The coefficient of x^1012 in the expansion of (1+x^n+x^253)^10, (where n22 is any positive…

Question

The coefficient of $x^{1012}$ in the expansion of $(1+x^n+x^{253})^{10}$, (where $n\le22$ is any positive integer), is:

Accepted Answer

Write $x^{1012}=1^a(x^n)^b(x^{253})^c$ with $a+b+c=10$, $bn+253c=1012$. For $n\le22$ and $c<4$: $bn=253(4-c)$ requires $b>10$, impossible. So $c=4,b=0,a=6$:

$$\text{Coefficient}=\frac{10!}{6!0!4!}={}^{10}C_4$$

Solutions · 1