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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 $f(2016)?$

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    $\begingroup$ Then use $2017 = 2016+1$, and put $x = 2016$. $\endgroup$ – Teresa Lisbon May 24 '20 at 16:23
  • $\begingroup$ Hint: Let $x = 2016$. Then $$x^{15} = 2016x^{14} = (2017 - 1)x^{14} = 2017 x^{14} - x^{14}.$$ I.e., $$x^{15} - 2017x^{14} = -x^{14}\ldots$$ $\endgroup$ – Good Boy May 24 '20 at 16:36
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    $\begingroup$ Is the constant term instead supposed to be $-2017$? $\endgroup$ – RobPratt May 24 '20 at 16:49
  • $\begingroup$ I checked the question again and made sure I was not mistaken, and that the constant term is indeed +2017. Thanks for all the replies here $\endgroup$ – jhuang2004 May 25 '20 at 13:15
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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 $x=2016$.

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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.

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To better see the principle, set $a=2016.$

Then $\begin{aligned}f(2016)=f(a)&=a^{15}-\underbrace{(a+1)\times a^{14}}_{a^{15}+a^{14}}+\underbrace{(a+1)\times a^{13}}_{a^{14}+a^{13}}-\dots-(a+1)\times a^{2}+(a+1)\times a+(a+1)\\ &=a^{15}-a^{15}-a^{14}+a^{14}+ a^{13}-a^{13}\dots-a^{3}- a^{2}+a^{2}+ a+a+1\\&=2a+1\\&=4033\end{aligned}$

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