7. Given $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are arithmetic sequences (i.e. $a_{2}-a_{1}=a_{3}-a_{2}=a_{4}-a_{3}=\cdots$ and $\left.b_{2}-b_{1}=b_{3}-b_{2}=b_{4}-b_{3}=\cdots\right)$ such that $(3 n+1) a_{n}=(2 n-1) b_{n}$ for all positive integers $n$. If $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$ and $T_{n}=b_{1}+b_{2}+\cdots+b_{n}$, find $\frac{S_{9}}{T_{6}}$.
(1 mark
已知 $\left\{a_{n}\right\}$ 和 $\left\{b_{n}\right\}$ 都 是 等差 數 列 （即 $a_{2}-a_{1}=a_{3}-a_{2}=a_{4}-a_{3}=\cdots$ 、 $\left.b_{2}-b_{1}=b_{3}-b_{2}=b_{4}-b_{3}=\cdots\right)$ ，而且對所有正整數 $n$ 皆有 $(3 n+1) a_{n}=(2 n-1) b_{n}$ 。若 $S_{n}=a_{1}+a_{2}+\cdots+a_{n} \backslash T_{n}=b_{1}+b_{2}+\cdots+b_{n}$, 求 $\frac{S_{9}}{T_{6}}$ 。
7. $\frac{27}{23}$
7. Note that $S_{n}$ and $T_{n}$, both being the sum of the first $n$ terms of an arithmetic sequence, are both quadratic polynomials with constant term 0 . Hence we may assume $S_{n}=A n^{2}+B n$ and $T_{n}=C n^{2}+D n$. Consequently, $a_{n}=S_{n}-S_{n-1}=2 A n+(B-A)$. Similarly $b_{n}=2 C n+(D-C)$. Plugging these into the given equation, we have
$$
\frac{2 A n+(B-A)}{2 C n+(D-C)}=\frac{2 n-1}{3 n+1},
$$
which gives $6 A n^{2}+(3 B-A) n+(B-A)=4 C n^{2}+(2 D-4 C) n+(C-D)$. As this is true for all $n$, we equate the coefficients to solve for $A, B, C, D$. We get $B=0, C=\frac{3 A}{2}$ and $D=\frac{5 A}{2}$. Hence we have
$$
\frac{S_{9}}{T_{6}}=\frac{A(9)^{2}+(0)(9)}{\left(\frac{3 A}{2}\right)(6)^{2}+\left(\frac{5 A}{2}\right)(6)}=\frac{27}{23} .
$$

Remark. It is easy to guess the answer if one assumes $a_{n}=2 n-1$ and $b_{n}=3 n+1$.