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Question Number 186468 by aba last updated on 04/Feb/23

$$\mathrm{Simplify}$$$$\frac{1^2\cdot 2!+2^2\cdot 3!+3^2\cdot 4!+\cdot \cdot \cdot +n^2(n+1)!-2}{(n+1)!}$$$$\mathrm{to}$$$${\mathrm{n}}^2+\mathrm{n}-2$$

Answered by SEKRET last updated on 07/Feb/23

$$1^2\cdot 2=2$$$$2^2\cdot 3!=4!$$$$3^2\cdot 4!=2\cdot 5!-4!$$$$4^2\cdot 5!=3\cdot 6!-2\cdot 5!$$$$5^2\cdot 6!=4\cdot 7!-3\cdot 6!$$$$.......$$$$............$$$${\mathbf{n}}^2(\mathbf{n}+1)!=(\mathbf{n}-1)\cdot (\mathbf{n}+2)!-(\mathbf{n}-2)\cdot (\mathbf{n}+1)!$$$$(+)\mathbf{all}$$$$2+(\mathbf{n}-1)\cdot (\mathbf{n}+2)!$$$$\frac{2+(\mathbf{n}-1)\cdot (\mathbf{n}+2)!-2}{(\mathbf{n}+1)!}=\frac{(\mathbf{n}-1)\cdot (\mathbf{n}+2)(\mathbf{n}+1)!}{(\mathbf{n}+1)!}$$$$=(\mathbf{n}-1)(\mathbf{n}+2)={\mathbf{n}}^2+\mathbf{n}-2$$$$\mathit{A}\mathit{B}\mathit{D}\mathit{U}\mathit{L}\mathit{A}\mathit{Z}\mathit{I}\mathit{Z}\mathit{A}\mathit{B}\mathit{D}\mathit{U}\mathit{V}\mathit{A}\mathit{L}\mathit{I}\mathit{Y}\mathit{E}\mathit{V}$$$$$$

Commented by aba last updated on 11/Feb/23

thank a lot��

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