求级数 (1×3×5)／(3^6-64)+(3×5×7)／(5^6-64)+…+(19×21×23)／(21^6-64) 之和

wintex

請問數學

Treenewbee

\[a_n=\frac{(2n-1)(2n+1)(2n+3)}{(2n+1)^6-2^6}\]
\[=\frac{(2n-1)(2n+1)(2n+3)}{(2n-1) (2 n+3) (3 + 4 n^2) (7 + 8 n + 4 n^2)}\]
\[=\frac{(2n+1)}{(3 + 4 n^2) (7 + 8 n + 4 n^2)}\]
\[=\frac{1}{4(4 n^2+3)}-\frac{1}{4 (4(n+1)^2+3)}\]
故
\[S_n=\frac{1}{4(4*1^2+3)}-\frac{1}{4 (4(n+1)^2+3)}\]
\[S_{10}=\frac{1}{4(4*1^2+3)}-\frac{1}{4 (4(10+1)^2+3)}=\frac{120}{3409}\]

luyuanhong

楼上 Treenewbee 的解答很好！已收藏。

王守恩

这样也不是不可以。《从问题出发——关于解题的理解和讲授 》

\(\frac{3}{7*19},\frac{8}{7*39},\frac{15}{7*67},\frac{24}{7*103},\frac{35}{7*147},\frac{48}{7*199},\frac{63}{7*259},\frac{80}{7*327},\frac{99}{7*403},\frac{120}{7*487},...\frac{n^2-1}{7(4n^2+3)}\)

王守恩

求证:\(\displaystyle\sum_{k=1}^{\infty}\frac{\big(n(k-1)+1\big)\big(nk+1\big)\big(n(k+1)+1\big)}{(nk+1)^6-n^6}=\frac{1}{2(n^3+n^2+n)}\)

要不分开是这样:

\(\frac{1×2×3}{2^6-1^6}+\frac{2×3×4}{3^6-1^6}+\frac{3×4×5}{4^6-1^6}+\frac{4×5×6}{4^6-1^6}+\frac{5×6×7}{6^6-1^6}+…=\frac{1}{6}\)

\(\frac{1×3×5}{3^6-2^6}+\frac{3×5×7}{5^6-2^6}+\frac{5×7×9}{7^6-2^6}+\frac{7×9×11}{9^6-2^6}+\frac{9×11×13}{11^6-2^6}+…=\frac{1}{28}\)

\(\frac{1×4×7}{4^6-3^6}+\frac{4×7×10}{7^6-3^6}+\frac{7×10×13}{10^6-3^6}+\frac{10×13×16}{13^6-3^6}+\frac{13×16×19}{16^6-3^6}+…=\frac{1}{78}\)

\(\frac{1×5×9}{5^6-4^6}+\frac{5×9×13}{9^6-4^6}+\frac{9×13×17}{13^6-4^6}+\frac{13×17×21}{17^6-4^6}+\frac{17×21×25}{21^6-4^6}+…=\frac{1}{168}\)

\(\frac{1×6×11}{6^6-5^6}+\frac{6×11×16}{11^6-5^6}+\frac{11×16×21}{16^6-5^6}+\frac{16×21×26}{21^6-5^6}+\frac{21×26×31}{26^6-5^6}+…=\frac{1}{310}\)

\(\frac{1×7×13}{7^6-6^6}+\frac{7×13×19}{13^6-6^6}+\frac{13×19×25}{19^6-6^6}+\frac{19×25×31}{25^6-6^6}+\frac{25×31×37}{31^6-6^6}+…=\frac{1}{516}\)

王守恩

5楼的题目可以用4楼的方法。《从问题出发——关于解题的理解和讲授 》。

1,\(\frac{1×2×3}{2^6-1^6}+\frac{2×3×4}{3^6-1^6}+\frac{3×4×5}{4^6-1^6}+\frac{4×5×6}{4^6-1^6}+\frac{5×6×7}{6^6-1^6}+…=\frac{1}{6}\)

  \(\frac{2}{3*7},\frac{5}{3*13},\frac{9}{3*21},\frac{14}{3*31},\frac{20}{3*43},\frac{27}{3*57},\frac{35}{3*73},\frac{44}{3*91},\frac{54}{3*111},\frac{65}{3*133},...\frac{n(n+1)/2-1}{3(n(n+1)+1)}=\frac{1}{6}\)

2,\(\frac{1×3×5}{3^6-2^6}+\frac{3×5×7}{5^6-2^6}+\frac{5×7×9}{7^6-2^6}+\frac{7×9×11}{9^6-2^6}+\frac{9×11×13}{11^6-2^6}+…=\frac{1}{28}\)

  \(\frac{3}{7*19},\frac{8}{7*39},\frac{15}{7*67},\frac{24}{7*103},\frac{35}{7*147},\frac{48}{7*199},\frac{63}{7*259},\frac{80}{7*327},\frac{99}{7*403},\frac{120}{7*487},...\frac{n^2-1}{7(4n^2+3)}=\frac{1}{28}\)

3,类似的题。 有这样一串数:  \(a(n)=n^4+n^2+1\)   

   3, 21, 91, 273, 651, 1333, 2451, 4161, 6643, 10101, 14763, 20881, 28731, ......

  求证:\(\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}=\frac{1}{2}\)

  \(\frac{1}{3},\frac{3}{7},\frac{6}{13},\frac{10}{21},\frac{15}{31},\frac{21}{43},\frac{28}{57},\frac{36}{73},\frac{45}{91},\frac{55}{111},\frac{66}{133},\frac{78}{157},...\frac{n(n+1)/2}{n(n+1)+1}=\frac{1}{2}\)