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Question Number 7824 by ridwan balatif last updated on 17/Sep/16

Commented by Yozzia last updated on 17/Sep/16

$$n\times n!=(n+1-1)n!=(n+1)!-n!$$$$\Rightarrow \underset{i=1}{\sum ^n}r!r=(n+1)!-1$$

Commented by prakash jain last updated on 17/Sep/16

$$\mathrm{English}\mathrm{translation},\mathrm{please}$$

Commented by ridwan balatif last updated on 17/Sep/16

$$youneedtocount$$$$1.1!+2.2!+...+n.n!$$

Commented by FilupSmith last updated on 17/Sep/16

$$1\times 1!+2\times 2!+...+n\times n!$$$$=\underset{k=1}{\sum ^n}k(k!)$$$$=\underset{k=1}{\sum ^n}{k}^2(k-1)!$$

Answered by prakash jain last updated on 02/Oct/16

$$\mathrm{see}\mathrm{comments}$$

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