- #1

- 1,235

- 1

How does one deduce the following:

We are given

[tex] \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!} [/tex]. How do we get from here to:

[tex] \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} [/tex]?

Thanks

We are given

[tex] \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!} [/tex]. How do we get from here to:

[tex] \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} [/tex]?

Thanks

Last edited: