Question

Number of terms having rational coefficients in the expansion of \((3+\sqrt{3}x-5^{1/3}x^2)^{18}\) is

Answer 1

Hint :

It would be same as the number of whole number solutions to
\[a + 2b + 3c = 18\]

Answer 2

How did you get this equation ?

Answer 3

Maybe let's look at a more simple expression \[(\sqrt{2} + \sqrt[3]{5})^{18}\]

Answer 4

\((\sqrt{2} + \sqrt[3]{5})^{18} =(\sqrt{2} + \sqrt[3]{5})(\sqrt{2} + \sqrt[3]{5}) \cdots (\sqrt{2} + \sqrt[3]{5}) \)

You're multiplying the binomial \((\sqrt{2} + \sqrt[3]{5}) \) with itself 18 times. For each term in the final result, you will need to carry out exacty 18 multiplications. Also notice that in each factor, you have a choice : You can choose either \(\sqrt{2}\) or you can choose \(\sqrt[3]{5}\).

Answer 5

You will end up with an "integer" term only if :

1) you choose "even" number of \(\sqrt{2}\) s ; because multiplying \(\sqrt{2}\) with itself even number of times gives you an integer. And,
2) the number of \(\sqrt[3]{5}\) that you choose must be a multiple of \(3\); because multiplying \(\sqrt[3]{5}\) with itself 3 times gives you an integer.

Answer 6

Then the problem of finding the number of ways you get an integer can be translated to finding the number of nonnegative integer solutions to
\[2a+3b=18\]

Answer 7

\(2a\) : choosing even number of \(\sqrt{2}\)s
\(3b\) : number of \(\sqrt[3]{5}\) s that you choose must be a multiple of 3

Answer 8

oh now i gt it
thanx a lot sir :)

Answer 9

Np :) Do you know how to find the number of whole number solutions to
\[a + 2b + 3c = 18\]

?

It may not be that straightforward..

Answer 10

I may try a hit & trial method here. the value of c can't e larger than 6 & the value of b can't be larger than 9.