二项式系数级数连带奇数倒数平方和
The Binomial Coefficient Series Is Associated with a Sum of Odd Reciprocal Squares

作者: 薛坐远 , 及万会 :银川能源学院电力学院,宁夏 银川; 薛 雪 :滨河科技学院,宁夏 银川;

关键词: 二项式系数奇数倒数平方和微分裂项级数Binomial Coefficients Sum of Odd Reciprocal Squares Differential Split Terms Series

摘要:
根据一个已知级数,使用裂项方法得到分母含奇偶性不定因子 1个,2个,3个,4个,5个线性因子的二项式系数级数连带奇数倒数平方和。利用反正弦与反双曲正弦关系给出交错二项式系数级数连带奇数倒数平方和。所给出级数的和式是封闭形的。并给出二项式系数级数连带奇数倒数平方和数值恒等式。

Abstract: Using one known series, we can structure several new binomial coefficient series which is associated with a sum of odd reciprocal squares. Their denominator has parity indefinite linear factors 1, 2, 3, 4, 5. Using relation of inverse Trigonometric and Hyperbolic function, we get that alternating the binomial coefficient series is associated with a sum of odd reciprocal squares. The numerical identities of binomial coefficient series with odd reciprocal square are given.

1. 引言及引理

二项式系数在数论,图论,统计和概率等数学分支扮演重要角色。二项式系数变换研究有大量文献

[1] - [10] 。文献 [2] [3] [4] [5] 中都提到被称为Lehmer级数恒等式 n = 1 ( 2 x ) 2 n n ( 2 n n ) = 2 x arcsin x 1 x 2 | x | < 1 ,一些作者

使用微分,积分,发生函数,白塔–伽马函数,递推等数学工具得到二项式系数倒数级数的重要结果。文 [3] 给出二项式系数倒数数值级数用积分表示,并给出递推公式。文 [4] 给出二项式系数倒数数值级数用反三角函数表示的公式;文 [5] 给出二项式系数倒数数值级数用积分表示。文 [5] [6] [7] [8] 利已知级数裂项构造出一批新二项式系数的倒数级数。文 [9] 利用白塔函数建立非中心型二项式倒数级数。文 [10] 利用“变换核”函数导出无穷级数恒等式。我们利用一个已知级数,用微分裂项法,将分式的化成部分分式经过一定程序转化分母含奇偶性不定的1个,2个,3个,4个,5个线性因子的二项式系数级数连带奇数倒数平方和。利用反正弦与反双曲正弦关系给出交错二项式系数级数连带奇数倒数平方和。所给出级数是封闭形的。并给出二项式系数级数连带奇数倒数平方和数值恒等式。

引理1 [11] k = 1 ( 2 k k ) 2 2 k ( j = 1 k 1 ( 2 j 1 ) 2 ) x 2 k + 1 2 k + 1 = 1 6 ( arcsin 3 x )

2. 主要结果和证明

2.1. 定理

B = arcsin 2 x 2 1 x 2 ,则二项式系数级数连带续奇数倒数平方和

1) 分母含有1个线性因子的二项式系数级数连带奇数倒数平方和

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( j = 1 m + 1 1 ( 2 j 1 ) 2 ) x 2 m = ( 2 x 2 + 2 ) B + 1 (1)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 2 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) x 2 m = ( 4 3 x 4 + 2 3 x 2 + 2 3 ) B + 2 3 x 2 + 2 9 (2)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 3 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) x 2 m = ( 16 15 x 6 + 8 15 x 4 + 2 15 x 2 + 2 5 ) B + 8 15 x 4 + 8 45 x 2 + 64 675 (3)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 32 35 x 8 + 16 35 x 6 + 4 35 x 4 + 2 35 x 2 + 2 7 ) B + 16 35 x 6 + 16 105 x 4 + 128 1575 x 2 + 64 1225 (4)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 5 ) ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m = ( 256 315 x 10 + 128 315 x 8 + 32 315 x 6 + 16 315 x 4 + 2 63 x 2 + 2 9 ) B + 128 315 x 10 + 128 945 x 6 + 1024 14175 x 4 + 512 11025 x 2 + 16384 496125 (5)

2) 分母含有2个线性因子的二项式系数级数连带奇数倒数平方和

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) x 2 m = ( 4 3 x 4 2 3 x 2 2 3 ) B 2 3 x 2 + 7 9 (6)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 3 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) x 2 m = ( 8 15 x 6 4 15 x 4 16 15 x 2 + 4 5 ) B 4 15 x 4 4 45 x 2 + 611 1350 (7)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 2 ) ( m + 3 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) x 2 m = ( 16 15 x 6 28 15 x 4 + 8 15 x 2 + 4 15 ) B 8 15 x 4 + 22 45 x 2 + 86 675 (8)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 32 105 x 8 16 105 x 6 4 105 x 4 24 35 x 2 + 4 7 ) B 16 105 x 6 16 315 x 4 128 4725 x 2 + 387 1225 (9)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 2 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 16 35 x 8 8 35 x 6 76 105 x 4 + 32 105 x 2 + 4 21 ) B 8 35 x 6 8 105 x 4 + 461 1575 x 2 + 937 11025 (10)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 3 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 32 35 x 8 32 21 x 6 + 44 105 x 4 + 8 105 x 2 + 4 35 ) B 16 35 x 6 + 8 21 x 4 + 152 1575 x 2 + 1408 33075 (11)

3) 分母含有3个线性因子的二项式系数级数连带奇数倒数平方和

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) x 2 m = ( 8 15 x 6 + 8 5 x 4 8 5 x 2 + 8 15 ) B + 4 15 x 4 26 45 x 2 + 439 1350 (12)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 16 105 x 8 + 8 105 x 6 + 24 35 x 4 104 105 x 2 + 8 21 ) B + 8 105 x 6 + 568 105 x 4 1511 4725 x 2 + 2546 11025 (13)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 3 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 32 105 x 8 + 24 35 x 6 8 35 x 4 8 21 x 2 + 8 35 ) B + 16 105 x 6 68 315 x 4 292 4725 x 2 + 9041 66150 (14)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 2 ) ( m + 3 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 16 35 x 8 + 136 105 x 6 8 7 x 4 + 8 35 x 2 + 8 105 ) B + 8 35 x 6 16 35 x 4 + 103 525 x 2 + 1403 33075 (15)

4) 分母含有4个线性因子的二项式系数级数连带奇数倒数平方和

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 16 105 x 8 64 105 x 6 + 32 35 x 4 64 105 x 2 + 16 105 ) B 8 105 x 6 + 76 315 x 4 1219 4725 x 2 + 1247 13230 (16)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m = ( 64 945 x 10 176 945 x 8 + 64 945 x 6 + 32 135 x 4 256 945 x 2 + 16 189 ) B 32 945 x 8 + 184 2835 x 6 + 828 42525 x 4 671 6615 x 2 + 311081 5953500 (17)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 3 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m = ( 32 315 x 10 16 45 x 8 + 128 315 x 6 32 315 x 4 32 315 x 2 + 16 315 ) B 16 315 x 8 + 128 945 x 6 1298 14175 x 4 766 33075 x 2 + 124031 3969000 (18)

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 2 ) ( m + 3 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m = ( 128 945 x 10 496 945 x 8 + 704 945 x 6 416 945 x 4 + 64 945 x 2 + 16 945 ) B 64 945 x 8 + 584 2835 x 6 8612 42525 x 4 + 1823 33075 x 2 + 2179 212625 (19)

5) 分分母含有5个线性因子的二项式系数级数连带奇数倒数平方和

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m = ( 32 945 x 10 + 32 189 x 8 64 189 x 6 + 64 189 x 4 32 189 x 2 + 32 945 ) B + 16 945 x 8 40 567 x 6 + 674 6075 x 4 863 11025 x 2 + 250069 11907000 (20)

推论1 设 b = ln 2 ( x + 1 + x 2 ) 2 1 + x 2 ,则交错二项式系数连带续奇数倒数平方和级数

1) 分母含有1个因子交错的二项式系数级数连带奇数倒数平方和

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( j = 1 m + 1 1 ( 2 j 1 ) 2 ) x 2 m = ( 2 x 2 + 2 ) b + 1 (21)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 2 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) x 2 m = ( 4 3 x 4 2 3 x 2 + 2 3 ) b 2 3 x 2 + 2 9 (22)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 3 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) x 2 m = ( 16 15 x 6 + 8 15 x 4 2 15 x 2 + 2 5 ) b + 8 15 x 4 8 45 x 2 + 64 675 (23)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 32 35 x 8 16 35 x 6 + 4 35 x 4 2 35 x 2 + 2 7 ) b 16 35 x 6 + 16 105 x 4 128 1575 x 2 + 64 1225 (24)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 5 ) ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m = ( 256 315 x 10 + 128 315 x 8 32 315 x 6 + 16 315 x 4 2 63 x 2 + 2 9 ) b + 128 315 x 8 128 945 x 6 + 1024 14175 x 4 512 11025 x 2 + 16384 496125 (25)

2) 母含有2个因子交错的二项式系数级数连带奇数倒数平方和

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) x 2 m = ( 4 3 x 4 + 2 3 x 2 2 3 ) b + 2 3 x 2 + 7 9 (26)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 3 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) x 2 m = ( 8 15 x 6 4 15 x 4 + 16 15 x 2 + 4 5 ) b 4 15 x 4 + 4 45 x 2 + 611 1350 (27)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 2 ) ( m + 3 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) x 2 m = ( 16 15 x 6 28 15 x 4 8 15 x 2 + 4 15 ) b 8 15 x 4 22 45 x 2 + 86 675 (28)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 32 105 x 8 + 16 105 x 6 4 105 x 4 + 24 35 x 2 + 4 7 ) b + 16 105 x 6 16 315 x 4 + 128 4725 x 2 + 387 1225 (29)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 2 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 16 35 x 8 + 8 35 x 6 76 105 x 4 32 105 x 2 + 4 21 ) b + 8 35 x 6 8 105 x 4 461 1575 x 2 + 937 11025 (30)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 3 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 32 35 x 8 + 32 21 x 6 + 44 105 x 4 8 105 x 2 + 4 35 ) b + 16 35 x 6 + 8 21 x 4 152 1575 x 2 + 1408 33075 (31)

3) 母含有3个因子交错的二项式系数级数连带奇数倒数平方和

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) x 2 m = ( 8 15 x 6 + 8 5 x 4 + 8 5 x 2 + 8 15 ) b + 4 15 x 4 + 26 45 x 2 + 439 1350 (32)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 16 105 x 8 8 105 x 6 + 24 35 x 4 + 104 105 x 2 + 8 21 ) b 8 105 x 6 + 568 105 x 4 + 1511 4725 x 2 + 2546 11025 (33)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 3 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 32 105 x 8 24 35 x 6 8 35 x 4 + 8 21 x 2 + 8 35 ) b 16 105 x 6 68 315 x 4 + 292 4725 x 2 + 9041 66150 (34)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 2 ) ( m + 3 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 16 35 x 8 136 105 x 6 8 7 x 4 8 35 x 2 + 8 105 ) b 8 35 x 6 16 35 x 4 103 525 x 2 + 1403 33075 (35)

4) 分母含有4个因子交错的二项式系数级数连带奇数倒数平方和

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = ( 16 105 x 8 + 64 105 x 6 + 32 35 x 4 + 64 105 x 2 + 16 105 ) b + 8 105 x 6 + 76 315 x 4 + 1219 4725 x 2 + 1247 13230 (36)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m = ( 64 945 x 10 + 176 945 x 8 64 945 x 6 + 32 135 x 4 + 256 945 x 2 + 16 189 ) b 32 945 x 8 184 2835 x 6 + 828 42525 x 4 + 671 6615 x 2 + 311081 5953500 (37)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 3 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m = ( 32 315 x 10 16 45 x 8 128 315 x 6 32 315 x 4 + 32 315 x 2 + 16 315 ) b 16 315 x 8 128 945 x 6 1298 14175 x 4 + 756 33075 x 2 + 124031 3969000 (38)

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 2 ) ( m + 3 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m = ( 128 945 x 10 496 945 x 8 704 945 x 6 416 945 x 4 64 945 x 2 + 16 945 ) b 64 945 x 8 584 2835 x 6 8612 42525 x 4 1823 33075 x 2 + 2179 212625 (39)

5) 分母含有5个因子交错的二项式系数级数连带奇数倒数平方和

m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m = ( 32 945 x 10 + 32 189 x 8 + 64 189 x 6 + 64 189 x 4 + 32 189 x 2 + 32 945 ) b + 16 945 x 8 + 40 567 x 6 + 674 6075 x 4 + 863 11025 x 2 + 250069 11907000 (40)

2.2. 定理的证明

由文 [11] 公式 k = 1 ( 2 k k ) 2 2 k ( j = 1 k 1 ( 2 j 1 ) 2 ) x 2 k + 1 2 k + 1 = 1 6 ( arcsin 3 x ) 两端关于x微分

k = 1 ( 2 k k ) x 2 k 2 2 k ( j = 1 k 1 ( 2 j 1 ) 2 ) = 1 2 arcsin 2 x 1 x 2 = B (41)

1) 对(41)式左端裂项

x 2 2 + k = 1 ( 2 k 2 ) ! ( 2 k 1 ) 2 k 2 2 k ( ( k 1 ) ! ) 2 ( k ) 2 ( j = 1 k 1 ( 2 j 1 ) 2 ) x 2 k = B k 1 = m ,化成

x 2 2 + m = 1 ( 2 m ) ! ( 2 m + 1 ) ( 2 m + 2 ) 2 2 m + 2 ( m ! ) 2 ( m + 1 ) 2 ( j = 1 m + 1 1 ( 2 j 1 ) 2 ) x 2 m + 2 = B ,两端同乘以 2 x 2 ,得出

1 + m = 1 ( 2 m ) ! ( 2 m + 1 ) 2 2 m ( m ! ) 2 ( m + 1 ) ( j = 1 m + 1 1 ( 2 j 1 ) 2 ) x 2 m = 2 x 2 B

1 + m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( 2 1 m + 1 ) ( j = 1 m + 1 1 ( 2 j 1 ) 2 ) x 2 m = 2 x 2 B ,整理得到如下(1)式,令其为 B 1

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( j = 1 m + 1 1 ( 2 j 1 ) 2 ) x 2 m = ( 2 x 2 + 2 ) B + 1

2) 对(41)式左端裂项

x 2 2 + 5 x 4 12 + k = 1 ( 2 k 4 ) ! ( 2 k 3 ) ( 2 k 2 ) ( 2 k 1 ) ( 2 k ) 2 2 k ( ( k 2 ) ! ) 2 ( k 1 ) 2 ( k ) 2 ( j = 1 k 1 ( 2 j 1 ) 2 ) x 2 k = B ,令 k 2 = m

x 2 2 + 5 x 4 12 + m = 1 ( 2 m ) ! ( 2 m + 1 ) ( 2 m + 2 ) ( 2 m + 3 ) ( 2 m + 4 ) 2 2 m + 4 ( m ! ) 2 ( m + 1 ) 2 ( m + 2 ) 2 ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) x 2 m + 4 = B

两端同乘以 4 x 4

2 x 2 + 5 3 + m = 1 ( 2 m ) ! ( 2 m + 1 ) ( 2 m + 3 ) 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) x 2 m = 4 x 4 B

2 x 2 + 5 3 + m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( 2 1 m + 1 ) ( 2 1 m + 2 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) x 2 m = 4 x 4 B

2 x 2 + 5 3 + m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( 4 2 m + 1 2 m + 2 + 1 ( m + 1 ) ( m + 2 ) ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) x 2 m = 4 x 4 B (42)

a) (42)式化成部分分式,得出

2 x 2 + 5 3 + m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( 4 1 m + 1 3 m + 2 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) x 2 m = 4 x 4 B

由于 B , B 1 已知,计算得出下面(2)式,并令其为 B 2

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 2 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) x 2 m = ( 4 3 x 4 + 2 3 x 2 + 2 3 ) B + 2 3 x 2 + 2 9

b) 在(42)式, B , B 1 , B 2 已知,易得2个因子乘积的二项式系数级数连带奇数倒数平方和公式(6)。

3) 对(41)式左端裂项

x 2 2 + 5 x 4 12 + 259 x 6 720 + k = 1 ( 2 k 6 ) ! ( 2 k 5 ) ( 2 k 1 ) ( 2 k ) 2 2 k ( ( k 3 ) ! ) 2 ( k 2 ) 2 ( k 1 ) 2 ( k ) 2 ( j = 1 k 1 ( 2 j 1 ) 2 ) x 2 k = B

k 2 = m

x 2 2 + 5 x 4 12 + 259 x 6 720 + k = 1 ( 2 m ) ! ( 2 m + 1 ) ( 2 m + 2 ) ( 2 m + 3 ) ( 2 m + 4 ) ( 2 m + 5 ) ( 2 m + 6 ) 2 2 k + 6 ( m ! ) 2 ( m + 1 ) 2 ( m + 2 ) 2 ( m + 3 ) 2 ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) x 2 m + 6 = B

两端同乘以 8 x 6

4 x 4 + 10 3 x 2 + 259 90 + k = 1 ( 2 m ) ! ( 2 m + 1 ) ( 2 m + 3 ) ( 2 m + 5 ) 2 2 k ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) x 2 m = 8 x 6 B

4 x 4 + 10 3 x 2 + 259 90 + m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( 2 1 m + 1 ) ( 2 1 m + 2 ) ( 2 1 m + 3 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) x 2 m = 8 x 6 B

4 x 4 + 10 3 x 2 + 259 90 + m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( 8 4 m + 1 4 m + 2 4 m + 3 + 2 ( m + 1 ) ( m + 2 ) + 2 ( m + 1 ) ( m + 3 ) + 2 ( m + 2 ) ( m + 3 ) 1 ( m + 1 ) ( m + 2 ) ( m + 3 ) ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) x 2 m = 8 x 6 B (43)

a) 将上式所有分式化成部分分式,得出

4 x 4 + 10 3 x 2 + 259 90 + m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( 8 3 / 2 m + 1 3 m + 2 15 / 2 m + 3 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) x 2 m = 8 x 6 B

由于 B , B 1 , B 2 已知,计算得出下面(3)式,并令其为 B 3

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 3 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) x 2 m = ( 16 15 x 6 + 8 15 x 4 + 2 15 x 2 + 2 5 ) B + 8 15 x 4 + 8 45 x 2 + 64 675

为行文简便,今后将 m = 1 2 m ! x 2 m 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 )

m = 1 2 m ! x 2 m 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) 等用符号 B 12 B 123 表示。

B 3 证明级数连带奇数倒数平方和上标增加1项或几项不改变其和式收敛性。

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 3 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) x 2 m = ( 16 15 x 6 + 8 15 x 4 + 2 15 x 2 + 2 5 ) B + 8 15 x 4 + 8 45 x 2 + 64 675

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 3 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) x 2 m = m = 1 ( 2 m ) ! x 2 m 2 2 m ( m ! ) 2 ( m + 3 ) ( j = 1 m 1 ( 2 j 1 ) 2 + 1 [ 2 ( m + 1 ) 1 ] 2 + 1 [ 2 ( m + 2 ) 1 ] 2 + 1 [ 2 ( m + 3 ) 1 ] 2 ) = m = 1 ( 2 m ) ! x 2 m 2 2 m ( m ! ) 2 ( m + 3 ) ( j = 1 m 1 ( 2 j 1 ) 2 ) + 2 m ! x 2 m 2 2 m ( m ! ) 2 ( m + 3 ) ( 1 [ 2 ( m + 1 ) 1 ] 2 + 1 [ 2 ( m + 2 ) 1 ] 2 + 1 [ 2 ( m + 3 ) 1 ] 2 )

前和式为二项式系数级数连带奇数倒数平方和表达式,后式极限趋于0。

根据阶乘斯特林渐进公式 n ! 2 n π n n e n ,注意到变量 | x | < 1

2 m ! ( m ! ) 2 4 π m ( 2 m ) 2 m e 2 m 2 m π m m e m m m e m ( 2 m ) 2 m m π ( m ) 2 m = 2 2 m m π

lim m 2 m ! x 2 m 2 2 m ( m ! ) ( m + 3 ) ( 1 [ 2 ( m + 1 ) 1 ] 2 + 1 [ 2 ( m + 2 ) 1 ] 2 + 1 [ 2 ( m + 3 ) 1 ] 2 ) = lim m x 2 m m π ( m + 3 ) ( 1 [ 2 ( m + 1 ) 1 ] 2 + 1 [ 2 ( m + 2 ) 1 ] 2 + 1 [ 2 ( m + 3 ) 1 ] 2 ) = 0

因此 m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 3 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) x 2 m = m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 3 ) ( j = 1 m 1 ( 2 j 1 ) 2 + 0 ) x 2 m = m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 3 ) ( j = 1 m 1 ( 2 j 1 ) 2 ) x 2 m

所以,级数连带奇数倒数平方和上标增加1项或几项不改变其和式收敛性。

b) 在(43)保留2个因子分式,(在(41)式中出现的2个因子分式不再计算),对这些2个因子的分式,每次保留1个,其余化成部分分式,得到:

4 x 4 + 10 3 x 2 + 259 90 + B 13 + 8 B 2 B 1 3 B 2 7 B 3 = 8 x 6 B

4 x 4 + 10 3 x 2 + 259 90 + B 23 + 8 B 4 B 1 3 B 2 13 2 B 3 = 8 x 6 B

由于 B , B 1 , B 2 , B 3 已知,计算得出公式(7)~(8)

c) 在(43)保留3个因子分式,其他分式化成部分分式得到:

4 x 4 + 10 3 x 2 + 259 90 + B 123 + 8 B 2 B 1 2 B 2 8 B 3 = 8 x 6 B

由于 B , B 1 , B 2 , B 3 已知,计算得出公式(12)式。

4) 对(41)式左端裂项

x 2 2 + 5 x 4 12 + 259 x 6 720 + 3229 x 8 10080 + k = 1 ( 2 k 8 ) ! ( 2 k 7 ) ( 2 k 1 ) ( 2 k ) 2 2 k ( ( k 4 ) ! ) 2 ( k 3 ) 2 ( k 2 ) 2 ( k 1 ) 2 ( k ) 2 ( j = 1 k 1 ( 2 j 1 ) 2 ) x 2 k = B k 4 = m

x 2 2 + 5 x 4 12 + 259 x 6 720 + 3229 x 8 10080 + m = 1 ( 2 m ) ! ( 2 m + 1 ) ( 2 m + 2 ) ( 2 m + 3 ) ( 2 m + 6 ) ( 2 m + 7 ) ( 2 m + 8 ) 2 2 m + 8 ( m ! ) 2 ( m + 1 ) 2 ( m + 2 ) 2 ( m + 3 ) 2 ( m + 4 ) 2 ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m + 8 = B

x 2 2 + 5 x 4 12 + 259 x 6 720 + 3229 x 8 10080 + m = 1 8 ( 2 m ) ! ( 2 m + 1 ) ( 2 m + 3 ) ( 2 m + 5 ) ( 2 m + 7 ) 2 2 m + 8 ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m + 8 = B ,两端同乘以 16 x 8

8 x 6 + 20 3 x 4 + 259 45 x 2 + 3229 630 + m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( 2 1 m + 1 ) ( 2 1 m + 2 ) ( 2 1 m + 3 ) ( 2 1 m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = 16 x 8 B

8 x 6 + 20 3 x 4 + 259 45 x 2 + 3229 630 + m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( 16 8 m + 1 8 m + 2 8 m + 3 8 m + 4 + 4 ( m + 1 ) ( m + 2 ) + 4 ( m + 1 ) ( m + 3 ) + 4 ( m + 2 ) ( m + 3 ) + 4 ( m + 1 ) ( m + 4 ) + 4 ( m + 2 ) ( m + 4 ) + 4 ( m + 3 ) ( m + 4 ) + 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) + 2 ( m + 1 ) ( m + 2 ) ( m + 4 ) + 2 ( m + 1 ) ( m + 3 ) ( m + 4 ) + 2 ( m + 2 ) ( m + 3 ) ( m + 4 ) + 1 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( m + 4 ) ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) x 2 m = 16 x 8 B (44)

a) 将上式所有分式化成部分分式,得出

8 x 6 + 20 3 x 4 + 259 45 x 2 + 3229 630 + m = 1 ( 2 m ) ! x 2 m 2 2 m ( m ! ) 2 ( 16 5 / 2 m + 1 9 / 2 m + 2 15 / 2 m + 3 35 / 2 m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) = 16 x 8 B

由于 B , B 1 , B 2 , B 3 已知,计算得出下面(4)式,并令其为 B 4

m = 1 ( 2 m ) ! x 2 m 2 2 m ( m ! ) 2 ( m + 4 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) = ( 32 35 x 8 + 16 35 x 6 + 4 35 x 4 + 2 35 x 2 + 2 7 ) B + 16 35 x 6 + 16 105 x 4 + 128 1575 x 2 + 64 1225

b) 在(44)式,其余保留2个因子分式,(在(43)式中出现的2个因子分式不再计算)然后对这些2个因子的分式,每次保留1个,其余化成部分分式,得到:

8 x 6 + 20 3 x 4 + 259 45 x 2 + 3229 630 + 16 B + B 14 17 5 B 1 9 2 B 2 15 2 B 3 103 6 B 4 = 16 x 8 B

8 x 6 + 20 3 x 4 + 259 45 x 2 + 3229 630 + 16 B + B 24 5 2 B 1 5 B 2 15 2 B 3 17 B 4 = 16 x 8 B

8 x 6 + 20 3 x 4 + 259 45 x 2 + 3229 630 + 16 B + B 34 5 2 B 1 9 2 B 2 17 2 B 3 33 2 B 4 = 16 x 8 B

由于 B , B 1 , B 2 , B 3 , B 4 已知,计算得出公式(9)~(11)式。

c) 在(44)保留3个因子分式(在(43)式出现的3个因子分式不再计算),然后对这些3个因子的分式,每次保留1个,其余化成部分分式,得到:

8 x 6 + 20 3 x 4 + 259 45 x 2 + 3229 630 + 16 B + B 124 17 6 B 1 4 B 2 15 2 B 3 53 3 B 4 = 16 x 8 B

8 x 6 + 20 3 x 4 + 259 45 x 2 + 3229 630 + 16 B + B 134 8 3 B 1 9 2 B 2 7 B 3 107 6 B 4 = 16 x 8 B

8 x 6 + 20 3 x 4 + 259 45 x 2 + 3229 630 + 16 B + B 234 5 2 B 1 5 B 2 13 2 B 3 18 B 4 = 16 x 8 B

由于 B , B 1 , B 2 , B 3 , B 4 , B 5 已知,计算得出(13)~(15)式。

d) 在(44)保留4个因子分式,然后对这些2个因子的分式,每次保留1个,其余化成部分分式,得到:

8 x 6 + 20 3 x 4 + 259 45 x 2 + 3229 630 + 16 B + B 1234 8 3 B 1 4 B 2 8 B 3 52 3 B 4 = 16 x 8 B

由于 B , B 1 , B 2 , B 3 , B 4 ,已知,计算得出公式(16)式。

5) 对(41)式左端裂项

x 2 2 + 5 x 4 12 + 259 x 6 720 + 3229 x 8 10080 + 117469 403200 x 10 + k = 1 ( 2 k 10 ) ! ( 2 k 9 ) ( 2 k 1 ) ( 2 k ) 2 2 k ( ( k 5 ) ! ) 2 ( k 4 ) 2 ( k 3 ) 2 ( k 2 ) 2 ( k 1 ) 2 ( k ) 2 ( j = 1 k 1 ( 2 j 1 ) 2 ) x 2 k = B k 5 = m

x 2 2 + 5 x 4 12 + 259 x 6 720 + 3229 x 8 10080 + 117469 403200 x 10 + m = 1 ( 2 m ) ! ( 2 m + 1 ) ( 2 m + 2 ) ( 2 m + 3 ) ( 2 m + 8 ) ( 2 m + 9 ) ( 2 m + 10 ) 2 2 m + 10 ( m ! ) 2 ( m + 1 ) 2 ( m + 2 ) 2 ( m + 3 ) 2 ( m + 4 ) 2 ( m + 5 ) 2 ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m + 10 = B

两端同乘以 32 x 10

16 x 8 + 40 3 x 6 + 518 45 x 4 + 3229 315 x 2 + 117469 12600 + m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( 2 1 m + 1 ) ( 2 1 m + 2 ) ( 2 1 m + 3 ) ( 2 1 m + 4 ) ( 2 1 m + 5 ) ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m = 32 x 10 B

展开乘积表达式

16 x 8 + 40 3 x 6 + 518 45 x 4 + 3229 315 x 2 + 117469 12600 + m = 1 ( 2 m ) ! x 2 m 2 2 m ( m ! ) 2 [ 32 16 m + 1 16 m + 2 16 m + 3 16 m + 4 16 m + 5 + 8 ( m + 1 ) ( m + 2 ) + 8 ( m + 1 ) ( m + 3 ) + 8 ( m + 2 ) ( m + 3 ) + 8 ( m + 1 ) ( m + 4 ) + 8 ( m + 2 ) ( m + 4 ) + 8 ( m + 3 ) ( m + 4 ) + 8 ( m + 1 ) ( m + 5 ) + 8 ( m + 2 ) ( m + 5 ) + 8 ( m + 3 ) ( m + 5 ) + 8 ( m + 4 ) ( m + 5 ) + 4 ( m + 1 ) ( m + 2 ) ( m + 3 ) + 4 ( m + 1 ) ( m + 2 ) ( m + 4 ) + 4 ( m + 1 ) ( m + 3 ) ( m + 4 ) + 4 ( m + 2 ) ( m + 3 ) ( m + 4 ) + 4 ( m + 1 ) ( m + 2 ) ( m + 5 ) + 4 ( m + 1 ) ( m + 3 ) ( m + 5 )

(45)

在(45)式有1个因子分式,2个因子分式4个,3个因子分式6个,4个因子分式4个,5个因子分式1个。为使论文简短写,我们仅选1个因子分式,4个因子分式,5个因子分式,通过如下操作程序:将分式化成分母为1个因子。4个因子(在(44)式出现的4个因子分式不再计算),5个因子乘积的二项式系数连带连续奇数倒数平方和级数

a) 将(45)式所有分式化成部分分式,得出:

16 x 8 + 40 3 x 6 + 518 45 x 4 + 3229 315 x 2 + 117469 12600 + m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( 32 35 / 8 m + 1 15 / 2 m + 2 45 / 4 m + 3 35 / 2 m + 4 315 / 8 m + 5 ) ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m = 32 x 10 B

128 315 x 8 + 320 945 x 6 + 4144 14175 x 4 + 25832 99225 x 2 + 117469 496125 + m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( 256 315 1 / 9 m + 1 4 / 21 m + 2 2 / 7 m + 3 4 / 9 m + 4 1 m + 5 ) ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m = 256 315 x 10 B

由于 B , B 1 , B 2 , B 3 , B 4 已知,计算得出下面(5)式,并令其为 B 5

m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 5 ) ( j = 1 m + 5 1 ( 2 j 1 ) 2 ) x 2 m = ( 256 315 x 10 + 128 315 x 8 + 32 315 x 6 + 16 315 x 4 + 2 63 x 2 + 2 9 ) B + 128 315 x 10 + 128 945 x 6 + 1024 14175 x 4 + 512 11025 x 2 + 16384 496125

b) 在(45)式保留4个因子分式,然后对这些4个因子的分式,每次保留1个,其余化成部分分式,得到:

16 x 8 + 40 3 x 6 + 518 45 x 4 + 3229 315 x 2 + 117469 12600 + 32 B + B 1245 107 24 B 1 22 3 B 2 45 4 B 3 53 3 B 4 943 24 B 5 = 32 x 10 B

16 x 8 + 40 3 x 6 + 518 45 x 4 + 3229 315 x 2 + 117469 12600 + 32 B + B 1345 53 12 B 1 15 2 B 2 11 B 3 107 6 B 4 157 4 B 5 = 32 x 10 B

16 x 8 + 40 3 x 6 + 518 45 x 4 + 3229 315 x 2 + 117469 12600 + 32 B + B 2345 35 8 B 1 23 3 B 2 43 4 B 3 18 B 4 941 24 B 5 = 32 x 10 B

由于 B , B 1 , B 2 , B 3 , B 4 , B 5 已知,计算得出(17)~(19)式。

c) 在(45)式保留5个因子分式,其余化成部分分式,得到:

16 x 8 + 40 3 x 6 + 518 45 x 4 + 3229 315 x 2 + 117469 12600 + 32 B + B 12345 53 12 B 1 22 3 B 2 23 2 B 3 52 3 B 4 473 12 B 5 = 32 x 10 B

由于 B , B 1 , B 2 , B 3 , B 4 , B 5 已知,计算得出(20)式。定理证毕。

2.3. 推论的证明

在定理的公式用ix代替x,注意到 arcsin ( i x ) = i arcsin h x

arcsin h x = ln ( x + 1 + x 2 ) arcsin 2 ( i x ) 2 1 ( i x ) 2 = ln 2 ( x + 1 + x 2 ) 2 1 + x 2 = b ,推论成立。

3. 一些数值级数

3.1. 数值级数

在定理公式(1)~(20),令 x = 1 / 2 = 2 / 2 arcsin x = π 4 B = 2 32 π 2

则分母为奇偶性不定线性因子乘积的二项式系数级数连带奇数倒数平方和数值恒等式

1) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 1 ) ( j = 1 m 1 ( 2 j 1 ) 2 ) = 2 16 π 2 + 1

2) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 2 ) ( j = 1 m + 1 1 ( 2 j 1 ) 2 ) = 2 16 π 2 + 14 9

3) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 3 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) = 43 2 240 π 2 + 1729 675

4) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 177 2 560 π 2 + 49408 11025

5) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 5 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) = 2867 2 5040 π 2 + 1323008 165375

6) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( j = 1 m + 1 1 ( 2 j 1 ) 2 ) = 5 2 48 π 2 5 9

7) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 1 ) ( m + 3 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) = 7 2 120 π 2 1069 1350

8) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 2 ) ( m + 3 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) = 3 2 40 π 2 694 675

9)

10) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 2 ) ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 73 2 840 π 2 16229 11025

11) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 3 ) ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 23 2 168 π 2 62768 33075

12) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) = 2 60 π 2 + 319 1350

13) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 3 2 140 π 2 + 722324 33075

14) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 1 ) ( m + 3 ) ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 11 2 420 π 2 + 4877 13230

15) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 2 ) ( m + 3 ) ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 13 2 240 π 2 + 44621 33075

16) m = 1 2 m ! 2 3 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 2 210 π 2 1459 22050

17) m = 1 ( 2 m ) ! 2 3 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) = 13 2 1890 π 2 587439 5953500

18) m = 1 ( 2 m ) ! 2 3 m ( m ! ) 2 ( m + 1 ) ( m + 3 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) = 2 126 π 2 146123 1323000

19) m = 1 ( 2 m ) ! 2 3 m ( m ! ) 2 ( m + 2 ) ( m + 3 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) = 17 2 1890 π 2 62119 496125

20) m = 1 ( 2 m ) ! 2 3 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) = 2 945 π 2 + 8369 567000

3.2. 数值级数

在公式(1)~(20),令 x = 1 arcsin x = π 2 ;计算公式中的项 ( a 1 b 1 + a 2 b 2 + + a t b t ) B = 0

因为 ( a 1 b 1 + a 2 b 2 + + a t b t ) = 0 ,而 B = π 2 / 8 1 x 2 ( x 1 )

这时定理计算公式(1)~(20)中, ( a 1 b 1 + a 2 b 2 + + a t b t ) 与B相乘的项的均为0

则分母为奇偶性不定线性因子乘积的二项式系数级数连带奇数倒数平方和数值恒等式

1) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 1 ) ( j = 1 m 1 ( 2 j 1 ) 2 ) = 1

2) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 2 ) ( j = 1 m + 1 1 ( 2 j 1 ) 2 ) = 8 9

3) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 3 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) = 544 675

4) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 8192 11025

5) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 5 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) = 16384 23625

6) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( j = 1 m + 1 1 ( 2 j 1 ) 2 ) = 1 9

7) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 3 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) = 131 1350

8) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 2 ) ( m + 3 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) = 56 675

9) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 2833 33075

10) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 2 ) ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 268 3675

11) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 3 ) ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 416 6615

12) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) = 19 1350

13) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 178501 33075

14) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 3 ) ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 251 22050

15) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 2 ) ( m + 3 ) ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 332 33075

16) m = 1 2 m ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 89 66150

17) m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) = 780701 5953500

18) m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 3 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) = 13 147000

19) m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 2 ) ( m + 3 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) = 556 496125

20) m = 1 ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( m + 2 ) ( m + 3 ) ( m + 4 ) ( m + 5 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) = 223 3969000

3.3. 数值级数

在推论公式(1)~(5)令 x = 1 b = 2 4 ln 2 ( 1 + 2 )

则交错的分母为奇偶性不定因子乘积二项式系数级数连带奇数倒数平方和数值恒等式

1) m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 1 ) ( j = 1 m 1 ( 2 j 1 ) 2 ) = 2 ln 2 ( 1 + 2 ) + 1

2) m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 2 ) ( j = 1 m + 1 1 ( 2 j 1 ) 2 ) = 3 2 ln 2 ( 1 + 2 ) 4 9

3) m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 3 ) ( j = 1 m + 2 1 ( 2 j 1 ) 2 ) = 7 15 2 ln 2 ( 1 + 2 ) + 304 675

4) m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 4 ) ( j = 1 m + 3 1 ( 2 j 1 ) 2 ) = 9 35 2 ln 2 ( 1 + 2 ) 736 2205

5) m = 1 ( 1 ) m ( 2 m ) ! 2 2 m ( m ! ) 2 ( m + 5 ) ( j = 1 m + 4 1 ( 2 j 1 ) 2 ) = 107 315 2 ln 2 ( 1 + 2 ) + 18176 55125

文章引用: 薛坐远 , 薛 雪 , 及万会 (2019) 二项式系数级数连带奇数倒数平方和。 理论数学, 9, 818-835. doi: 10.12677/PM.2019.97108

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