The number of positive integers k such that the constant term in the binomial expansion of…

Question

The number of positive integers $k$ such that the constant term in the binomial expansion of $\left(2x^3+\dfrac{3}{x^k}ight)^{12}$, $x
e0$ is $2^8\cdot l$, where $l$ is an odd integer, is\_\_\_

Answer: 2

Accepted Answer

$T_{r+1}=\binom{12}{r}2^{12-r}3^r\,x^{3(12-r)-kr}$; for the constant term, $36-3r-kr=0\Rightarrow k=\dfrac{36-3r}{r}$.

Checking the power of 2 in $\binom{12}{r}2^{12-r}$ for integer $k$: at $r=6,\,k=3$: $\binom{12}{6}2^6=924	imes64$, power of 2 in $924=4	imes231$ is $2$, total $=2+6=8$ . At $r=3,\,k=9$: power too high. At $r=9,\,k=1$: power too low.

So the valid values are $k=3,6$: 2 values.

Solutions · 1