Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.5 AIME, harder Find the answer

16. Given the sequence $\left\{a_{n}\right\}$ satisfies the condition $(n-1) a_{n+1}=(n+1)\left(a_{n}-1\right), a_{2}=6$ , let $b_{n}=$ $a_{n}+n, n \in \mathbf{N}_{+}$ .
（1）Write out the first 4 terms of the sequence $\left\{a_{n}\right\}$ ;（2）Find $b_{n}$ ;（3）Does there exist non-zero constants $p, q$ , such that the sequence $\left\{\frac{a_{n}}{p \cdot n+q}\right\}$ forms an arithmetic sequence? If so, find the relationship that $p$ and $q$ should satisfy; if not, explain the reason.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

16. (1) $a_{1}=1, a_{2}=6, a_{3}=15, a_{4}=28$ ;
(2) From the condition, we get $\frac{a_{n+1}}{n(n+1)}=\frac{a_{n}}{(n-1) n}-\frac{1}{(n-1) n}$ , then $\frac{a_{n+1}}{n(n+1)}+\frac{1}{n}=\frac{a_{n}}{(n-1) n}-\frac{1}{n-1}=\frac{a_{2}}{2}-1=2$ , so $a_{n}=2 n(n-1)+n=2 n^{2}-n$ for $n \geqslant 2$ . And $a_{1}=1$ also satisfies this formula, hence $a_{n}=2 n^{2}-n$ , thus $b_{n}=2 n^{2}$ .
(3) Since $\frac{a_{n}}{p n+q}=\frac{2 n^{2}-n}{p n+q}=n \cdot \frac{2 n-1}{p n+q}$ , we only need $\frac{2}{p}=\frac{-1}{q}$ , i.e., $p+2 q=0$ . When $p \neq 0$ , $\frac{a_{n}}{p n+q}=\frac{2}{p} n$ forms an arithmetic sequence.

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