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Question Number 143393 by ZiYangLee last updated on 13/Jun/21

Given that f(r)=(r+1)! ∙ r, show that  f(r)−f(r−1)=r!(r^2 +1).  Hence or otherwise, show that  2! ∙ 5+3! ∙ 10+4! ∙ 17+......n!(n^2 +1)=(n+1)! ∙ (n−2)

$$\mathrm{Given}\:\mathrm{that}\:{f}\left({r}\right)=\left({r}+\mathrm{1}\right)!\:\centerdot\:{r},\:\mathrm{show}\:\mathrm{that} \\ $$$${f}\left({r}\right)−{f}\left({r}−\mathrm{1}\right)={r}!\left({r}^{\mathrm{2}} +\mathrm{1}\right). \\ $$$$\mathrm{Hence}\:\mathrm{or}\:\mathrm{otherwise},\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{2}!\:\centerdot\:\mathrm{5}+\mathrm{3}!\:\centerdot\:\mathrm{10}+\mathrm{4}!\:\centerdot\:\mathrm{17}+......{n}!\left({n}^{\mathrm{2}} +\mathrm{1}\right)=\left({n}+\mathrm{1}\right)!\:\centerdot\:\left({n}−\mathrm{2}\right) \\ $$

Answered by TheHoneyCat last updated on 13/Jun/21

  f(r)−f(r−1)  =(r+1)!∙r−r!∙(r−1)  =(r+1)∙r!∙r−(r−1)∙r!  =r!∙(r^2 +r−r+1)  =r!∙(r^2 +1)

$$ \\ $$$${f}\left({r}\right)−{f}\left({r}−\mathrm{1}\right) \\ $$$$=\left({r}+\mathrm{1}\right)!\centerdot{r}−{r}!\centerdot\left({r}−\mathrm{1}\right) \\ $$$$=\left({r}+\mathrm{1}\right)\centerdot{r}!\centerdot{r}−\left({r}−\mathrm{1}\right)\centerdot{r}! \\ $$$$={r}!\centerdot\left({r}^{\mathrm{2}} +{r}−{r}+\mathrm{1}\right) \\ $$$$={r}!\centerdot\left({r}^{\mathrm{2}} +\mathrm{1}\right) \\ $$

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