you have a wrong question

it's divisible when it's remander 0.

I think you should say: f(x) is divided by ..., because f(x) is divisible when it doesn't have any remainder.

Thank you very much 

P(x) = a₀ + a₁x + a₂x² + ... + a₄₅x⁴⁵

=> P(1) = a₀ + a₁ + a₂ + ... + a₄₅ = (2 + 1 + 2)¹⁵ = 5¹⁵

=> P(-1) = a₀ - a₁ + a₂ - ... - a₄₅ = (2 - 1 - 2)¹⁵ = (-1)¹⁵

==> P(1) - P(-1) = 2a₁ + 2a₃ + ... + 2a₄₅ = 5¹⁵ + 1

        P(1) + P(-1) = 2a₀ + 2a₂ + ... + 2a₄₄ = 5¹⁵ - 1

=> (a₁ + a₃ + ... + a₄₅) - (a₀ + a₂ + ... + a₄₄) = \(\dfrac{5^{15}+1-5^{15}+1}{2}=1\)

=> S = -a₀ + a₁ - a₂ + .... - a₄₄ + a₄₅ = 1

P/s: Not S in the topic 

Let \(Q\left(x\right)=\left(x+1\right)P\left(x\right)-x\),  (*)

since \(P\left(n\right)=\dfrac{n}{n+1}\) we infer:

 \(Q\left(n\right)=\left(n+1\right)P\left(n\right)-n=0\) for \(n=0,1,..,2015\).

Because P is a polynomial of dgree 2016, Q is a polynomial of dgree 2016. Q has 2016 solutions (0, 1, .. , 2015) so Q can be expressed in a form:

  \(Q\left(x\right)=a\left(x-0\right)\left(x-1\right)...\left(x-2015\right)\)   (**)

In other hand, in (*) we set x = -1 then  \(Q\left(-1\right)=1\). And replace to (**) we have:

  \(1=Q\left(-1\right)=a\left(-1\right)\left(-2\right)...\left(-2016\right)\)

 \(\Rightarrow a=\dfrac{1}{2016!}\)

Finally \(Q\left(x\right)=\dfrac{1}{2016!}\left(x-0\right)\left(x-1\right)...\left(x-2015\right)\)

\(\Rightarrow Q\left(2016\right)=\dfrac{1}{2016!}2016.2015...1=1\)

from (*) \(Q\left(2016\right)=\left(2016+1\right)P\left(2016\right)-2016\)

\(\Rightarrow\left(2016+1\right)P\left(2016\right)-2016=1\)

\(\Rightarrow P\left(2016\right)=\dfrac{2017}{2017}=1\)

Let Q(x)=(x+1)P(x)−x

,  (*)

since P(n)=nn+1

 we infer:

 Q(n)=(n+1)P(n)−n=0

 for n=0,1,..,2015

.

Because P is a polynomial of dgree 2016, Q is a polynomial of dgree 2016. Q has 2016 solutions (0, 1, .. , 2015) so Q can be expressed in a form:

  Q(x)=a(x−0)(x−1)...(x−2015)

   (**)

In other hand, in (*) we set x = -1 then  Q(−1)=1

. And replace to (**) we have:

  1=Q(−1)=a(−1)(−2)...(−2016)

 ⇒a=12016!

Finally Q(x)=12016!(x−0)(x−1)...(x−2015)

⇒Q(2016)=12016!2016.2015...1=1

from (*) Q(2016)=(2016+1)P(2016)−2016

⇒(2016+1)P(2016)−2016=1

⇒P(2016)=20172017=1