[数列] 一数列题

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题目：已知数{$a_n$},$a_1=a_2=1,且 a_{n+2} = a_{n+1} + a_n$, $S_n$为{$ \frac  {a_n} {2^n}$}的前n项和，

求证：  $S_n < 2$


证明：$\frac {a_{n+2}}{2^{n+2}}$  =  $\frac 12     \frac {a_{n+1}}{2^{n+1}} + \frac 14     \frac {a_n}{2^n}$

$S_{n+2} - \frac {a_1}{2} - \frac {a_2}{4}$ = $\frac 12 ( S_{n+1} - \frac {a_1}{2}) + \frac 14 S_{n}$

$ S_{n+2} = \frac {1}{2}  S_{n+1} +\frac  {1}{4} S_n +\frac {1}{2} $

$ S _1= \frac {1}{2}$<2 ,    $ S _2=\frac {3}{4} < 2 $

假设：$ S_n<2, S_{n+1}<2 $

那么：$ S_{n+2} = \frac {1}{2}  S_{n+1} +\frac  {1}{4} S_n +\frac {1}{2} $ < $\frac 12  2  + \frac  14  2 +\frac  12  =  2$

由数学归纳法原理可知，$S_n <  2$

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