Question Number 194135 by cortano12 last updated on 28/Jun/23
$$\:\:\:\:\:\:\begin{array}{|c|}{\frac{\mathrm{23}!−\mathrm{23}}{\mathrm{1}.\mathrm{1}!+\mathrm{2}.\mathrm{2}!+\mathrm{3}.\mathrm{3}!+…+\mathrm{21}.\mathrm{21}!}\:=?}\\\hline\end{array} \\ $$
Answered by Frix last updated on 28/Jun/23
$$\mathrm{23} \\ $$$$\left[\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}{j}×{j}!=\left({n}+\mathrm{1}\right)!−\mathrm{1}\right] \\ $$
Answered by aba last updated on 28/Jun/23
$$\mathrm{1}.\mathrm{1}!+\mathrm{2}.\mathrm{2}!+…+\mathrm{21}.\mathrm{21}!=\left(\mathrm{2}−\mathrm{1}\right)\mathrm{1}!+\left(\mathrm{3}−\mathrm{1}\right)\mathrm{2}!+…+\left(\mathrm{22}−\mathrm{1}\right)\mathrm{21}! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\cancel{\mathrm{2}!}−\mathrm{1}!+\cancel{\mathrm{3}!}−\cancel{\mathrm{2}!}+\cancel{.}.\cancel{.}+\mathrm{22}!−\cancel{\mathrm{21}!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{22}!−\mathrm{1}! \\ $$$$\mathrm{A}=\frac{\mathrm{23}!−\mathrm{23}}{\mathrm{1}.\mathrm{1}!+\mathrm{2}.\mathrm{2}!+…+\mathrm{21}.\mathrm{21}!}=\frac{\mathrm{23}\left(\mathrm{22}!−\mathrm{1}\right)}{\mathrm{22}!−\mathrm{1}!}=\mathrm{23}\:\checkmark\: \\ $$$$ \\ $$
Answered by MM42 last updated on 28/Jun/23
$${n}×{n}!=\left({n}+\mathrm{1}\right)!−{n}! \\ $$$$\Rightarrow{s}=\mathrm{1}×\mathrm{1}!+\mathrm{2}×\mathrm{2}!+\mathrm{3}×\mathrm{3}!+…+\mathrm{21}×\mathrm{22}!= \\ $$$$\left(\mathrm{2}!−\mathrm{1}!\right)+\left(\mathrm{3}!−\mathrm{2}!\right)+\left(\mathrm{4}!−\mathrm{3}!\right)+…+\left(\mathrm{21}!−\mathrm{20}!\right)+\left(\mathrm{22}!−\mathrm{21}!\right)= \\ $$$$\mathrm{22}!−\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{23}!−\mathrm{23}}{{s}}=\frac{\mathrm{23}\left(\mathrm{22}!−\mathrm{1}\right)}{\mathrm{22}!−\mathrm{1}}=\mathrm{23}\:\checkmark \\ $$