Polynomial Equations of the Year!

$\begin{aligned} f(x) & = 1 - x + x^2 - x^3 + \cdots + x^{16} - x^{17} \\ & = \alpha_0 + \alpha_1(1+x) + \alpha_2(1+x)^2 + \cdots + \alpha_{17}(1+x)^{17} \end{aligned}$

Let $u = 1+x$ . Then we have:

$\begin{aligned} g(u) & = \alpha_0 + \alpha_1u + \alpha_2u^2 + \cdots + \alpha_{17}u^{17} \\ g'(u) & = \alpha_1 + 2\alpha_2u + 3\alpha_3u^2 + \cdots + 17\alpha_{17}u^{16} \\ g''(u) & = 2\alpha_2 + 6\alpha_3u + 12\alpha_4u^2 + \cdots + 272\alpha_{17}u^{15} \\ g''(0) & = 2\alpha_2 \\ \implies \alpha_2 & = \frac {g''(0)}2 \end{aligned}$

Note that $g''(u) = f''(x)\cdot \dfrac {dx}{du} = f''(x)$ . And we have:

$\begin{aligned} f(x) & = 1 - x + x^2 - x^3 + \cdots + x^{16} - x^{17} \\ f'(x) & = - 1 + 2x - 3x^2 + 4x^3 - \cdots + 16x^{15} - 17x^{16} \\ f''(x) & = 2 \cdot 1 - 3\cdot 2 x + 4\cdot 3 x^2 - 5\cdot 4x^3 + \cdots + 16\cdot 15 x^{14} - 17 \cdot 16 x^{16} & \small \color{#3D99F6} \text{when }u=0 \implies x = -1 \end{aligned}$

$\begin{aligned} \implies g''(0) & = f''(-1) = 2 \cdot 1 + 3\cdot 2 + 4\cdot 3 x^2 + 5\cdot 4 + \cdots + 16\cdot 15 + 17\cdot 16 \\ & = \sum_{n=1}^{16} n(n+1) = \sum_{n=1}^{16} \left(n^2+n\right) = \frac {16(17)(33)}6 + \frac {16(17)}2 = 1632 \\ \implies \alpha_2 & = \frac {g''(0)}2 = \frac {1632}2 = \boxed{816} \end{aligned}$

Use formula for GP to get $f(x)=\dfrac{1-x^{18}}{1+x}$ . Then substitute $1+x=t$ to get: $\dfrac{1-(t-1)^{18}}{t}=\displaystyle\sum_{i=0}^{17}\alpha_{i}t^i$ Use formula for binomial expansion on LHS to get: $\implies \sum_{r=0}^{17} \binom{18}{r+1}(-1)^{r} t^{r}=\displaystyle\sum_{i=0}^{17}\alpha_{i}t^i$ Compare coefficient of $t^2$ on both sides to get: $\alpha_2=\binom{18}3=\boxed{816}$ .