|
Author |
hbghlyj
Posted 2020-1-26 18:34
Last edited by hbghlyj 2020-1-27 10:44回复 2# tommywong \[{\left( {\frac{{10}}{3}} \right)^5} = \sum\limits_{i = 0}^\infty {\left( {111i + 411} \right) \cdot {{10}^{ - 3i}}} \]
使用\[\sum\limits_{i = 0}^\infty {\left( {ai + b} \right) \cdot {n^{ - i}}} = \frac{{{\rm{n}}\left( {a{\rm{ + }}b\left( {n - 1} \right)} \right)}}{{{{\left( {n - 1} \right)}^2}}}\]在各种进制下写出更多的类似式子 |
|