Simplifying $\left(\frac{1}{2}\right)^{-1/2}+\left(\frac{3}{2}\right)^{-3/2}+\left(\frac{5}{2}\right)^{-5/2}$

Question

The Art of Problem Solving: Volume 1 by Sandor Lehoczky and Richard Rusczyk - "Problems To Solve" for Chapter 1, No. 6:

We are asked to find the sum $$\left(\frac{1}{2}\right)^{-1/2}+\left(\frac{3}{2}\right)^{-3/2}+\left(\frac{5}{2}\right)^{-5/2}$$ with a rational common denominator.

After multiplying all the denominators together and simplifying, I currently have the answer $$16\sqrt{2}\left(2^{-\frac{8}{2}}+3^{-\frac{3}{2}}2^{-\frac{6}{2}}+5^{-\frac{5}{2}}2^{-\frac{4}{2}}\right)$$ which is equivalent to the original expression, but probably not the right answer.

How do I approach this problem? (I am probably overthinking it, a lot) What is the solution?

Thank you!

Answer 1

I'd think the $3^{-\frac{3}{2}}$ and $5^{-\frac{5}{2}}$ are still contributing to an irrational denominator. I'm sure there's a faster way to do this, since I get to a whole number denominator, not just a rational one.

Take your reciprocals first:

$$ \left(\frac{1}{2}\right)^{-\frac{1}{2}} + \left(\frac{3}{2}\right)^{-\frac{3}{2}} + \left(\frac{5}{2}\right)^{-\frac{5}{2}} $$

$$ = \left(2\right)^{\frac{1}{2}} + \left(\frac{2}{3}\right)^{\frac{3}{2}} + \left(\frac{2}{5}\right)^{\frac{5}{2}} $$

Factor out rational factors:

$$ = \left(2\right)^{\frac{1}{2}} + \left(\frac{2}{3}\right)^{\frac{2}{2}+\frac{1}{2}} + \left(\frac{2}{5}\right)^{\frac{4}{2}+\frac{1}{2}} $$

$$ = \left(2\right)^{\frac{1}{2}} + \frac{2}{3}*\left(\frac{2}{3}\right)^{\frac{1}{2}} + \frac{4}{25}*\left(\frac{2}{5}\right)^{\frac{1}{2}} $$

I generally prefer radicals, easier to see what's going on with fractions, so that's my next step:

$$ = \sqrt{2} + \frac{2\sqrt{2}}{3\sqrt{3}} + \frac{4\sqrt{2}}{25\sqrt{5}} $$

And find your rational common denominator:

$$ = \sqrt{2} + \frac{2\sqrt{6}}{9} + \frac{4\sqrt{10}}{125} $$

$$ = \frac{9*125*\sqrt{2}}{9*125} + \frac{125*2\sqrt{6}}{125*9} + \frac{9*4\sqrt{10}}{9*125} $$

$$ = \frac{1125\sqrt{2} + 250\sqrt{6} + 36\sqrt{10}}{1125} $$

Answer 2

Not the only way but

$\left(\frac{1}{2}\right)^{-1/2}+\left(\frac{3}{2}\right)^{-3/2}+\left(\frac{5}{2}\right)^{-5/2} =$

$2^{\frac 12} + (\frac23)^{\frac 32} + (\frac 25)^{\frac 52}=$

$\sqrt 2 + \frac {\sqrt{2^3}}{\sqrt{ 3^3}} +\frac {\sqrt {2^5}}{\sqrt {5^5}}=$

$\sqrt 2 + \frac {2\sqrt 2}{3\sqrt 3} + \frac {4\sqrt 2}{25\sqrt 5}=$

$\sqrt 2+ \frac {2\sqrt 2}{3\sqrt 3}\frac {\sqrt 3}{\sqrt 3} + \frac {4\sqrt 2}{25\sqrt 5}\frac {\sqrt 5}{\sqrt 5}=$

$\sqrt 2 + \frac {2\sqrt 2\sqrt 3}{9} + \frac {4\sqrt 2\sqrt 5}{125} =$

$\sqrt 2(\frac {9 + 250\sqrt 3 + 36\sqrt 5}{9\cdot 125})$

.... and.... I don't know what form is wanted for the final answer we can redistribute the $\sqrt 2$ again but I prefer my answers factored.

$= \frac {9\sqrt 2 + 250 \sqrt 6 + 36\sqrt {10}}{1125}$