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Question Number 161800 by mathlove last updated on 22/Dec/21

$$\frac{1^2\cdot 2!+2^2\cdot 3!+3^2\cdot 4!+\cdot \cdot \cdot +n^2(n+1)!-2}{(n+1)!}=108$$$$n=?$$

Commented by Rasheed.Sindhi last updated on 22/Dec/21

$$n=10$$

Commented by mathlove last updated on 23/Dec/21

$$howsolve?$$

Answered by Rasheed.Sindhi last updated on 24/Dec/21

$$\frac{1^2\cdot 2!+2^2\cdot 3!+3^2\cdot 4!+\cdot \cdot \cdot +n^2(n+1)!-2}{(n+1)!}=108$$$$\mathrm{LHS}\mathrm{can}\mathrm{be}\mathrm{proved}\mathrm{to}\mathrm{be}\mathrm{equal}\mathrm{to}$$$${\mathrm{n}}^2+\mathrm{n}-2$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\therefore {\mathrm{n}}^2+\mathrm{n}-2=108$$$${\mathrm{n}}^2+\mathrm{n}-110=0$$$$(\mathrm{n}-10)(\mathrm{n}+11)=0$$$$\mathrm{n}=10\mid \mathrm{n}=-11\left(\mathrm{rejeted}\right)$$$$$$

Commented by Rasheed.Sindhi last updated on 24/Dec/21

$$\text{You can't use 'macro parameter character \#' in math mode}$$

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