Question Number 196728 by Amidip last updated on 30/Aug/23
Commented by mr W last updated on 30/Aug/23
$${Q}#\mathrm{196544}? \\ $$
Answered by Frix last updated on 30/Aug/23
$${f}\left(\mathrm{4}\right)={f}\left(\mathrm{2}+\mathrm{2}\right)=\mathrm{2}{f}\left(\mathrm{2}\right)+\mathrm{4}=\mathrm{10}\:\Rightarrow\:{f}\left(\mathrm{2}\right)=\mathrm{3} \\ $$$${f}\left(\mathrm{2}\right)={f}\left(\mathrm{1}+\mathrm{1}\right)=\mathrm{2}{f}\left(\mathrm{1}\right)+\mathrm{1}=\mathrm{3}\:\Rightarrow\:{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${f}\left(\mathrm{3}\right)={f}\left(\mathrm{1}+\mathrm{2}\right)=\mathrm{1}+\mathrm{3}+\mathrm{2}=\mathrm{6} \\ $$$$\mathrm{1}\:\mathrm{3}\:\mathrm{6}\:\mathrm{10} \\ $$$$\:\:\mathrm{2}\:\:\mathrm{3}\:\:\mathrm{4} \\ $$$${f}\left({x}\right)=\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$${f}\left(\mathrm{2023}\right)=\mathrm{2047276} \\ $$