Question Number 377 by 123456 last updated on 25/Jan/15
![f_(n+1) (x)=((f_n (x))/(n+1))+n f_0 (x)=1 f_5 (x)=?](https://www.tinkutara.com/question/Q377.png)
$${f}_{{n}+\mathrm{1}} \left({x}\right)=\frac{{f}_{{n}} \left({x}\right)}{{n}+\mathrm{1}}+{n} \\ $$$${f}_{\mathrm{0}} \left({x}\right)=\mathrm{1} \\ $$$${f}_{\mathrm{5}} \left({x}\right)=? \\ $$
Answered by prakash jain last updated on 25/Dec/14
![f_0 (x)=1 f_1 (x)=(1/1)+0=1 f_2 (x)=(1/2)+1=(3/2) f_3 (x)=((3/2)/(2+1))+2=(1/2)+2=(5/2) f_4 (x)=((3/2)/(3+1))+3=(3/8)+3=((27)/8) f_5 (x)=(((27)/8)/(4+1))+4=((27)/(40))+4=((27+160)/(40))=((187)/(40))](https://www.tinkutara.com/question/Q378.png)
$${f}_{\mathrm{0}} \left({x}\right)=\mathrm{1} \\ $$$${f}_{\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}}+\mathrm{0}=\mathrm{1} \\ $$$${f}_{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${f}_{\mathrm{3}} \left({x}\right)=\frac{\frac{\mathrm{3}}{\mathrm{2}}}{\mathrm{2}+\mathrm{1}}+\mathrm{2}=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${f}_{\mathrm{4}} \left({x}\right)=\frac{\frac{\mathrm{3}}{\mathrm{2}}}{\mathrm{3}+\mathrm{1}}+\mathrm{3}=\frac{\mathrm{3}}{\mathrm{8}}+\mathrm{3}=\frac{\mathrm{27}}{\mathrm{8}} \\ $$$${f}_{\mathrm{5}} \left({x}\right)=\frac{\frac{\mathrm{27}}{\mathrm{8}}}{\mathrm{4}+\mathrm{1}}+\mathrm{4}=\frac{\mathrm{27}}{\mathrm{40}}+\mathrm{4}=\frac{\mathrm{27}+\mathrm{160}}{\mathrm{40}}=\frac{\mathrm{187}}{\mathrm{40}} \\ $$