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- #1

x(x-1) = 1x^2 - 1x

x(x-1)(x-2) = 1x^3 - 3x^2 + 2x

x(x-1)(x-2)(x-3) = 1x^4 - 6x^3 + 11x^2 - 6x

x(x-1)(x-2)(x-3)(x-4) = 1x^5 - 10x^4 + 35x^3 - 50x^2 + 24x

...

So, I am looking for a short method how to find these coefficients ahead of each raisings of x:

1

1,-1

1,-3,2

1,-6,11,-6

1,-10,35,-50,24

...

Is the only method just to perform repetative multiplications, or is there a time and energy saving method available?

I do can find some predictable features:

-The leftmost coefficient is always "1"

-The next coefficient comes from triangular numbers (1,3,6,10,15,21,28...that is an integer when we sum up successive positive integers; for example: 1+2+3+4+5=15) with negative sign.

- The last term is a factorial with alternate + and - sign (for example: 3 factorial = 3! = 1*2*3 = 6, 4! = 1*2*3*4 =24 etc.)

- Positive and negative coefficients in a row, when summoned up, equals to 0 (for example in the last row example: 1-10+35-50+24 = 0)

Please, help me.