Find the coefficient

$f(x)=1-x+x^2-x^3+...+x^{16}-x^{17}$ $xf(x)=x-x^2+x^3-...-x^{18}$ Adding these two we get $(1+x)f(x)= 1-x^{18}$ $f(x)=f(y-1)=\frac{1-(y-1)^{18}}{1+(y-1)}\\ =\frac{1-(y-1)^{18}}{y}$ $a_2$ is equal to coefficient of $y^3$ in the expansion of $1-(y-1)^{18}$ i,e, $a_2={18 \choose 3}=\boxed{816}$

Let $f(x)=1-x+x^2+...+x^{16}-x^{17}$ . This expression is the sum of the first 18 terms of a geometric sequence whose first term is 1 and common ratio $-x.$ Therefore, $f(x)=\frac{1-(-x)^{18}}{1-(-x)}=\frac{1-x^{18}}{1+x}\:\:\:\:\:\:\text{for} \:\:x\neq -1\:\:\:(*)$ In the other hand, if $f(x)=a_0 + a_1(x+1) + a_2(x+1)^2+...+a_{16}(x+1)^{16}+a_{17}(x+1)^{17},$ then differentiating both sides twice and evaluating at $x=-1,$ we get $f ''(-1)=2a_2.$ Now using $(*)$ and L'hopital Rule three times, we get that $2a_2=\lim_{x \to -1}{\frac{d^2}{dx^2}\frac{1-x^{18}}{1+x}}=1632.$ Then $a_2=816.$

The last part, where we find the second derivative of the function $\frac{1-x^{18}}{1+x}$ and apply L'hopital can be a tedious procedure. So another way of finding the value of $f ''(-1)$ is the following. We can differentiate the equality $f(x)=\sum_{n=0}^{n=17}(-1)^nx^{n}$ twice, so we get $f ''(x)=\sum_{n=2}^{17}(-1)^nn(n-1)x^{n-2}.$ Evaluating at $x=-1$ , $f ''(-1)=\sum_{n=2}^{17}n(n-1)=\sum_{n=1}^{17}n^2-\sum_{n=1}^{17}n=\frac{17(17+1)(2(17)+1)}{6}-\frac{17(17+1)}{2}=1632.$

$x=y-1$ $f(x)=1-x+x^2-x^3 +...+x^{16}-x^{17}$ $=1-(y-1)+(y-1)^2-(y-1)^3...+(y-1)^{16}-(y-1)^{17}$

Note that the coefficient of the $y^2$ term in the expansion of each power of $(y-1)$ is equal to $(-1)^n{n \choose 2}$ , where $n$ is the power of $(y-1)$ . And since the sign of each odd power of $(y-1)$ is negative, all $y^2$ terms in $f(y-1)$ are positive, and $a_2$ is their sum. Thus,

$a_2=\sum_{n=1}^{17}{n \choose 2}=\frac{(17)(16)+(16)(15)+(15)(14)+...+(3)(2)+(2)(1)}{(2)(1)}$ $=\frac{(16)(17+15)+(14)(15+13)+...+(2)(3+1)}{2}$ $=16^2+14^2+...+2^2=4\sum_{n=1}^8{n^2}=4\frac{(8)(9)(17)}{6}=\boxed{816}$