Question Number 120477 by help last updated on 31/Oct/20
Answered by Dwaipayan Shikari last updated on 31/Oct/20
$${f}\left({x}\right)−{f}\left({x}−\mathrm{2}\right)=\mathrm{6} \\ $$$${f}\left(\mathrm{2}\right)−{f}\left(\mathrm{0}\right)=\mathrm{6} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{7}\:\:\:,\:\:{f}\left(\mathrm{1}\right)=\mathrm{4} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1}\:\:\:\:\:\: \\ $$$${And} \\ $$$${f}\left(\mathrm{3}\right)−{f}\left(\mathrm{1}\right)=\mathrm{6}\:\:\Rightarrow{f}\left(\mathrm{3}\right)=\mathrm{10} \\ $$$${f}\left({n}\right)=\mathrm{1}+\left({n}\right)\mathrm{3} \\ $$$${f}\left(\mathrm{71}\right)=\mathrm{1}+\mathrm{71}.\mathrm{3}=\mathrm{214} \\ $$
Answered by Ar Brandon last updated on 31/Oct/20
$${f}\left(\mathrm{x}\right)={f}\left(\mathrm{x}−\mathrm{2}\right)+\mathrm{6},\:{f}\left(\mathrm{0}\right)=\mathrm{1},\:{f}\left(\mathrm{1}\right)=\mathrm{4} \\ $$$${f}\left(\mathrm{2}\right)={f}\left(\mathrm{0}\right)+\mathrm{6}=\mathrm{7},\:{f}\left(\mathrm{3}\right)={f}\left(\mathrm{1}\right)+\mathrm{6}=\mathrm{10} \\ $$$${f}\left(\mathrm{4}\right)={f}\left(\mathrm{2}\right)+\mathrm{6}=\mathrm{13},\:{f}\left(\mathrm{5}\right)={f}\left(\mathrm{3}\right)+\mathrm{6}=\mathrm{16} \\ $$$$\Rightarrow{f}\left(\mathrm{x}\right)=\mathrm{3x}+\mathrm{1}\:\Rightarrow\:{f}\left(\mathrm{71}\right)=\mathrm{3}\left(\mathrm{71}\right)+\mathrm{1}=\mathrm{214} \\ $$
Commented by Dwaipayan Shikari last updated on 31/Oct/20
$${Oh}\:{it}\:{was}\:{a}\:{mistake}\:{of}\:{mine} \\ $$
Commented by Ar Brandon last updated on 31/Oct/20
$${Oh}\:{it}\:{was}\:{a}\:{mistake}\:{from}\:{mine} \\ $$
Commented by Ar Brandon last updated on 31/Oct/20
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