How many rational numbers are in the expression

[Loki123]

Is this correct?

 

[lex]

Substitute in k=0, 1, 2, 3 and see which gives a rational element.

 

[Loki123]

Substitute in k=0, 1, 2, 3 and see which gives a rational element.

I did.

 

[lex]

And what did you find?

 

[Loki123]

And what did you find?

only 0 works

 

[lex]

You found that $\binom{50}{0} 3^{\tfrac{0}{2}} 5^{\tfrac{50-0}{4}}$ is rational?
This may be a problem with the large numbers in your calculator - if you are using an ordinary calculator.

You want k/2 to be an integer and (50-k)/4 to be an integer
You want k to be a multiple of 2 and (2-k) to be a multiple of 4 (Since 48 is a multiple of 4, we can subtract it).
k must be a multiple of 2 and k-2 must be a multiple of 4.
What values of k does that give you?

 

[pka]

For every $k\in\mathbb{N}$ the number $\dbinom{50}{k}$ is rational so no need to worry there.
So we look at $\left(\sqrt{3}\right)^{50-k}\left(\sqrt[4]{5}\right)^{k}$ those products are rational for $k=0,~4,~8,\cdots~44,~48.$
$$$$$$

 

[Loki123]

For every $k\in\mathbb{N}$ the number $\dbinom{50}{k}$ is rational so no need to worry there.
So we look at $\left(\sqrt{3}\right)^{50-k}\left(\sqrt[4]{5}\right)^{k}$ those products are rational for $k=0,~4,~8,\cdots~44,~48.$
$$$$$$

Oh, so if I just switch places it's a lot easier to calculate. Thanks!!