Question

The number of positive integers k such that the constant term in the binomial expansion of \(( 2x^3 + \frac {3}{x^k} )^{12}, x ≠ 0\) is \(2^8\) . l, where l is an odd integer, is______.

Answer 1

Correct Answer: 2

\(T_{r+1} = ^{12}C_r (2x^3)^{12-r}( \frac {3}{x^k })^r\)

\(T_{r+1}= ^{12}C_r2^{12-r}3^r X^{36-3r-kr}\)
For constant term \(36 – 3r – kr = 0\)
\(r = \frac {36}{3+k}\)
So, k can be \(1, 3, 6, 9, 15, 33\)
In order to get \(2^8\), check by putting values of k and corresponding in general term. By checking, it is possible only where k = \(3\) or \(6\)

So, the answer is \(2\).