Let $f(x)=x^{15}-2017x^{14}+2017x^{13}-2017x^{12}+2017x^{11}-...+2017x^{3}-2017x^{2}+2017x+2017$ What is $f(2016)?$ This is a question from a Taiwanese high school extension maths class entrance exam.
Let $f(x)=x^{15}-2017x^{14}+2017x^{13}-2017x^{12}+2017x^{11}-...+2017x^{3}-2017x^{2}+2017x+2017$ What is $f(2016)?$
 A: You can use polynomial long division of $f(x)$ by $(x-2016)$, which is usually compactly written using Horner's method:
$$\begin{array}{c|cccc}
& x^{15} & x^{14} & x^{13} & x^{12} &\dots &x^{2} &x^{1} &x^{0} \\ \hline
& 1 & -2017 & 2017 & -2017  &\dots   &-2017  &2017  &2017  \\
2016&   &   &    &    \\ \hline
& 1 & -1& 1 & -1  &\dots &-1  &1  &\color{red}{4033}
\end{array}$$
and thus $f(x)=q(x)(x-2016)+4033$ for some polynomial $q(x)$. Now it should be clear what is the value of $f(2016)$.
Another way to view this is to write
$$
f(x)=(((\dots(((((x-2017)x+2017)x-2017)x+2017)x)...\\+2017)x-2017)x+2017)x+2017
$$
and then evaluate from inside for $x=2016$.
A: HINT: We can compress the expression a bit:
$$f(x)=x^{15}-2017(\sum_{j=0}^{14}(-x)^j) = x^{15}-2017\frac{x^{15}+1}{x+1}$$
Now it is much more simple to evaluate $f(2016)$, since there will be some cancilation there.
A: To better see the principle, set $a=2016.$
Then $\begin{aligned}f(2016)=f(a)&=a^{15}-\underbrace{(a+1)\times a^{14}}_{a^{15}+a^{14}}+\underbrace{(a+1)\times a^{13}}_{a^{14}+a^{13}}-\dots-(a+1)\times a^{2}+(a+1)\times a+(a+1)\\
&=a^{15}-a^{15}-a^{14}+a^{14}+ a^{13}-a^{13}\dots-a^{3}- a^{2}+a^{2}+ a+a+1\\&=2a+1\\&=4033\end{aligned}$
