# Find $f(1)+\cdots+f(60)$

Let $f(x) = \dfrac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}}$. Find $f(1)+\cdots+f(60)$.

I considered rationalizing the denominator, but that seems to make the fraction more complicated. We get $$\dfrac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} = \dfrac{1}{2}\left(4x+\sqrt{4x^2-1}\right)\left(\sqrt{2x+1}-\sqrt{2x-1}\right).$$ is there an easier way?

• We have $\frac{1}{2}(a^2 + b^2 + ab)(a-b) = \frac{1}{2}(a^3-b^3)$. The sum ends up telescoping. – Winther Nov 13 '16 at 18:24
• Try factoring $4x^2-1$... – D Wiggles Nov 13 '16 at 18:25

$$f(x) = \dfrac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}}$$ $$= \dfrac{1}{2}\left(4x+\sqrt{4x^2-1}\right)\left(\sqrt{2x+1}-\sqrt{2x-1}\right)$$

Now say $a=\sqrt{2x+1}$ and $b=\sqrt{2x-1}$

Then $f(x)=\dfrac{1}{2}\left(4x+\sqrt{4x^2-1}\right)\left(\sqrt{2x+1}-\sqrt{2x-1}\right)=\frac{1}{2}\left(a^2+b^2+ab\right)\left(a-b\right)=\frac{1}{2}(a^3-b^3)$

Therefore, $$\boxed{f(x)=\frac{1}{2}\left[(2x+1)^\frac{3}{2}-(2x-1)^\frac{3}{2}\right]}$$

So, $f(1)+\cdots+f(60)=\frac{1}{2}\left[(2\cdot 60+1)^\frac{3}{2}-(2\cdot 1-1)^\frac{3}{2}\right]=665$

Hope this helps you.

• Nicely done. +1 – Mark Viola Nov 13 '16 at 18:42

Using the fact that $(\sqrt{2x+1}+\sqrt{2x-1})^2=4x+2\sqrt{4x^2-1}$, you can produce : \begin{align} \frac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} &= \frac{(\sqrt{2x+1}+\sqrt{2x-1})^2-\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} \\ &= \sqrt{2x+1}+\sqrt{2x-1} - \frac{\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} \\ &= \sqrt{2x+1}+\sqrt{2x-1} - \frac{\sqrt{4x^2-1}(\sqrt{2x+1}-\sqrt{2x-1})}{2} \\ &= \sqrt{2x+1}+\sqrt{2x-1} - \frac{(2x+1)}{2}\sqrt{2x-1} + \frac{(2x-1)}{2}\sqrt{2x+1} \\ &= \frac{(2x+1)^{3/2}}{2} - \frac{(2x-1)^{3/2}}{2} \end{align} So your sum is telescopic, and you find : $$\sum_{k=1}^{60} f(k) = \frac{(2\times 60+1)^{3/2}}{2} - \frac{(2\times 1-1)^{3/2}}{2} = 665$$ Note : I'm sure there's a much cleaner way to obtain this, but I don't see how :-)

• Nicely done. +1 – Mark Viola Nov 13 '16 at 18:42

Maybe it's easier if you think that

\begin{align} A = \sqrt{2x+1}&\qquad B = \sqrt{2x-1}\\ AB &= \sqrt{4x^2-1}\\ A^2 + B^2 &= 4x \end{align}

When you split them you get $$A+B -\frac{AB}{A+B}$$

Then conjugate the right term with $A-B$

$$A+B - \frac{AB(A-B)}2\\ A+B-\frac{A^2B}2+\frac A{2B^2}$$

Now it looks more nicely seperated.