Question Number 190704 by mathlove last updated on 09/Apr/23
$${f}\left({x}+{y}\right)={f}\left({x}\right)+{f}\left({y}\right)+{x}\centerdot{y} \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{10}\:\:{finde}\:{f}\left(\mathrm{2022}\right)=? \\ $$
Answered by mahdipoor last updated on 09/Apr/23
![f(n+1)=[f(n)]+f(1)+n= [f(n−1)+f(1)+(n−1)]+f(1)+n=... =(n+1)f(1)+((n(n+1))/2) ⇒f(4)=4f(1)+((3×4)/2)=10⇒f(1)=1 f(2022)=2022f(1)+((2021×2022)/2)=1011×2023](https://www.tinkutara.com/question/Q190709.png)
$${f}\left({n}+\mathrm{1}\right)=\left[{f}\left({n}\right)\right]+{f}\left(\mathrm{1}\right)+{n}= \\ $$$$\left[{f}\left({n}−\mathrm{1}\right)+{f}\left(\mathrm{1}\right)+\left({n}−\mathrm{1}\right)\right]+{f}\left(\mathrm{1}\right)+{n}=… \\ $$$$=\left({n}+\mathrm{1}\right){f}\left(\mathrm{1}\right)+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{f}\left(\mathrm{4}\right)=\mathrm{4}{f}\left(\mathrm{1}\right)+\frac{\mathrm{3}×\mathrm{4}}{\mathrm{2}}=\mathrm{10}\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${f}\left(\mathrm{2022}\right)=\mathrm{2022}{f}\left(\mathrm{1}\right)+\frac{\mathrm{2021}×\mathrm{2022}}{\mathrm{2}}=\mathrm{1011}×\mathrm{2023} \\ $$