If the constant term in the expansion of (3x^3-2x^2+5x)^10 is 2^k l, where l is an odd int…

Question

If the constant term in the expansion of $\left(3x^3-2x^2+\dfrac5x\right)^{10}$ is $2^k\cdot l$, where $l$ is an odd integer, then the value of $k$ is equal to:

Accepted Answer

$\left(3x^3-2x^2+\dfrac5x\right)^{10}=\dfrac{(3x^8-2x^7+5)^{10}}{x^{50}}$; need coefficient of $x^{50}$ in $(3x^8-2x^7+5)^{10}$, i.e. $8n_1+7n_2=50$ with $n_1+n_2+n_3=10$.

The valid solution is $n_1=1,\,n_2=6,\,n_3=3$:

$$\frac{10!}{1!\,6!\,3!}(3)^1(-2)^6(5)^3=840\times3\times64\times125=7\times5^4\times3^2\times2^9$$

So $k=9$.

Solutions · 1