Đặt g(x)=f(x)-x-1 vì f(x) bậc 3 nên g(x) cũng bậc ba. Ta có g(2015)=g(2016)=0 Nên g(x)=(x-2015)(x-2016)(ax+b) suy ra f(x)=(x-2015)(x-2016)+x+1. Từ điều kiện f(2014)-f(2017)=3 suy ra a=-1, b tùy ý

Ta có

\(F\left(0\right)=2016\)

\(\Leftrightarrow a\cdot0^2+b\cdot0+c=2016\)

\(\Leftrightarrow0+0+c=2016\)

\(\Leftrightarrow c=2016\)

\(F\left(1\right)=2016\)

\(\Leftrightarrow a\cdot1^2+b\cdot1+c=2017\)

\(\Leftrightarrow a+b+c=2017\)

\(\Leftrightarrow a+b+2016=2017\)

\(\Leftrightarrow a+b=1\)       \(\left(1\right)\)

\(F\left(-1\right)=2018\)

\(\Leftrightarrow a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+c=2018\)

\(\Leftrightarrow a-b+c=2018\)

\(\Leftrightarrow a-b+2016=2018\)

\(\Leftrightarrow a-b=2\)       \(\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow a=\left(1+2\right)\div2=3\div2=1.5\)

\(\Rightarrow b=1-1.5=-0.5\)

Vậy \(F\left(x\right)=1.5x^2-0.5x+2016\)

\(\Leftrightarrow F\left(2\right)=1.5\cdot2^2-0.5\cdot2+2016\)

\(=1.5\cdot4-0.5\cdot2+2016\)

\(=6-1+2016=2021\)

Vậy \(F\left(2\right)=2021\)

nhớ k nha

Xét đa thức \(F\left(x\right)=ax^2+bx+c\)

\(F\left(0\right)=c=2016\)

\(F\left(1\right)=a+b+c=2017\Rightarrow a+b=1\)  (1)

\(F\left(-1\right)=a-b+c=2018\Rightarrow a-b=2\)  (2)

Từ (1), (2)

\(\Rightarrow\hept{\begin{cases}a+b-a+b=-1\\a+b+a-b=3\end{cases}}\Rightarrow\hept{\begin{cases}2b=-1\\2a=3\end{cases}}\Rightarrow\hept{\begin{cases}b=-0,5\\a=1,5\end{cases}}\)

\(\Rightarrow F\left(2\right)=1,5.2^2-0,5.2+2016=2021\)

Vậy \(F\left(2\right)=2021\).

Ta có : \(P\left(13\right)=\left(a+2015\right).13^3+\left(b+2016\right).13+2017=14\)

<=> \(\left(a+2015\right).13^3+\left(b+2016\right).13=14-2017=-2003\)

Mặt khác ta có : \(P\left(-13\right)=\left(a+2015\right).\left(-13\right)^3+\left(b+2016\right).\left(-13\right)+2017\)

=> \(P\left(-13\right)=-\left[\left(a+2015\right).13^3+\left(b+2016\right).13\right]+2017=-\left(-2003\right)+2017=4020\)

Với x = 0, ta có:

0²⁰¹⁶. f(0-2016) = (0 - 2017) . f(0)

=> 0. f(-2016) = - 2017. f(0)

=> 0 = - 2017. f(0) => f(0) = 0 (1)

Với x = 2017, ta có: 

2017²⁰¹⁶ . f(2017 - 2016) = (2017 -2017) . f(2017)

=> 2017²⁰¹⁶ . f(1) = 0. f(2017)

=>2017²⁰¹⁶ . f(1) = 0 => f(1) = 0 (2)

(1), (2) => (đpcm)

f(x)= x^2017 - 2016.x^2016 - 2016.x^2015 - ... - 2016x + 1

f(x)= x^2017 - (2017 - 1)x^2016 - (2017 - 1)x^2015 - ... - (2017 - 1)x +1

Với x=2017 ta có :

f(x)= x^2017 - (x - 1)x^2016 - (x-1)x^2015 - ... - (x - 1)x +1

f(x)= x^2017 - x^2017 +x^2016 - x^2016 +...+ x^2 - x^2 + x + 1

f(x)= x + 1

Thay x =2017 vào f(x) ta có :

f(2017) = 2017 +1 = 2018

\(f\left(1\right)=a_{2017}+a_{2016}+...+a_3+a_2+a_1+a_0\)

\(f\left(-1\right)=-a_{2017}+a_{2016}+...-a_3+a_2-a_1+a_0\)

\(f\left(1\right)+f\left(-1\right)=2\left(a_{2016}+a_{2014}+...+a_2+a_0\right)\)

\(S=\frac{f\left(1\right)+f\left(-1\right)}{2}=\frac{3^{2017}+1}{2}\)