For x ∈ R, x ≠ 0, x ≠ 1, let f₀(x) = 11 - x and fₙ₊₁(x) = f₀(fₙ(x)), n = 0, 1, 2, .... Then the value of f₁₀₀(3) + f₁(23) + f₂(32) is equal to:

f₁(x) = f₀(f₀(x)) = 11 - f₀(x) = 11 - (11 - x) = x
f₂(x) = f₀(f₁(x)) = f₀(x) = 11 - x
f₃(x) = f₀(f₂(x)) = f₀(f₀(x)) = f₁(x) = x
Thus, fₙ(x) = x if n is odd, and fₙ(x) = 11 - x if n is even.
Therefore,
f₁₀₀(3) = 11 - 3 = 8
f₁(23) = 23
f₂(32) = 11 - 32 = -21
f₁₀₀(3) + f₁(23) + f₂(32) = 8 + 23 + (-21) = 10