Bertie

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 #8
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$$\\\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)(N-\imath)^{k}.$$

 

The object of the exercise is to show that this is equal to zero if N>k and equal to N! if N=k.

 

Begin with

$$(x-y)^{N}=\left(\begin{array}{c} N \\ 0 \end{array} \right)x^{N}-\left(\begin{array}{c} N \\ 1 \end{array} \right)x^{N-1}y+\left(\begin{array}{c} N \\ 2 \end{array} \right)x^{N-2}y^{2}- \dots (-1)^{N}\left(\begin{array}{c} N \\ N \end{array} \right)y^{N}=\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)x^{N-\imath}y^{\imath}$$

Differentiating this wrt x,

$$N(x-y)^{N-1}=\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)(N-\imath)x^{N-\imath-1}y^{\imath}..............(1)$$

and substituting x =  y = 1,

$$0=\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)(N-\imath)$$                 This proves the result for k = 1.

Multiplying (1) by x and differentiating again,

$$N(x-y)^{N-1}+N(N-1)x(x-y)^{N-2}=\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)(N-\imath)^{2}x^{N-\imath-1}y^{\imath}..............(2)$$

and substituting x = y = 1,

$$0=\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)(N-\imath)^{2}$$ which is the result for k=2.

 

Proceeding in this way, (multiply by x, differentiate wrt x and substitute x = y = 1), gets the results for k = 3 ,4,...,  until we arrive at k = N. For that, we find

$$\text{ (a whole load of terms, each containing the factor (x-y))} +N!x=\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)(N-\imath)^{N}x^{N-\imath-1}y^{\imath}$$

And now, putting x = y = 1,

$$N!=\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)(N-\imath)^{N}$$ which is the N=k result.

Apr 5, 2015