# binomial expansion of $(1+x)^n \left(\dfrac{1}{x} -1\right)^m$

Let $n$ and $m$ be positive integers with $n \gt m$. Can you show that the constant term of $${(1+x)^n}\left(\frac{1}{x} - 1\right)^m$$ is not equal to zero?

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I tried $\LaTeX$-ifying your equation; please check that I did this correctly (I did not know if you meant $(\frac{1}{x}-1)^m$ or $(\frac{1}{x-1})^m$, but I'm guessing the former). – Arturo Magidin Mar 30 '12 at 20:40
Thanks for the editing! Your writing is correct. – hkju Mar 30 '12 at 20:52
You can expand : \begin{align*} (1+x)^n \left( \frac 1x - 1 \right)^m & = \left( \sum_{i=0}^n \binom ni x^i\right) \left( \sum_{j=0}^m \binom mj x^{-j} (-1)^{m-j}\right) \\ & = \sum_{i=0}^n \sum_{j=0}^m \binom ni \binom mj x^{i-j} (-1)^{m-j} \\ \end{align*} so that the constant term is $$\sum_{j=0}^{m} \binom nj \binom mj (-1)^{m-j}$$ (let $i=j$ to find the possible coefficients that belong to this sum). Does that help? – Patrick Da Silva Mar 30 '12 at 21:01
Yes, it is. Also the case when n=m. If n is odd, it is zero. But for even n it is ((-1)^(n/2))(n choose n/2) – hkju Mar 30 '12 at 21:18
For the cases $n=m+k$, where $k=1,2,3,4$, it is done. – hkju Mar 31 '12 at 21:45