If f(x)=(2^(2x))/(2^(2x)+2),x in z, then f((1)/(2023))+...+f((2022)/(2

To solve the problem, we need to evaluate the sum f(12023)+f(22023)+…+f(20222023) where f(x)=22x22x+2.

Step 1: Simplify the function f(x)

We start with the function:
f(x)=22x22x+2

We can rewrite 22x as 4x:
f(x)=4x4x+2

Step 2: Find f(1−x)

Next, we will find f(1−x):
f(1−x)=41−x41−x+2=44x44x+2=44+2⋅4x

Step 3: Compute f(x)+f(1−x)

Now we will compute f(x)+f(1−x):
f(x)+f(1−x)=4x4x+2+44+2⋅4x

To combine these fractions, we find a common denominator:
=4x(4+2⋅4x)+4(4x+2)(4x+2)(4+2⋅4x)

Step 4: Simplify the numerator

Expanding the numerator:
=4x+1+2⋅42x+4x+1+8(4x+2)(4+2⋅4x)
=2⋅4x+1+2⋅42x+8(4x+2)(4+2⋅4x)
Factoring out 2:
=2(4x+1+42x+4)(4x+2)(4+2⋅4x)

Step 5: Evaluate f(x)+f(1−x)

Notice that f(x)+f(1−x)=1:
f(x)+f(1−x)=1

Step 6: Sum f(k2023)

Now, we can compute the sum:
f(12023)+f(22023)+…+f(20222023)

We can pair the terms:
f(k2023)+f(2023−k2023)=1

Since there are 2022 terms, we can pair them up:
Number of pairs=20222=1011

Step 7: Final Result

Thus, the total sum is:
1011⋅1=1011

Conclusion

The final answer is:
1011