1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ We have $\frac{1}{2}(a^2 + b^2 + ab)(a-b) = \frac{1}{2}(a^3-b^3)$. The sum ends up telescoping. $\endgroup$ – Winther Nov 13 '16 at 18:24
  • $\begingroup$ Try factoring $4x^2-1$... $\endgroup$ – D Wiggles Nov 13 '16 at 18:25
6
$\begingroup$

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

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.

$\endgroup$
  • $\begingroup$ Nicely done. +1 $\endgroup$ – Mark Viola Nov 13 '16 at 18:42
6
$\begingroup$

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 :-)

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.