How many rational numbers are in the expression

Substitute in k=0, 1, 2, 3 and see which gives a rational element.
 
You found that [imath]\binom{50}{0} 3^{\tfrac{0}{2}} 5^{\tfrac{50-0}{4}}[/imath] 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?
 
For every [imath]k\in\mathbb{N}[/imath] the number [imath]\dbinom{50}{k}[/imath] is rational so no need to worry there.
So we look at [imath]\left(\sqrt{3}\right)^{50-k}\left(\sqrt[4]{5}\right)^{k}[/imath] those products are rational for [imath]k=0,~4,~8,\cdots~44,~48.[/imath]
[imath][/imath][imath][/imath][imath][/imath]
 
pka said:
For every [imath]k\in\mathbb{N}[/imath] the number [imath]\dbinom{50}{k}[/imath] is rational so no need to worry there.
So we look at [imath]\left(\sqrt{3}\right)^{50-k}\left(\sqrt[4]{5}\right)^{k}[/imath] those products are rational for [imath]k=0,~4,~8,\cdots~44,~48.[/imath]
[imath][/imath][imath][/imath][imath][/imath]
Click to expand...
Oh, so if I just switch places it's a lot easier to calculate. Thanks!!
 
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