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I want to rewrite the series $1 + a + a(a-1) + a(a-1) (a-2) +\cdots+a(a-1)\cdots(a-(n-1))$ as $(a^n-1)Y$ or $(a^{n-1}-1)Y$

Short-form: $\{1+\sum_{i=1}^{n} \prod_{j=0}^{i-1}(a-j)\}$ as $(a^n-1)Y$ or $(a^{n-1}-1)Y$

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    Y is some random number2012-06-04

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I don't think you can: $(a^n-1)Y$ is zero at $a=1$ but this is not true for $1 + a + a(a-1) + a(a-1) (a-2) + a(a-1)(a-2)(a-3)\cdots+a(a-1)\cdots (a-(n-1))$.

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    @viji, if you want an algebraic formula then it'll probably be continuous and so applicable to $a=1$, so no.2012-06-04
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You can get a kind of recursive stuff: $K(a):=1+a+a(a-1)+....+a(a-1)\cdot ...\cdot \left(a-(n-1)\right)=$$=1+a\left\{1+(a-1)\left[1+(a-2)\left(...(1+a-(n-1)\right)\right]\right\}$This formula allows you to see, perhaps a little clearer than before, that$K(1)=2\,\,,\,\,K(2)=5\,\,,\,K(3)=16\,\,,\,K(4)=66\,,...$According to OEIS this is sequence $\,A000522\,$, which, pretty surprisingly for me, I must say, equals $n!\sum_{k=0}^n\frac{1}{k!}$with $\,K(0):=1\,$

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    @viji: $\lim_{n \to \infty} \sum_{k=0}^n \frac{1}{k!} = e$ might help.2012-06-04