三角錐数
∑ k = 1 ∞ 1 k ( k + 1 ) ( k + 2 ) 6 = 6 ∑ k = 1 ∞ 1 2 ( 1 k − 2 k + 1 + 1 k + 2 ) = 3 { ( ⧸ 1 ⧸ 1 − ⧸ 2 ⧸ 2 + ⧸ 1 ⧸ 3 ) + ( 1 2 − ⧸ 2 ⧸ 3 + ⧸ 1 ⧸ 4 ) + ( ⧸ 1 ⧸ 3 − ⧸ 2 ⧸ 4 + 1 5 ) + ( ⧸ 1 ⧸ 4 − 2 5 + 1 6 ) + ⋯ } = 3 2 {\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{\frac {k(k+1)(k+2)}{6}}}&=6\sum _{k=1}^{\infty }{\frac {1}{2}}\left({\frac {1}{k}}-{\frac {2}{k+1}}+{\frac {1}{k+2}}\right)\\&=3{\bigg \{}\left({\frac {\color {Green}\not 1}{\color {Green}\not 1}}-{\frac {\color {Green}\not 2}{\color {Green}\not 2}}+{\frac {\not 1}{\not 3}}\right)+\left({\frac {1}{2}}-{\frac {\not 2}{\not 3}}+{\frac {\color {Red}\not 1}{\color {Red}\not 4}}\right)+\left({\frac {\not 1}{\not 3}}-{\frac {\color {Red}\not 2}{\color {Red}\not 4}}+{\frac {1}{5}}\right)+\left({\frac {\color {Red}\not 1}{\color {Red}\not 4}}-{\frac {2}{5}}+{\frac {1}{6}}\right)+\cdots {\bigg \}}\\&={\frac {3}{2}}\end{aligned}}}