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

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  Can you evaluate S?

$${S}=\underset{{x}_{\mathrm{2}} =\mathrm{1}} {\overset{{x}_{\mathrm{1}} } {\sum}}\underset{{x}_{\mathrm{3}} =\mathrm{1}} {\overset{{x}_{\mathrm{2}} } {\sum}}\centerdot\centerdot\centerdot\underset{{x}_{{n}} =\mathrm{1}} {\overset{{x}_{{n}−\mathrm{1}} } {\sum}}\underset{{t}=\mathrm{1}} {\overset{{x}_{{n}} } {\sum}}{t} \\ $$$$\mathrm{Can}\:\mathrm{you}\:\mathrm{evaluate}\:{S}? \\ $$

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)!))

$$\:\:\:\:\underset{{t}=\mathrm{1}} {\overset{{x}_{{n}} } {\sum}}{t} \\ $$$$=\frac{\left({x}_{{n}} +\mathrm{1}\right){x}_{{n}} }{\mathrm{2}!} \\ $$$$\:\:\:\:\underset{{x}_{{n}} =\mathrm{1}} {\overset{{x}_{{n}−\mathrm{1}} } {\sum}}\underset{{t}=\mathrm{1}} {\overset{{x}_{{n}} } {\sum}}{t} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}_{{n}} =\mathrm{1}} {\overset{{x}_{{n}−\mathrm{1}} } {\sum}}\left({x}_{{n}} +\mathrm{1}\right){x}_{{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}_{{n}} =\mathrm{1}} {\overset{{x}_{{n}−\mathrm{1}} } {\sum}}\frac{\mathrm{1}}{\mathrm{3}}\left\{\left({x}_{{n}} +\mathrm{2}\right)\left({x}_{{n}} +\mathrm{1}\right){x}_{{n}} −\left({x}_{{n}} +\mathrm{1}\right){x}_{{n}} \left({x}_{{n}} −\mathrm{1}\right)\right\} \\ $$$$=\frac{\left({x}_{{n}−\mathrm{1}} +\mathrm{2}\right)\left({x}_{{n}−\mathrm{1}} +\mathrm{1}\right){x}_{{n}−\mathrm{1}} }{\mathrm{3}!} \\ $$$$.... \\ $$$${S}=\underset{{x}_{\mathrm{2}} =\mathrm{1}} {\overset{{x}_{\mathrm{1}} } {\sum}}\underset{{x}_{\mathrm{3}} =\mathrm{1}} {\overset{{x}_{\mathrm{2}} } {\sum}}\centerdot\centerdot\centerdot\underset{{x}_{{n}} =\mathrm{1}} {\overset{{x}_{{n}−\mathrm{1}} } {\sum}}\underset{{t}=\mathrm{1}} {\overset{{x}_{{n}} } {\sum}}{t} \\ $$$$\:\:\:=\frac{\left({x}_{\mathrm{1}} +{n}\right)\left({x}_{\mathrm{1}} +{n}−\mathrm{1}\right)...\left({x}_{\mathrm{1}} +\mathrm{1}\right){x}_{\mathrm{1}} }{\left({n}+\mathrm{1}\right)!} \\ $$

Commented by mrW1 last updated on 16/May/17

Nice!

$$\mathscr{N}{ice}! \\ $$

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