S=Σ_(x_2 =1) ^x_1 Σ_(x_3 =1) ^x_2 ∙∙∙Σ_(x_n =1) ^x_(n−1) Σ_(t=1) ^x


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Question Number 13097 by FilupS last updated on 14/May/17

$S = \underset{x_{2} = 1}{\sum^{x_{1}}} \underset{x_{3} = 1}{\sum^{x_{2}}} \cdot \cdot \cdot \underset{x_{n} = 1}{\sum^{x_{n - 1}}} \underset{t = 1}{\sum^{x_{n}}} t$ $Can you evaluate S ?$
	Answered by nume1114 last updated on 15/May/17
	$Σ_(t=1) ^x_n t =(((x_n +1)x_n )/(2!)) Σ_(x_n =1) ^x_(n−1) Σ_(t=1) ^x_n t =(1/2)Σ_(x_n =1) ^x_(n−1) (x_n +1)x_n =(1/2)Σ_(x_n =1) ^x_(n−1) (1/3){(x_n +2)(x_n +1)x_n −(x_n +1)x_n (x_n −1)} =(((x_(n−1) +2)(x_(n−1) +1)x_(n−1) )/(3!)) .... S=Σ_(x_2 =1) ^x_1 Σ_(x_3 =1) ^x_2 ∙∙∙Σ_(x_n =1) ^x_(n−1) Σ_(t=1) ^x_n t =(((x_1 +n)(x_1 +n−1)...(x_1 +1)x_1 )/((n+1)!))$
	$\underset{t = 1}{\sum^{x_{n}}} t$ $= \frac{(x_{n} + 1) x_{n}}{2!}$ $\underset{x_{n} = 1}{\sum^{x_{n - 1}}} \underset{t = 1}{\sum^{x_{n}}} t$ $= \frac{1}{2} \underset{x_{n} = 1}{\sum^{x_{n - 1}}} (x_{n} + 1) x_{n}$ $= \frac{1}{2} \underset{x_{n} = 1}{\sum^{x_{n - 1}}} \frac{1}{3} {(x_{n} + 2) (x_{n} + 1) x_{n} - (x_{n} + 1) x_{n} (x_{n} - 1)}$ $= \frac{(x_{n - 1} + 2) (x_{n - 1} + 1) x_{n - 1}}{3!}$ $. . . .$ $S = \underset{x_{2} = 1}{\sum^{x_{1}}} \underset{x_{3} = 1}{\sum^{x_{2}}} \cdot \cdot \cdot \underset{x_{n} = 1}{\sum^{x_{n - 1}}} \underset{t = 1}{\sum^{x_{n}}} t$ $= \frac{(x_{1} + n) (x_{1} + n - 1) . . . (x_{1} + 1) x_{1}}{(n + 1)!}$
	Commented by mrW1 last updated on 16/May/17

	$N i c e!$
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