Question

(a) For what positive integers $n$ does $\left(x^2+\frac{1}{x}\right)^n$ have a nonzero constant term?

Answer 1

a. By the binomial theorem,

$\displaystyle\left(x^2+\frac1x\right)^n=\sum_{k=0}^n\binom nk(x^2)^{n-k}\left(\frac1x\right)^k=\sum_{k=0}^n\binom nkx^{2n-3k}$

which produces a constant term when $2n-3k=0$, or $2n=3k$. $2n$ is even for any choice of $n$, so $n$ must be any positive integer for which $2n$ is an even multiple of 3 i.e. a multiple of 6. Then $n$ can be any of the integers in the sequence {6, 12, 18, ...}.

b. Let $n=6m$ for some positive integer $m$. Then

$\displaystyle\left(x^2+\frac1x\right)^{6m}=\sum_{k=0}^{6m}\binom {6m}k(x^2)^{6m-k}\left(\frac1x\right)^k=\sum_{k=0}^{6m}\binom{6m}kx^{12m-3k}$

and the constant term occurs when $12m-3k=0$, or $4m-k=0$, or $k=4m$. At this value of $k$, we get the term

$\dbinom{6m}{4m}x^{12m-3(4m)}=\dfrac{(6m)!}{(4m)!(6m-4m)!}=\dfrac{(6m)!}{(4m)!(2m)!}$