题目 （12分）数列 $\{a_{n}\}$ 的前 $n$ 项和为 $S_{n}$，满足：$a_{1}=1$，$3tS_{n}-(2t+3)S_{n-1}=3t$，其中 $t>0,n\in\mathbb N^{*}$ 且 $n\geqslant 2$． 问题1 求证：数列 $\{a_{n}\}$ 是等比数列； 答案1 略 解析1 当 $n\geqslant 2$ 时，\[\begin{split}3tS_{n}-(2t+3)S_{n-1}=3t\cdots\cdots\text{ ① }\\ 3tS_{n+1}-(2t+3)S_{n}=3t.\cdots\cdots\text{ ② }\end{split}\]② $-$ ① 得：$3ta_{n+1}-(2t+3)a_{n}=0$，所以 $\dfrac{a_{n+1}}{a_{n}}=\dfrac{2t+3}{3t}(n\geqslant 2)$．又 $a_{1}=1$，$3t(a_{1}+a_{2})-(2t+3)a_{1}=3t$，解得：$a_{2}=\dfrac{2t+3}{3t}$，\[\dfrac{a_{2}}{a_{1}}=\dfrac{a_{3}}{a_{2}}=\cdots=\dfrac{a_{n+1}}{a_{n}}=\dfrac{2t+3}{2t}.\]所以 $\{a_{n}\}$ 是首项为 $1$，公比为 $\dfrac{2t+3}{3t}$ 的等比数列． 备注1 问题2 设数列 $\{a_{n}\}$ 的公比为 $f(t)$，数列 $\{b_{n}\}$ 满足 $b_{1}=1$，$b_{n}=f\left(\dfrac{1}{b_{n-1}}\right)(n\geqslant 2)$，求 $b_{n}$ 的通项公式； 答案2 $b_{n}=\dfrac{2}{3}n+\dfrac{1}{3}$ 解析2 $f(t)=\dfrac{2t+3}{3t}$ $b_{n}=\dfrac{\dfrac{2}{b_{n-1}}+3}{\dfrac{3}{b_{n-1}}}=\dfrac{3b_{n-1}+2}{3}=b_{n-1}+\dfrac{2}{3}$．所以 $b_{n}-b_{n-1}=\dfrac{2}{3}$，则\[b_{n}=1+(n-1)\cdot \dfrac{2}{3}=\dfrac{2}{3}n+\dfrac{1}{3}.\] 备注2 问题3 记 $T_{n}=b_{1}b_{2}-b_{2}b_{3}+b_{3}b_{4}-b_{4}b_{5}+\cdots+b_{2n-1}b_{2n}-b_{2n}b_{2n+1}$，求证：$T_{n}\leqslant -\dfrac{2}{9}$． 答案3 略 解析3 \[\begin{split}T_{n}&=b_{2}(b_{1}-b_{3})+b_{4}(b_{3}-b_{5})+\cdots+b_{2n}(b_{2n-1}-b_{2n+1})\\&=-\dfrac{4}{3}(b_{2}+b_{4}+\cdots+b_{2n})\\&=-\dfrac{4}{3}\cdot \dfrac{n\left(\dfrac{5}{3}+\dfrac{4n+1}{3}\right)}{2}\\&=-\dfrac{2n(4n+6)}{9}\\&=-\dfrac{4}{9}(2n^{2}+3n).\end{split}\]当 $n\geqslant 1$ 时，$2n^{2}+3n$ 递增，所以 $T_{n}\leqslant -\dfrac{4\times 5}{9}=-\dfrac{20}{9}$． 备注3