s=pi ?
正整数中:使得十进制表示的非零子序列不能被 3 整除。1, 2, 4, 5, 7, 8, 10, 11, 14, 17, 20, 22, 25, 28, 40, 41, 44, 47, 50, 52, 55, 58, 70, 71,
74, 77, 80, 82, 85, 88, 100, 101, 104, 107, 110, 140, 170, 200, 202, 205, 208, 220,
250, 280, 400, 401, 404, 407, 410, 440, 470, 500, 502, 505, 508, 520, 550, 580,....
s=1+1/2+1/4+1/5+1/7+1/8+1/10+1/11+1/14+1/17+1/20+1/22+1/25+........
求证: \(s=\pi\) 谢谢 northwolves！
a=With[{k=3},Select,0],Mod[#,k]==0&]&]];s=Total@(1/a);{Length@a,a,s,N}
{123, {1, 2, 4, 5, 7, 8, 10, 11, 14, 17, 20, 22, 25, 28, 40, 41, 44, 47, 50, 52, 55, 58, 70, 71, 74, 77, 80, 82, 85, 88, 100, 101, 104, 107, 110, 140, 170, 200, 202, 205, 208, 220, 250, 280, 400, 401, 404, 407, 410, 440,
470, 500, 502, 505, 508, 520, 550, 580, 700, 701, 704, 707, 710, 740, 770, 800, 802, 805, 808, 820, 850, 880, 1000, 1001, 1004, 1007, 1010, 1040, 1070, 1100, 1400, 1700, 2000, 2002, 2005, 2008, 2020, 2050, 2080,
2200, 2500, 2800, 4000, 4001, 4004, 4007, 4010, 4040, 4070, 4100, 4400, 4700, 5000, 5002, 5005, 5008, 5020, 5050, 5080, 5200, 5500, 5800, 7000, 7001, 7004, 7007, 7010, 7040, 7070, 7100, 7400, 7700, 8000}, 2175919945339027979739835924147121496956377212405817719/692584026612285902036909113880759445107872284941840000, 3.141741452} s=(1 + 1/2 + 1/4 + 1/5 + 1/7 + 1/8)
+(1/10 + 1/11 + 1/14 + 1/17 + 1/20 +1/22 +1/25 +1/28 +1/40 +1/41 +1/44 +1/47+ 1/50 +1/52 +1/55 +1/58 +1/70 +1/71 +1/74 +1/77 +1/80 +1/82 +1/85 +1/88)
+(1/100 +1/101 +1/104 +1/107 +1/110 +1/140 +1/170 +1/200 +1/202+1/205 + 1/208 +1/220 +1/250 +1/280 +1/400 +1/401 +1/404 +1/407 +1/410+1/440+1/470
+ 1/500 +1/502 +1/505 +1/508 +1/520 +1/550 +1/580 +1/700 +1/701+1/704 + 1/707 +1/710 +1/740 +1/770 +1/800 +1/802 +1/805 +1/808 +1/820+1/850+1/880)
+......
<(1 + 1/2 + 1/4 + 1/5 + 1/7 + 1/8)*1/1
+(1 + 1/2 + 1/4 + 1/5 + 1/7 + 1/8)*4/10
+(1 + 1/2 + 1/4 + 1/5 + 1/7 + 1/8)*7/100
+(1 + 1/2 + 1/4 + 1/5 + 1/7 + 1/8)*10/1000
+ ......
=(1 + 1/2 + 1/4 + 1/5 + 1/7 + 1/8)*(1/1 + 4/10 + 7/100 + 10/1000 + 13/10000 + 16/100000 + 19/1000000 + 22/10000000 + ......)
=621/280*40/27=23/7
即:s<23/7=22/7=pi a=With[{k=2},Select,0],Mod[#,k]==0&]&]];s=Total@(1/a);{Length@a,a,s,N}
{82030, {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 31, 33, 35, 37, 39, 51, 53, 55, 57, 59, 71, 73, 75, 77, 79, 91, 93, 95, 97, 99, 111, 113, 115, 117, 119, 131, 133, 135, 137, 139, 151, 153, 155, 157, 159, 171,
173, 175, 177, 179, 191, 193, 195, 197, 199, 311, 313, 315, 317, 319, 331, 333, 335, 337, 339, 351, 353, 355, 357, 359, 371, 373, 375, 377, 379,{"81880"},7999511, 7999513, 7999515, 7999517,
7999519, 7999531, 7999533, 7999535, 7999537, 7999539, 7999551, 7999553, 7999555, 7999557, 7999559, 7999571, 7999573, 7999575, 7999577, 7999579, 7999591, 7999593, 7999595, 7999597,
7999599, 7999711, 7999713, 7999715, 7999717, 7999719, 7999731, 7999733, 7999735, 7999737, 7999739, 7999751, 7999753, 7999755, 7999757, 7999759, 7999771, 7999773, 7999775, 7999777,
7999779, 7999791, 7999793, 7999795, 7999797, 7999799, 7999911, 7999913, 7999915, 7999917, 7999919, 7999931, 7999933, 7999935, 7999937, 7999939, 7999951, 7999953, 7999955, 7999957,
7999959, 7999971, 7999973, 7999975, 7999977, 7999979, 7999991, 7999993, 7999995, 7999997, 7999999},3.14877765722813074857426878424994826998},
求证:
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{20}+\frac{1}{30}+\frac{1}{40}+\frac{1}{50}+\frac{1}{60}+\frac{1}{70}+\frac{1}{80}+\frac{1}{90}+\frac{1}{100}+\frac{1}{200}+\frac{1}{300}+\frac{1}{400}+\frac{1}{500}+\frac{1}{600}+\frac{1}{700}+\frac{1}{800}+\frac{1}{900}+\frac{1}{1000}+\cdots+\)\(\frac{1}{10^{\lfloor n/9\rfloor}*(1+n-9*\lfloor n/9\rfloor)}=\frac{7129}{2268}\) 王守恩 发表于 2024-4-6 17:38
求证:
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac ...
$\sum_{k=1}^9\frac1k *\sum_{i=0}^{\infty}\frac{1}{10^i}=\frac{7129}{2520}*\frac{10}{9}=\frac{7129}{2268}$ 我就好奇:改一下, 就出不来了？
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{19}+\frac{1}{29}+\frac{1}{39}+\frac{1}{49}+\frac{1}{59}+\frac{1}{69}+\frac{1}{79}+\frac{1}{89}+\frac{1}{99}+\frac{1}{199}+\frac{1}{299}+\frac{1}{399}+\frac{1}{499}+\frac{1}{599}+\frac{1}{699}+\frac{1}{799}+\frac{1}{899}+\frac{1}{999}+\frac{1}{1999}+\cdots+\)\(\frac{1}{10^{\lfloor n/9\rfloor}*(1+n-9*\lfloor n/9\rfloor)-1}=?\) 加深印象。这个还是可以有的。
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{20}-\frac{1}{30}+\frac{1}{40}-\frac{1}{50}+\frac{1}{60}-\frac{1}{70}+\frac{1}{80}-\frac{1}{90}+\frac{1}{100}-\frac{1}{200}+\frac{1}{300}-\frac{1}{400}+\frac{1}{500}-\frac{1}{600}+\frac{1}{700}-\frac{1}{800}+\frac{1}{900}-\frac{1}{1000}+\cdots-\)\(\frac{1}{10^{\lfloor n/9\rfloor}*(1+n-9*\lfloor n/9\rfloor)\cos(n\pi)}=\frac{1879}{2772}\)
王守恩 发表于 2024-4-7 08:19
我就好奇:改一下, 就出不来了？
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\fr ...
N, {k, 0, Infinity}], 20]
3.0490575908610721956 嗨！A037124——用你这通项还不就简单些？
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 200, 300, 400, 500, 600, 700, 800, 900, 1000, 2000, 3000, 4000, 5000, 6000, 7000, 8000, 9000, 10000, 20000, 30000, 40000, 50000, 60000, 70000, 80000, 90000, 100000
Table(好像缺点什么)
页:
[1]
2