[数列] 昨天早上粉丝群这道数列题怎么玩[6#反证应该没问题]

唯因却恩(7643*****)  8:28:35

(14.15 KB)

2013-2-27 08:47

第二问帮看看

$a_1=1$, $a_{2k}=-a_k$, $a_{2k-1}=(-1)^{k+1}a_k$，证 $S_n\geqslant0$。

昨天看了一下，不会，后来研究别的东西去了，刚才再看，还是不会
感觉有点意思，先转发上来先。

用程序列了一下 an 和 Sn
a[1] = 1;
Do[Which[EvenQ[n], a[n] = -a[n/2], OddQ[n],
  a[n] = (-1)^((n + 3)/2) a[(n + 1)/2]], {n, 2, 1000}]
S[n_] := Sum[a[k], {k, 1, n}]

Table[a[n], {n, 1, 100}]
{1, -1, 1, 1, 1, -1, -1, -1, 1, -1, 1, 1, -1, 1, 1, 1, 1,
-1, 1, 1, 1, -1, -1, -1, -1, 1, -1, -1, 1, -1, -1, -1, 1, -1, 1, 1,
1, -1, -1, -1, 1, -1, 1, 1, -1, 1, 1, 1, -1, 1, -1, -1, -1, 1, 1, 1,
1, -1, 1, 1, -1, 1, 1, 1, 1, -1, 1, 1, 1, -1, -1, -1, 1, -1, 1, 1,
-1, 1, 1, 1, 1, -1, 1, 1, 1, -1, -1, -1, -1, 1, -1, -1, 1, -1, -1,
-1, -1, 1, -1, -1}

Table[S[n], {n, 1, 600}]
{1, 0, 1, 2, 3, 2, 1, 0, 1, 0, 1, 2, 1, 2, 3, 4, 5, 4, 5, 6,
7, 6, 5, 4, 3, 4, 3, 2, 3, 2, 1, 0, 1, 0, 1, 2, 3, 2, 1, 0, 1, 0, 1,
2, 1, 2, 3, 4, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 5, 6, 5, 6, 7, 8, 9, 8,
9, 10, 11, 10, 9, 8, 9, 8, 9, 10, 9, 10, 11, 12, 13, 12, 13, 14, 15,
14, 13, 12, 11, 12, 11, 10, 11, 10, 9, 8, 7, 8, 7, 6, 5, 6, 7, 8, 7,
8, 7, 6, 7, 6, 5, 4, 5, 4, 5, 6, 7, 6, 5, 4, 3, 4, 3, 2, 3, 2, 1, 0,
1, 0, 1, 2, 3, 2, 1, 0, 1, 0, 1, 2, 1, 2, 3, 4, 5, 4, 5, 6, 7, 6, 5,
4, 3, 4, 3, 2, 3, 2, 1, 0, 1, 0, 1, 2, 3, 2, 1, 0, 1, 0, 1, 2, 1, 2,
3, 4, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 5, 6, 5, 6, 7, 8, 7, 8, 7, 6, 5,
6, 7, 8, 7, 8, 7, 6, 7, 6, 5, 4, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 5, 6,
5, 6, 7, 8, 9, 8, 9, 10, 11, 10, 9, 8, 9, 8, 9, 10, 9, 10, 11, 12,
11, 12, 11, 10, 9, 10, 11, 12, 13, 12, 13, 14, 13, 14, 15, 16, 17,
16, 17, 18, 19, 18, 17, 16, 17, 16, 17, 18, 17, 18, 19, 20, 21, 20,
21, 22, 23, 22, 21, 20, 19, 20, 19, 18, 19, 18, 17, 16, 17, 16, 17,
18, 19, 18, 17, 16, 17, 16, 17, 18, 17, 18, 19, 20, 19, 20, 19, 18,
17, 18, 19, 20, 21, 20, 21, 22, 21, 22, 23, 24, 25, 24, 25, 26, 27,
26, 25, 24, 25, 24, 25, 26, 25, 26, 27, 28, 29, 28, 29, 30, 31, 30,
29, 28, 27, 28, 27, 26, 27, 26, 25, 24, 23, 24, 23, 22, 21, 22, 23,
24, 23, 24, 23, 22, 23, 22, 21, 20, 21, 20, 21, 22, 23, 22, 21, 20,
19, 20, 19, 18, 19, 18, 17, 16, 15, 16, 15, 14, 13, 14, 15, 16, 15,
16, 15, 14, 15, 14, 13, 12, 11, 12, 11, 10, 9, 10, 11, 12, 13, 12,
13, 14, 13, 14, 15, 16, 15, 16, 15, 14, 13, 14, 15, 16, 15, 16, 15,
14, 15, 14, 13, 12, 13, 12, 13, 14, 15, 14, 13, 12, 11, 12, 11, 10,
11, 10, 9, 8, 9, 8, 9, 10, 11, 10, 9, 8, 9, 8, 9, 10, 9, 10, 11, 12,
13, 12, 13, 14, 15, 14, 13, 12, 11, 12, 11, 10, 11, 10, 9, 8, 7, 8,
7, 6, 5, 6, 7, 8, 7, 8, 7, 6, 7, 6, 5, 4, 5, 4, 5, 6, 7, 6, 5, 4, 3,
4, 3, 2, 3, 2, 1, 0, 1, 0, 1, 2, 3, 2, 1, 0, 1, 0, 1, 2, 1, 2, 3, 4,
5, 4, 5, 6, 7, 6, 5, 4, 3, 4, 3, 2, 3, 2, 1, 0, 1, 0, 1, 2, 3, 2, 1,
0, 1, 0, 1, 2, 1, 2, 3, 4, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 5, 6, 5, 6,
7, 8, 9, 8, 9, 10, 11, 10, 9, 8, 9, 8, 9, 10, 9, 10, 11, 12, 13, 12,
13, 14, 15, 14, 13, 12}

看上去很复杂，我感觉是不是用二进制来解释的？

$a_n$的通项数字全是$1$，$-1$的，realnumber的观察归纳能力强，realnumber出马试试归纳看？

搞不定

问题:$a_1=1$, $a_{2k}=-a_k$, $a_{2k-1}=(-1)^{k+1}a_k$，证 $S_n\geqslant0$。
解:$a_{4k}=-a_{2k}=a_k,a_{4k-2}=-a_{2k-1},a_{4k-3}=a_{2k-1},a_{4k-1}=-a_{2k}=a_k$,得到$S_{4k}=2S_k$.
易得$S_1,S_2,S_3,S_4\ge0$.
假设使得$S_n<0$的最小$n=n_0$(容易得出$S_{n_0}=-1,S_{n_0-1}=0,S_{n_0-2}=1,a_{n_0}=-1,a_{n_0-1}=-1$),
1.若$n_0=4k$,由$S_{4k}=2S_k<0$,与$4k$最小矛盾.
2.若$n_0=4k+2$,$S_{4k+2}=S_{4k}+a_{4k+1}+a_{4k+2}=S_{4k}<0$,这与$4k+2$最小矛盾.
3.若$n_0=4k+3$,$S_{4k+3}=S_{4k}+a_{4k+1}+a_{4k+2}+a_{4k+3}=S_{4k}+a_{4k+3}=2S_{k}+a_{k+1}=S_{k+1}<0$,这与$4k+3$最小矛盾.(说明$a_{4k+1}+a_{4k+2}=-a_{2k-1}+a_{2k-1}=0,a_{4k+3}=a_{k+1}=-1,S_{4k}=0=S_k$.)
4.若$n_0=4k+1$,则$a_{4k+1}=-1,S_{4k+1}=-1,S_{4k}=0=S_{k}$,如此,推测出$k$必定为偶数(因为{$a_n$}由$1,-1$组成,若$k$为奇数,则$S_{k}$不为零.ps,也可以用这个办法来处理1.2.两条.),$a_{4k+1}=-1=a_{2k+1}=(-1)^ka_{k+1}=a_{k+1}$,那么$S_{k+1}=S_{k}+a_{k+1}=-1<0$,与$4k+1$最小矛盾.
---又,其实k发的时候就看到,没想到怎么入手.

4# yes94
终于解决,请检查~~.

初步看了下，没看出问题来。明天再仔细看，先闪了

7# realnumber
虽然看起来吓人，但要赞一个先，这种专研精神值得学习！

9# yes94

你看有没有问题？