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Question Number 72047 by aliesam last updated on 23/Oct/19

Commented by mind is power last updated on 23/Oct/19

$${\mathrm{x}}_1=1,{\mathrm{x}}_2=4,{\mathrm{x}}_3=9,{\mathrm{x}}_4=16$$$${\mathrm{\forall }}_{\mathrm{n}}\in {\mathbb{N}}^{\ast },{\mathrm{X}}_{\mathrm{n}}={\mathrm{n}}^2$$$$\mathrm{for}\mathrm{n}=1\mathrm{we}\mathrm{have}{\mathrm{X}}_1=1=1^2$$$$\mathrm{lets}\mathrm{assume}\mathrm{\forall }\mathrm{n}\in {\mathbb{N}}^{\ast }{\mathrm{x}}_{\mathrm{n}}={\mathrm{n}}^2$$$${\mathrm{X}}_{\mathrm{n}+1}=\frac{\mathrm{n}+2}{\mathrm{n}}.{\mathrm{n}}^2+1={\mathrm{n}}^2+2\mathrm{n}+1={(\mathrm{n}+1)}^2\Rightarrow {\mathrm{\forall }}_{\mathrm{n}}\in \mathbb{N}{}^{\ast }{\mathrm{x}}_{\mathrm{n}}={\mathrm{n}}^2$$$$2\mathrm{nd}$$$${\mathrm{x}}_{\mathrm{n}}={\mathrm{U}}_{\mathrm{n}}+\mathrm{p}\left(\mathrm{n}\right)$$$${\mathrm{U}}_{\mathrm{n}+1}+\mathrm{p}(\mathrm{n}+1)=\frac{\mathrm{n}+2}{\mathrm{n}}.({\mathrm{u}}_{\mathrm{n}}+\mathrm{p}\left(\mathrm{n}\right))+1$$$$\mathrm{p}(\mathrm{n}+1)=\left(\frac{\mathrm{n}+2}{\mathrm{n}}\right)\mathrm{p}\left(\mathrm{n}\right)+1$$$$\Rightarrow \mathrm{np}(\mathrm{n}+1)=(\mathrm{n}+2)\mathrm{p}\left(\mathrm{n}\right)+\mathrm{n}$$$$\mathrm{p}\left(\mathrm{n}\right)=\mathrm{an}+\mathrm{b}$$$$\Rightarrow \mathrm{n}(\mathrm{a}(\mathrm{n}+1)+\mathrm{b})=(\mathrm{n}+2)(\mathrm{an}+\mathrm{b})+\mathrm{n}$$$$\Rightarrow {\mathrm{an}}^2+\mathrm{n}(\mathrm{a}+\mathrm{b})={\mathrm{an}}^2+(\mathrm{b}+2\mathrm{a}+1)\mathrm{n}+2\mathrm{b}$$$$\mathrm{b}=0,\mathrm{a}=-1$$$$\mathrm{p}\left(\mathrm{n}\right)=-\mathrm{n}$$$$\Rightarrow {\mathrm{U}}_{\mathrm{n}+1}=\frac{\mathrm{n}+2}{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}}$$$$\Rightarrow \frac{{\mathrm{u}}_{\mathrm{n}+1}}{{\mathrm{u}}_{\mathrm{n}}}=\frac{\mathrm{n}+2}{\mathrm{n}}\Rightarrow \underset{\mathrm{k}=1}{\prod ^{\mathrm{n}-1}}\frac{{\mathrm{u}}_{\mathrm{n}+1}}{{\mathrm{u}}_{\mathrm{n}}}=\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}-1}}\frac{\mathrm{n}+2}{\mathrm{n}}$$$$\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}-1}}\frac{{\mathrm{u}}_{\mathrm{k}+1}}{{\mathrm{u}}_{\mathrm{k}}}=\frac{{\mathrm{u}}_2}{{\mathrm{u}}_1}.\frac{{\mathrm{u}}_3}{{\mathrm{u}}_2}....\frac{{\mathrm{u}}_{\mathrm{n}}}{{\mathrm{u}}_{\mathrm{n}-1}}=\frac{{\mathrm{u}}_{\mathrm{n}}}{{\mathrm{u}}_1}$$$$\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}-1}}\frac{\mathrm{n}+2}{\mathrm{n}}=\frac{3}{1}.\frac{4}{2}.\frac{5}{3}.\frac{6}{4}.....\frac{\mathrm{n}+1}{\mathrm{n}}=\frac{\left(\mathrm{n}\right)(\mathrm{n}+1)}{1.2}$$$$\Rightarrow {\mathrm{u}}_{\mathrm{n}}={\mathrm{u}}_1.\frac{(\mathrm{n}+1)\left(\mathrm{n}\right)}{2}$$$${\mathrm{X}}_{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}+\mathrm{p}\left(\mathrm{n}\right)=\Rightarrow {\mathrm{u}}_1+\mathrm{p}\left(1\right)={\mathrm{x}}_1=1\Rightarrow {\mathrm{u}}_1-1=1\Rightarrow {\mathrm{u}}_1=2$$$$\Rightarrow {\mathrm{u}}_{\mathrm{n}}=(\mathrm{n}+1)\left(\mathrm{n}\right)={\mathrm{n}}^2+\mathrm{n}$$$${\mathrm{x}}_{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}+\mathrm{p}\left(\mathrm{n}\right)={\mathrm{n}}^2+\mathrm{n}-\mathrm{n}={\mathrm{n}}^2$$$$$$$$$$

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