#32 Hua Luo Geng.

The first few $x_n$ indicate the probable claim that $\displaystyle x_n = (-1)^n{225 \choose n}$ . Let us proof by induction that the claim is true by $n \ge 0$ .

Proof:

For $n=0$ , $\displaystyle x_0 = (-1)^0{225 \choose 0} = 1$ , which is as given. Therefore, the claim is true for $n=0$ .

Assuming the claim is true for $n$ , then we have:

$\begin{aligned} x_{\color{#D61F06}n+1} & = - \frac {225}{n+1} \sum_{k=0}^n x_k \\ & = - \frac {225}{n+1} \left(\sum_{k=0}^{n-1} x_k + x_n \right) \\ & = \frac n{n+1} x_n - \frac {225}{n+1} x_n \\ & = \frac {n-225}{n+1} \color{#3D99F6} x_n & \small \color{#3D99F6} \text{Note that } x_n = (-1)^n{225 \choose n} \\ & = \frac {n-225}{n+1} \times \color{#3D99F6} (-1)^n{225 \choose n} \\ & = (-1)^{n+1} \times \frac {225-n}{n+1} \times \frac {225!}{n!(225-n)!} \\ & = (-1)^{\color{#D61F06}n+1}{225 \choose {\color{#D61F06}n+1}} \end{aligned}$

Therefore, the claim is also true for $n+1$ and it is true for all $n \ge 0$ . $\square$

Then, we have:

$\begin{aligned} X & = x_0 + 2x_1 + 2^2x_2 + \cdots + 2^{225}x_{225} \\ & = \sum_{k=0}^{225} 2^k x_k \\ & = \sum_{k=0}^{225} (-1)^k{225 \choose k}2^k & \small \color{#3D99F6} \text{Note that } (1-u)^n = \sum_{k=0}^n (-1)^k{n \choose k}u^k \\ & = (1-2)^{225} \\ & = \boxed{-1} \end{aligned}$