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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
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Simplify} \\ $$$$\frac{\mathrm{1}^{\mathrm{2}} \centerdot\mathrm{2}!+\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{3}!+\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{4}!+\centerdot\centerdot\centerdot+{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)!−\mathrm{2}}{\left({n}+\mathrm{1}\right)!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{to} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{n}^{\mathrm{2}} +\mathrm{n}−\mathrm{2} \\ $$
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
$${Statement}: \\ $$$${n}^{\mathrm{2}} +{n}−\mathrm{2}=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}\left(\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}^{\mathrm{2}} \left({i}+\mathrm{1}\right)!\:−\:\mathrm{2}\right) \\ $$$${We}'{ll}\:{use}\:{proof}\:{by}\:{induction}: \\ $$$${for}\:{n}=\mathrm{0}: \\ $$$$\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}\left(\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}^{\mathrm{2}} \left({i}+\mathrm{1}\right)!\:−\:\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{1}!}\left(\mathrm{0}−\mathrm{2}\right)=−\mathrm{2} \\ $$$$\mathrm{0}^{\mathrm{2}} +\mathrm{0}−\mathrm{2}=−\mathrm{2} \\ $$$${for}\:{n}=\mathrm{1}: \\ $$$$\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}\left(\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}^{\mathrm{2}} \left({i}+\mathrm{1}\right)!\:−\:\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}!}\left(\mathrm{1}^{\mathrm{2}} ×\mathrm{2}!−\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}=\mathrm{0} \\ $$$${for}\:{n}={k}: \\ $$$$\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)!}\left(\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}{i}^{\mathrm{2}} \left({i}+\mathrm{1}\right)!\:−\:\mathrm{2}\right)={k}^{\mathrm{2}} +{k}−\mathrm{2} \\ $$$${for}\:{n}={k}+\mathrm{1}: \\ $$$$\frac{\mathrm{1}}{\left({k}+\mathrm{2}\right)!}\left(\underset{{i}=\mathrm{1}} {\overset{{k}+\mathrm{1}} {\sum}}{i}^{\mathrm{2}} \left({i}+\mathrm{1}\right)!\:−\:\mathrm{2}\right)= \\ $$$$=\frac{\mathrm{1}}{{k}+\mathrm{2}}\left(\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)!}\left(\underset{{i}=\mathrm{1}} {\overset{{k}+\mathrm{1}} {\sum}}{i}^{\mathrm{2}} \left({i}+\mathrm{1}\right)!\:−\:\mathrm{2}\right)\right)= \\ $$$$=\frac{\mathrm{1}}{{k}+\mathrm{2}}\left(\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)!}\left(\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}{i}^{\mathrm{2}} \left({i}+\mathrm{1}\right)!\:−\:\mathrm{2}\right)+\frac{\left({k}+\mathrm{1}\right)^{\mathrm{2}} \left({k}+\mathrm{2}\right)!}{\left({k}+\mathrm{1}\right)!}\right)= \\ $$$$=\frac{\mathrm{1}}{{k}+\mathrm{2}}\left({k}^{\mathrm{2}} +{k}−\mathrm{2}+\left({k}+\mathrm{2}\right)\left({k}+\mathrm{1}\right)^{\mathrm{2}} \right)= \\ $$$$=\frac{{k}^{\mathrm{2}} +{k}−\mathrm{2}}{{k}+\mathrm{2}}+\left({k}+\mathrm{1}\right)^{\mathrm{2}} = \\ $$$$=\frac{\left({k}+\mathrm{2}\right)\left({k}−\mathrm{1}\right)}{{k}+\mathrm{2}}+\left({k}+\mathrm{1}\right)^{\mathrm{2}} = \\ $$$$=\left({k}+\mathrm{1}\right)^{\mathrm{2}} +{k}−\mathrm{1}= \\ $$$$=\left({k}+\mathrm{1}\right)^{\mathrm{2}} +\left({k}+\mathrm{1}\right)−\mathrm{2} \\ $$$${QED} \\ $$
Commented by Rasheed.Sindhi last updated on 24/Dec/21
Nice proof sir! But the question is about simplification, not about proof.
$$\mathcal{N}{ice}\:{proof}\:{sir}!\: \\ $$$$\mathcal{B}{ut}\:{the}\:{question}\:{is}\:{about}\:{simplification}, \\ $$$${not}\:{about}\:{proof}. \\ $$

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