Question Number 155842 by zainaltanjung last updated on 05/Oct/21
$$\mathrm{Prof}\:\mathrm{that}: \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\:\mathrm{n}.\mathrm{n}!=\mathrm{11}!−\mathrm{1} \\ $$
Commented by puissant last updated on 05/Oct/21
$${Q}\mathrm{155803} \\ $$
Answered by mr W last updated on 05/Oct/21
![Σ_(n=1) ^k n×n! =Σ_(n=1) ^k [(n+1)×n!−n!] =Σ_(n=1) ^k [(n+1)!−n!] =(k+1)!−1! =(k+1)!−1](https://www.tinkutara.com/question/Q155850.png)
$$\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}{n}×{n}! \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\left[\left({n}+\mathrm{1}\right)×{n}!−{n}!\right] \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\left[\left({n}+\mathrm{1}\right)!−{n}!\right] \\ $$$$=\left({k}+\mathrm{1}\right)!−\mathrm{1}! \\ $$$$=\left({k}+\mathrm{1}\right)!−\mathrm{1} \\ $$