\sqrt[3]{16}}+\frac{1}{\sqrt[3]{16}+\sqrt[3]{20}+\sqrt[3]{25}}. \tag{\circ}$$

To evaluate this, as sos440 did in the comments, one notes the structure of the denominators are apparent as $a^2+ab+b^2$, which appears in the factorization in a difference of cubes, $b^3-a^3$. More generally, this sort of homogeneous polynomial results from using the geometric series formula on the common ratio $b/a$, but I digress. It is clear that we have

$$\frac{1}{a^2+ab+b^2}=\frac{b~-~a}{b^3-a^3} \tag{\bullet}$$

as the individual terms in $(\circ)$, with $b=\sqrt[3]{k+1}$ and $a=\sqrt[3]{k}$, for $k=2,3,4$. The denominators will all simply be $1$, and the three terms (the numerators) subsequently telescope:

$$\big(\sqrt[3]3-\sqrt[3]2\big)+\big(\sqrt[3]4-\sqrt[3]3\big)+\big(\sqrt[3]5-\sqrt[3]4\big)=\sqrt[3]5-\sqrt[3]2. \tag{\square}$$

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Wolfram Alpha provides the alternative form

$$\frac{25725}{42883}-\frac{2 \sqrt3}{37}-\frac{2 \sqrt5}{61}-\frac{\sqrt6}{19}$$

by rationalizing the denominator of each term. This is approximately $$.3040294826$$

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For your original question, it might be worth noting that

$$\frac{1}{\sqrt{k^2}+\sqrt{k(k+1)}+\sqrt{(k+1)^2}}=\frac{1}{2k+1+\sqrt{k(k+1)}}=\frac{2k+1-\sqrt{k(k+1)}}{4k^2+4k+1-k^2-k}=\frac{2k+1-\sqrt{k(k+1)}}{3k^2+3k+1}$$

Thus

$$\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{16}+\sqrt{20}+\sqrt{25}}=\frac{5-\sqrt{6}}{19}+\frac{7-2\sqrt{3}}{37}+\frac{9-2\sqrt{5}}{61}$$

As a side not, a much nicer expression is:

$$\frac{1}{\sqrt{4}+2\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+2\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{16}+2\sqrt{20}+\sqrt{25}}$$

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