f(0) = a.02 + b. 0 + c = 2016
<=> c =2016
f (1) = a.12 + b.1 + c =2017
<=> a + b =1 (1)
f ( -1 ) = a (-1)2 + b . (-1) +c =2018
<=> a -b =2 (2)
Từ (1),(2) <=> a = 1,5 ; b = -0,5
=> F(x) = 1,5x2 -0,5 x + 2016
F (2) = 1,5 . 22 -0,5 .2 +2016
= 6 -1 +2016 =2021
Ta có:
\(F\left(0\right)=a.0^2+b.0+c=2016\)
\(\Rightarrow c=2016\)
\(F\left(1\right)=a.1^2+b.1+c=2017\)
\(\Rightarrow a+b=1\)
\(F\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=2018\)
\(\Rightarrow a-b=2\)
Vì a + b =1 và a - b = 2 nên \(\Rightarrow a=\frac{3}{2};b=\frac{-1}{2}\)
Vậy \(F\left(2\right)=\frac{3}{2}.2^2-\left(\frac{-1}{2}\right).2+2016=2023\)
