AIME 1986

First note that the polynomial can be written as

$1 + (-x) + (-x)^{2} + (-x)^{3} + .... + (-x)^{17} = \dfrac{1 - (-x)^{18}}{1 - (-x)} = \dfrac{1 - x^{18}}{1 + x}$ .

Now let $y = 1 + x$ . The polynomial can now be written as

$\dfrac{1 - (y - 1)^{18}}{y}$ .

So to find the coefficient of $y^{2}$ we will first need to find the coefficient of $y^{3}$ in the expansion of $(y - 1)^{18}$ . This will be

$\dbinom{18}{3} * (-1)^{15} = -\dbinom{18}{3}$ ,

and so the coefficient of $y^2$ in $\dfrac{1 - (y - 1)^{18}}{y}$ will be

$\dbinom{18}{3} = \boxed{816}$ .

Let $f(x) = 1 - x + x^2 ... -x^{17}$ , and $g(y) = a_0 + a_{1}y ... + a_{17}y^{17}$ .

$g(x + 1) = f(x)$ , so $g''(x+1) = f''(x)$ . Putting $x = -1$ , $g''(0) = f''(-1)$ .

So $2a_2 = 2.1 + 3.2 + ... + 17.16 = \sum_{i=1}^{16}(i^2 + i) = \frac{16.17.33}{6} + \frac{16.17}{2} = 8.17.11 + 8.17 = 8.17.12$

So $a_2 = 8.17.6 = 816$ .

This problem was posted before by Sharky Kesa . See here . I provided solution to it two days ago. See below.

Let $\quad \displaystyle f(x) = \sum_{n=0}^{17} {(-x)^n} = g(x) = \sum_{n=0}^{17} {a_ny^n}$ .

Then $\quad \dfrac {df(x)}{dx} = \dfrac {dg(y)}{dx} = \dfrac {dg(y)}{dy} \dfrac {dy}{dx} = \dfrac {dg(y)}{dy}$ .

Similarly, $\quad \dfrac {d^2f(x)}{dx^2} = \dfrac {d^2g(y)}{dy^2}$

We note that $\displaystyle \quad \dfrac {df(x)}{dx} = \sum_{n=1}^{17} {[(-1)^nnx^{n-1}]} \quad$ and $\displaystyle \quad \dfrac {d^2f(x)}{dx^2} = \sum_{n=2}^{17} {[(-1)^nn(n-1)x^{n-2}]}$

We also know that:

$g''(0) = \left[ \dfrac {d^2g(y)}{dy^2} \right]_{y=0} = 2a_2 = f''(-1) = \left[ \dfrac {d^2f(x)}{dx^2} \right]_{x=-1}$

$= \displaystyle \sum_{n=2}^{17} {n(n-1)} = \sum_{n=1}^{17} {n^2} - \sum_{n=1}^{17} {n} = \dfrac {(17)(18)(35)}{6}-\dfrac {(17)(18)}{2} = 1632$

$\Rightarrow a_2 = \dfrac {1632}{2} = \boxed{816}$

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

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

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

Thus

$a_{2}=\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+...\binom{17}{2}=\binom{18}{3}=\boxed{816}$

@Rishabh Jain , I posted the exact same problem AIME 1986 Problem .