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Question Number 205919 by Red1ight last updated on 03/Apr/24

$$\mathrm{write}\mathrm{the}\mathrm{following}\mathrm{recursive}\mathrm{function}\mathrm{in}\mathrm{explicit}\mathrm{form}$$$$f\left(1\right)=1$$$$f(n+1)=(n+1)f\left(n\right)+n!$$

Answered by Tinku Tara last updated on 03/Apr/24

$$f\left(n\right)=\left(n\right)f(n-1)+(n-1)!$$$$f\left(n\right)=nf(n-1)+\frac{n!}{n}$$$$=n[(n-1)f(n-2)+\frac{(n-1)!}{(n-1)}]+\frac{n!}{n}$$$$=n(n-1)f(n-2)+\frac{n!}{(n-1)}+\frac{n!}{n}$$$$\mathrm{continue}$$$$f\left(n\right)=n!+\underset{i=2}{\sum ^n}\frac{n!}{i}$$

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