Question Number 26405 by math solver last updated on 25/Dec/17
Answered by prakash jain last updated on 25/Dec/17
$${r}^{\mathrm{2}} +\mathrm{1}=\left({r}+\mathrm{1}\right){r}−{r}+\mathrm{1} \\ $$$$\left({r}^{\mathrm{2}} +\mathrm{1}\right){r}!={r}\left({r}+\mathrm{1}\right)!−\left({r}−\mathrm{1}\right){r}! \\ $$$${sum}=\mathrm{1}\centerdot\mathrm{2}!−\mathrm{0}\centerdot\mathrm{1}! \\ $$$$\:\:\:\:\:\:\:\:\:\:+\mathrm{2}\centerdot\mathrm{3}!−\mathrm{1}\centerdot\mathrm{2}! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3}\centerdot\mathrm{4}!−\mathrm{2}\centerdot\mathrm{3}! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+{n}\centerdot\left({n}+\mathrm{1}\right)!−\left({n}−\mathrm{1}\right)\centerdot{n}! \\ $$$$={n}\centerdot\left({n}+\mathrm{1}\right)! \\ $$
Commented by math solver last updated on 25/Dec/17
$${thank}\:{you}\:{sir}! \\ $$