I saw another post with the same problem but it didn't really answer my question.
This problem is Problem 11 from AIME 1986:
The polynomial $$1-x+x^2-x^3+\cdots+x^{16}-x^{17}$$ may be written in the form $$a_0+a_1y+a_2y^2+\cdots+a_{16}y^{16}+a_{17}y^{17}$$ where $y=x+1$ and the $a_i$ are constants. Find $a_2$.
Here's my attempt:
Starting from $1-x + x^2 + … -x^{17} = a_0 + a_1y + … +a_{17}y^{17}$, let $x = -1$:
$$ 1 + 1 + 1 + \dotsb + 1 = a_0 + a_1\cdot 0 + \dotsb + a_{17}\cdot 0.$$
Hence, $18 = a_0$.
Letting $x = 9$:
$$1 – 9 + 9^2 +\dotsb – 9^{17} = 18 + a_1\cdot 10 + \dotsb +a_{17}\cdot 10^{17}.$$
The LHS is a geometric sequence with $r = -9$
$$-15,009,463,529,699,912 = 18 + a_1\cdot 10 + \dotsb +a_{17}\cdot 10^{17}.$$
Hence, $a_2 = -9$.
However, this isn't the answer and I don't understand why. I feel like I must be making a stupid mistake somewhere but I cannot find it…
Best Answer
Your method has some merit, but fails with too big coefficients. For example $$ (10+1)^4=11^4=14641$$ corresponds nicely with $$ (x+1)^4=x^4+4x^3+6x^2+4x+1,$$ but $11^5=161051$ does not imply that $$ (x+1)^5=x^6+6x^4+x^3+5x+1$$ instead of correctly $$ (x+1)^5=x^6+5x^4+10x^3+10x^2+5x+1.$$ You could find $a_1$ by manipulating the polynomial such that it becomes the constant term and then plugging in $y=0$ again. Namely, for $y\ne0$ (equivalently, $x\ne -1$), we have $$\begin{align}a_1+a_2y+a_3y^3+\cdots a_{17}y^{16} &=\frac{(a_0+a_1y+\cdots+a_{17}y^{17})-a_0}{y}\\ &=\frac{(1-x+x^2\mp\cdots -x^{17})-18}{x+1}\\ &=\frac{(x+1)(-x^{16}+2x^{15}-3x^{14}\pm\cdots+16x-17)}{x+1}\\ &=-x^{16}+2x^{15}-3x^{14}\pm\cdots+16x-17 \end{align} $$ The first and least expression are also equal when $y=0$ / $x=-1$ and by pluggin in, you obtain $$a_1=-1-2-3-\cdots -17. $$
You might repeat the method to find $a_2$. But actually, you could, perhaps more simply, perform a polynomial division of $1-x\pm\cdots -x^{17}$ by $x^3+3x^2+3x+1$ ($=(x+1)^3$) right away and have a look and the quadratic polynomial that you obtain as remainder!