So really, my question has more to do with a very specific step in an example of Newton's Binomial Theorem in my textbook's answer key, and doesn't have much to do with the theorem itself. In any case, The example goes like this:

$$(x^2-x^5)\frac{1}{(1-x)^3} \\= (x^2-x^5)\sum_{i=0}^\infty {-3 \choose i}(-x)^i \\= (x^2-x^5)\sum_{i=0}^\infty {-3 \choose i}(-1)^ix^i \\= (x^2-x^5)\sum_{i=0}^\infty {3+i-1 \choose i}x^i $$

So, I'm sure I'm just overlooking something relatively simple, but can anyone explain that last step to me? Specifically, how do we go from having ${-3 \choose i}(-1)^i$ to ${3+i-1\choose i}$?



(I'll use $k$ instead of $i$.)

Recall that $\displaystyle \binom{n}k$ means $\dfrac{n(n-1)\dots(n-k+1)}{k!}$. So

$$\binom{-3}k = \frac{(-3)(-4)\dots(-3-k+1)}{k!} = \frac{(-1)^k(3+k-1)\dots(4)(3)}{k!} = (-1)^k\binom{3+k-1}{k}$$ and therefore $$(-1)^k\binom{-3}{k} = \binom{3 + k - 1}{k}.$$

In general, by the same reasoning, $$\binom{-n}{k} = (-1)^k \binom{n + k - 1}{k}.$$

  • $\begingroup$ Ok, I got it now. Thank you! $\endgroup$ – kirky Jul 9 '13 at 18:40

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