If f(n)=αn+βn and ∣∣ ∣ ∣∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣∣ ∣ ∣∣=k(1−α)2(1−β)2(α−β)2, then k is equal to
f(x)=2x2+bx+c and f(0)=3 and f(2)=1, then f(1) is equal to