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Posted

I'm trying to prove that, hopefully this comes out right,

 

[math]\sum_{r=0}^{n}{(-1)}^{r}{_n}C{_r}{(n+1)}^n = n![/math]

 

Can anyone help?

 

The pattern is from

 

1 4 9

3 5

2

 

1 8 27 64

7 19 37

12 18

6

 

1 2

1

 

etc.

Posted

I'm not sure about the sum part, but the others come from taking differences. Here iswhy it works:

((x+1)^n-x^n) = nx^(n-1)+n*(n-1)*x^(n-2)/2 + ...

So the coefficient of the first term will be n.

You then subtract again, because you eventually want to get a constant number times the first term without the x value. This leaves you with n*(n-1).

Then you subtract again, leaving you with n*(n-1)*(n-2)

.

.

.

Then you subtract again, leaving you with n*(n-1)*(n-2)*....*3*2*1 = n!

All the other coefficients cancel out.

Hope this helps.

-Uncool-

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