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Simplify-1-2-2-2-2-3-3-2-4-n-2-n-1-2-n-1-to-n-2-n-2-

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Question Number 161860 by Rasheed.Sindhi last updated on 23/Dec/21
Simplify ((1^2 ∙2!+2^2 ∙3!+3^2 ∙4!+∙∙∙+n^2 (n+1)!−2)/((n+1)!)) to n^2 +n−2
Simplify12⋅2!+22⋅3!+32⋅4!+⋅⋅⋅+n2(n+1)!−2(n+1)!ton2+n−2
Answered by aleks041103 last updated on 23/Dec/21
Statement: n^2 +n−2=(1/((n+1)!))(Σ_(i=1) ^n i^2 (i+1)! − 2) We′ll use proof by induction: for n=0: (1/((n+1)!))(Σ_(i=1) ^n i^2 (i+1)! − 2)=(1/(1!))(0−2)=−2 0^2 +0−2=−2 for n=1: (1/((n+1)!))(Σ_(i=1) ^n i^2 (i+1)! − 2)=(1/(2!))(1^2 ×2!−2)=0 1^2 +1−2=0 for n=k: (1/((k+1)!))(Σ_(i=1) ^k i^2 (i+1)! − 2)=k^2 +k−2 for n=k+1: (1/((k+2)!))(Σ_(i=1) ^(k+1) i^2 (i+1)! − 2)= =(1/(k+2))((1/((k+1)!))(Σ_(i=1) ^(k+1) i^2 (i+1)! − 2))= =(1/(k+2))((1/((k+1)!))(Σ_(i=1) ^k i^2 (i+1)! − 2)+(((k+1)^2 (k+2)!)/((k+1)!)))= =(1/(k+2))(k^2 +k−2+(k+2)(k+1)^2 )= =((k^2 +k−2)/(k+2))+(k+1)^2 = =(((k+2)(k−1))/(k+2))+(k+1)^2 = =(k+1)^2 +k−1= =(k+1)^2 +(k+1)−2 QED
Statement:n2+n−2=1(n+1)!(∑ni=1i2(i+1)!−2)We′lluseproofbyinduction:forn=0:1(n+1)!(∑ni=1i2(i+1)!−2)=11!(0−2)=−202+0−2=−2forn=1:1(n+1)!(∑ni=1i2(i+1)!−2)=12!(12×2!−2)=012+1−2=0forn=k:1(k+1)!(∑ki=1i2(i+1)!−2)=k2+k−2forn=k+1:1(k+2)!(∑k+1i=1i2(i+1)!−2)==1k+2(1(k+1)!(∑k+1i=1i2(i+1)!−2))==1k+2(1(k+1)!(∑ki=1i2(i+1)!−2)+(k+1)2(k+2)!(k+1)!)==1k+2(k2+k−2+(k+2)(k+1)2)==k2+k−2k+2+(k+1)2==(k+2)(k−1)k+2+(k+1)2==(k+1)2+k−1==(k+1)2+(k+1)−2QED
Commented by Rasheed.Sindhi last updated on 24/Dec/21
Nice proof sir! But the question is about simplification, not about proof.
Niceproofsir!Butthequestionisaboutsimplification,notaboutproof.

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