Question Number 205919 by Red1ight last updated on 03/Apr/24
$$\mathrm{write}\:\mathrm{the}\:\mathrm{following}\:\mathrm{recursive}\:\mathrm{function}\:\mathrm{in}\:\mathrm{explicit}\:\mathrm{form} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${f}\left({n}+\mathrm{1}\right)=\left({n}+\mathrm{1}\right){f}\left({n}\right)+{n}! \\ $$
Answered by Tinku Tara last updated on 03/Apr/24
![f(n)=(n)f(n−1)+(n−1)! f(n)=nf(n−1)+((n!)/n) =n[(n−1)f(n−2)+(((n−1)!)/((n−1)))]+((n!)/n) =n(n−1)f(n−2)+((n!)/((n−1)))+((n!)/n) continue f(n)=n!+Σ_(i=2) ^n ((n!)/i)](https://www.tinkutara.com/question/Q205923.png)
$${f}\left({n}\right)=\left({n}\right){f}\left({n}−\mathrm{1}\right)+\left({n}−\mathrm{1}\right)! \\ $$$${f}\left({n}\right)={nf}\left({n}−\mathrm{1}\right)+\frac{{n}!}{{n}} \\ $$$$={n}\left[\left({n}−\mathrm{1}\right){f}\left({n}−\mathrm{2}\right)+\frac{\left({n}−\mathrm{1}\right)!}{\left({n}−\mathrm{1}\right)}\right]+\frac{{n}!}{{n}} \\ $$$$={n}\left({n}−\mathrm{1}\right){f}\left({n}−\mathrm{2}\right)+\frac{{n}!}{\left({n}−\mathrm{1}\right)}+\frac{{n}!}{{n}} \\ $$$$\mathrm{continue} \\ $$$${f}\left({n}\right)={n}!+\underset{{i}=\mathrm{2}} {\overset{{n}} {\sum}}\frac{{n}!}{{i}} \\ $$