数列 $u_1$、$u_2$、$u_3$、$\ldots$ 满足 $u_1=5$，且当 $n\geqslant1$ 时，$u_{n+1}=6u_n+5$。

(a) 用数学归纳法证明：对所有正整数 $n$，$u_n=6^n-1$。

(b) 由此推得，当 $n\geqslant1$ 时，$u_{2n}$ 能被 $u_n$ 整除。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

完整写出初项验证、归纳假设，以及代入递推式得到下一项。

For $n=1$, $6^1-1=5=u_1$, so the statement holds.

Assume that $u_k=6^k-1$ for some positive integer $k$. Then, using the recurrence,

Thus the result holds for $k+1$. By mathematical induction, $\boxed{u_n=6^n-1}$ for every positive integer $n$.

承接(a)，分子是平方差，因式分解后商为整数。

Using (a),

Since $6^n+1$ is an integer, $\boxed{u_n\text{ divides }u_{2n}}$ for every $n\ge1$.