Question Number 7824 by ridwan balatif last updated on 17/Sep/16
Commented by Yozzia last updated on 17/Sep/16
$${n}×{n}!=\left({n}+\mathrm{1}−\mathrm{1}\right){n}!=\left({n}+\mathrm{1}\right)!−{n}! \\ $$$$\Rightarrow\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{r}!{r}=\left({n}+\mathrm{1}\right)!−\mathrm{1} \\ $$
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
$${you}\:{need}\:{to}\:{count} \\ $$$$\mathrm{1}.\mathrm{1}!+\mathrm{2}.\mathrm{2}!+...+{n}.{n}! \\ $$
Commented by FilupSmith last updated on 17/Sep/16
$$\mathrm{1}×\mathrm{1}!+\mathrm{2}×\mathrm{2}!+...+{n}×{n}! \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\left({k}!\right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \left({k}−\mathrm{1}\right)! \\ $$
Answered by prakash jain last updated on 02/Oct/16
$$\mathrm{see}\:\mathrm{comments} \\ $$