Given,
f(x)=22x+222x,
so f(1−x)=22−2x+222−2x=22+22x+122=2+22x2
Now f(x)+f(1−x)=22x+222x+2+22x2=1
So by using the relation f(x)+f(1−x)=1 we will solve the given equation
f(20231)+f(20232)+f(20233).........f(20232022)
=f(20231)+f(20232022)+f(20232)+f(20232021)+f(20233)+f(20232020).........f(20231011)+f(20231012)
=1⏟f(20231)+f(20232022)+1⏟f(20232)+f(20232021)+1⏟f(20233)+f(20232020).........+1⏟f(20231011)+f(20231012)
=1+1+1.........1011times=1011