1.If f(x)=2f(x) = 2f(x)=2 for all x∈Rx \in \mathbb{R}x∈R, and f(1)+f(2)+f(3)+⋯+f(n)=2022f(1) + f(2) + f(3) + \dots + f(n) = 2022f(1)+f(2)+f(3)+⋯+f(n)=2022, then nnn is equal toLogin to continueOnly logged in users canattempt or see the solution.
