Question

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

(b) For the values of $n$ that you found in part (a), what is that constant term? (You can leave your answer in the form of a combination.)

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))=((6m)!)/((4m)!(6m-4m)!)=((6m)!)/((4m)!(2m)!)`