Benedict Irwin edited untitled.tex  over 9 years ago

Commit id: 2c9eb45c001d2557f736f442d03b813de230a24c

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This might not be so obvious and upon rearranging elucidates that \begin{equation}  n!=(2n-3)(n-1)!+(4n-3-n^2)(n-2)! n!=(2n-3)[(n-1)!]+(4n-3-n^2)[(n-2)!]  \end{equation}