(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 $n$

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.)

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Evelyn 2 weeks 2021-11-22T03:17:19+00:00 1 Answer 0

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    2021-11-22T03:19:08+00:00

    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)!}

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