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

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

$T_{r+1}= ^{12}C_r2^{12-r}3^r X^{36-3r-kr}$
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 $2$.