Recursive Sequence

Note that $U_0=0$ , $U_1=1$ , $U_2 = 3$ , $U_3=7$ ... It appears that

$\begin{aligned} U_n - U_{n-1} & = 2^{n-1} \\ \sum_{\color{#3D99F6}k=1}^{\color{#3D99F6}n} U_k - \sum_{\color{#3D99F6}k=1}^{\color{#3D99F6}n} U_{\color{#3D99F6}k-1} & = \sum_{\color{#3D99F6}k=1} ^{\color{#3D99F6}n} 2^{\color{#3D99F6}k-1} \\ \sum_{\color{#D61F06}k=0}^{\color{#3D99F6}n} U_k - \sum_{\color{#D61F06}k=0}^{\color{#D61F06}n-1} U_{\color{#D61F06}k} & = \sum_{\color{#D61F06}k=0} ^{\color{#D61F06}n-1} 2^{\color{#D61F06}k} \\ \implies U_n & = 2^n - 1 \end{aligned}$

Let us prove by induction that the claim $U_n = 2^n - 1$ is true for all $n \le 0$ .

Proof:

• For $n=0$ , $U_0 = 2^0 - 1 = 0$ as given. Therefore, the claim is true for $n=0$ .
• Assuming that the claim is true for $n$ , then $U_{\color{#D61F06}n+1} = 2U_n + 1 = 2\left(2^n-1\right) + 1 = 2^{\color{#D61F06}n+1} - 1$ . The claim is true for $n+1$ and therefore true for all $n \ge 0$ .

Therefore, we have $\dfrac {x^{2017}-1}{U_{2017}} = \dfrac {x^{2017}-1}{2^{2017}-1} = 1$ . $\implies x = \boxed{2}$ .