设数列 $\left\{ {{a_n}} \right\}$ 满足 ${a_1} = a$，${a_2} = b$，$2{a_{n + 2}} = {a_{n + 1}} + {a_n}$．
【难度】
【出处】
2011年清华大学夏令营试题
【标注】
-
设 ${b_n} = {a_{n + 1}} - {a_n}$，证明：若 $a \ne b $，则 $\left\{ {{b_n}} \right\}$ 是等比数列；标注答案略解析因为 $2{a_{n + 2}} = {a_{n + 1}} + {a_n}$，$2\left( {{a_{n + 2}} - {a_{n + 1}}} \right) = - \left( {{a_{n + 1}} - {a_n}} \right)$，所以 $2{b_{n + 1}} = - {b_n}$．
又因为 $a \ne b$，所以 ${b_1} \ne 0$，于是 ${b_n} \ne 0$，
因此 $\left\{ {{b_n}} \right\}$ 是公比为 $ - \dfrac{1}{2}$ 的等比数列． -
若 $\lim\limits_{n \to \infty } ({a_1} + {a_2} +\cdots + {a_n}) = 4$，求 $a,b$ 的值．标注答案$a = 6$，$b = - 3$解析因为 ${b_1} = b - a$，所以 ${b_n} = \left( {b - a} \right){\left( { - \dfrac{1}{2}} \right)^{n - 1}}$，
于是\[\begin{split}{a_n}& = {a_1} + \sum_{k=1}^n {{b_k}} \\&= a + \dfrac{{b - a}}{{1 - \left( { - \dfrac{1}{2}} \right)}} \cdot \left[ {1 - {{\left( { - \dfrac{1}{2}} \right)}^{n - 1}}} \right] \\& = \dfrac{{2b + a}}{3} - \dfrac{2}{3}\left( {b - a} \right){\left( { - \dfrac{1}{2}} \right)^{n - 1}}.\end{split}\]所以 $\left\{ {{a_n}} \right\}$ 的前 $n$ 项和\[\begin{split}{S_n}&= \dfrac{{2b + a}}{3} \cdot n - \dfrac{2}{3}\left( {b - a} \right) \cdot \dfrac{1}{{1 - \left( { - \dfrac{1}{2}} \right)}}\left[ {1 - {{\left( { - \dfrac{1}{2}} \right)}^n}} \right]\\& = \dfrac{{2b + a}}{3} \cdot n - \dfrac{4}{9}\left( {b - a} \right)\left[ {1 - {{\left( { - \dfrac{1}{2}} \right)}^n}} \right].\end{split}\]而 $\mathop {\lim }\limits_{n \to \infty } \left( {{a_1} + {a_2} + \cdots + {a_n}} \right) = 4$，所以$$\begin{cases}2b + a = 0 ,\\
- \dfrac{4}{9}\left( {b - a} \right) = 4 ,
\end{cases}$$解得 $a = 6$，$b = - 3$．
题目
问题1
答案1
解析1
备注1
问题2
答案2
解析2
备注2
