Let f(x)=2xn+λ,λ∈ℝ and n∈ℕf(4)=133 and f(5)=255, then sum of all the  positive integral divisors of (f(3)-f(2)) is

f(x)=2xn+λ,f(4)=133,f(5)=255f(4)=2.4n+λ=133  ………(i)f(5)=2.5n+λ=255  ………(ii)∴f(5)-f(4)=25n-4n=122⇒5n-4n=53-43⇒n=3∴f(4)=133=2×43+λ∴λ=5∴f(x)=2x3+5∴f(3)=59 and f(2)=21∴f(3)-f(2)=38=2×19=2a19b,(a=1,b=1) Sum of factors f(3)-f(2)=2a+1-12-1·19b+1-119-1=3×36018=60