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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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    What do you mean by .Y?2012-06-04
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    Y is some random number2012-06-04

2 Answers 2

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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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    Without the 1st 1, can you do ? a+a(a−1)+a(a−1)(a−2)+a(a−1)(a−2)(a−3)+...+a(a−1)⋯(a−(n−1))2012-06-04
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    @viji, no for the same reason.2012-06-04
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    Oops !! Sorry. Can you do the same for a > 1 ?2012-06-04
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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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    Is there anything that will get rid of the summation in the above formula ?2012-06-04
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    Why the surprise? $\frac{n!}{k!} = n \cdot (n-1) \cdot \dots \cdot (n-k+1)$ for natural numbers $k < n$.2012-06-04
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    The surprise came because I didn't expect it while developing the formula.2012-06-04
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    Too late to edit, but it should be $\frac{n!}{(n-k)!}$ in the above comment. Also more general: $\frac{\Gamma(z+1)}{\Gamma(z+1-k)} = z \cdot (z-1) \cdot \dots \cdot (z-k+1)$ for any natural number $k$, for which the above equation is defined.2012-06-04
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    @viji: $\lim_{n \to \infty} \sum_{k=0}^n \frac{1}{k!} = e$ might help.2012-06-04