# Binomial Series Question

How does one deduce the following:

We are given

$$\binom{-3}{n} = \frac{(-3)(-4)(-5)\cdot\cdot\cdot\cdot\cdot(-3-n+1)}{n!} = \frac{(-3)(-4)(-5)\cdot\cdot\cdot\cdot[-(n+2)]}{n!}$$. How do we get from here to:

$$\frac{(-1)^{n}\cdot2\cdot3\cdot4\cdot5\cdot\cdot\cdot\cdot(n+1)(n+2)}{2\cdot n!} = \frac{(-1)^{n}(n+1)(n+2)}{2}$$?

Thanks

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