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2015年浙江高考数学数列压轴题解答

2015年浙江高考理科数学最后一题是数列题,如下:

题目:已知数列$\{a_n\}$满足$a_1=\frac{1}{2}$且$a_{n+1}=a_n-a_n^2(n\in \mathbf{N}^{+})$.

  • 证明:$1\leq \frac{a_n}{a_{n+1}}\leq 2(n\in \mathbf{N}^{+})$.
  • 设数列$\{a_{n}^{2}\}$的前$n$项和为$S_{n}$,证明:$\frac{1}{2(n+2)}\leq \frac{S_{n}}{n}\leq \frac{1}{2(n+1)}(n\in \mathbf{N}^{+})$.

我们先用数学归纳法证明,

引理:对于一切$n\geq 2$,$0 < a_n\leq \frac{1}{4}$.

证明:首先,$0 < a_2=\frac{1}{4}\leq\frac{1}{4}$.假设当$n=k(k\geq 2)$时,有$0 < a_k\leq \frac{1}{4}$. 则当$n=k+1$时,
$$
0 < a_{k+1}=a_k-a_k^2=\frac{1}{4}-\left(\frac{1}{2}-a_{k}\right)^2\leq \frac{1}{4}.
$$
由数学归纳法,对一切$n\geq 2$,$0 < a_n\leq\frac{1}{4}$.

下面我们来证明题目第一步.

证明:由引理,可得对一切$n\geq 1$,$0 < a_n\leq \frac{1}{2}$.
\begin{align*}
1\leq\frac{a_n}{a_{n+1}}\leq 2&\iff 1\leq\frac{a_n}{a_n-a_n^2}\leq 2
\\&\iff 1\leq \frac{1}{1-a_n}\leq 2
\\&\iff 0\leq a_n\leq \frac{1}{2}.
\end{align*}
而$0\leq a_n\leq \frac{1}{2}$是显然的.于是$1\leq\frac{a_n}{a_{n+1}}\leq 2$.

下面证明题目的第二步.

证明:$a_n^2=a_n-a_{n+1}$.因此
$$
S_n=a_1^2+a_2^2+\cdots+a_n^2=(a_1-a_2)+(a_2-a_3)+\cdots+(a_n-a_{n+1})=a_1-a_{n+1}=\frac{1}{2}-a_{n+1}.
$$
因此,
\begin{align*}
\frac{1}{2(n+2)}\leq \frac{S_n}{n}\leq \frac{1}{2(n+1)}&\iff
\frac{1}{2(n+2)}\leq
\frac{\frac{1}{2}-a_{n+1}}{n}\leq
\frac{1}{2(n+1)}
\\&\iff \frac{1}{2(n+1)}\leq a_{n+1}\leq \frac{1}{n+2}.
\end{align*}
可见,为了证明第二步的结论,我们只用证明$\frac{1}{2(n+1)}\leq a_{n+1}\leq \frac{1}{n+2}$即可.即证明对于一切$n\geq 2$,\begin{equation}\label{eq:1}\frac{1}{2n}\leq a_n\leq \frac{1}{n+1}.\end{equation}
为此仍然采用数学归纳法.当$n=2$时,$\frac{1}{4}\leq a_2=\frac{1}{4}\leq\frac{1}{3}$,此时不等式\eqref{eq:1}成立.假设当$n=k(k\geq 2)$时,
$$
\frac{1}{2k}\leq a_k\leq \frac{1}{k+1},
$$
则当$n=k+1$时,由归纳假设可得
\begin{equation}\label{eq:2}
\frac{1}{2k}-\frac{1}{4k^{2}}=\frac{1}{4}-\left(\frac{1}{2}-\frac{1}{2k}\right)^{2}\leq
a_{k+1}=\frac{1}{4}-\left(\frac{1}{2}-a_{k}\right)^2\leq
\frac{1}{4}-\left(\frac{1}{2}-\frac{1}{k+1}\right)^2=\frac{1}{k+1}-\left(\frac{1}{k+1}\right)^2
\end{equation}
下面证明
$$
\frac{1}{k+1}-\left(\frac{1}{k+1}\right)^2\leq \frac{1}{k+2}.
$$
\begin{align*}
\frac{1}{k+1}-\left(\frac{1}{k+1}\right)^2\leq \frac{1}{k+2}&\iff
(k+1)(k+2)-(k+2)\leq (k+1)^2
\\&\iff k^2+3k+2-k-2\leq k^2+2k+1
\\&\iff 0\leq 1.
\end{align*}
再证明
$$
\frac{1}{2k}-\frac{1}{4k^2}\geq \frac{1}{2(k+1)}.
$$
\begin{align*}
\frac{1}{2k}-\frac{1}{4k^2}\geq \frac{1}{2(k+1)}&\iff
2k(k+1)-(k+1)\geq
2k^2
\\&\iff k\geq 1.
\end{align*}
这两个结果与不等式\eqref{eq:2}结合起来,可得
$$
\frac{1}{2(k+1)}\leq a_{k+1}\leq \frac{1}{k+2}.
$$
综上,由数学归纳法,对于一切$n\geq 2$,总有不等式\eqref{eq:1}成立.这样就证明了第二步的结论.