Changes

<math>\sum_{k=1}^{n} k = \frac{n(n+1)}{2}</math>  e teremos que descobrir o <math>\sum_{k=1}^{n} k!</math>  então <math>\sum_{k=1}^{n} k!+(n+1)! = \sum_{k=0}^{n} (k!+1)! </math> <math>1+\sum_{k=1}^{n} (k+1)! = 1+\sum_{k=1}^{n} (k+1)k!</math> <math>1+\sum_{k=1}^{n} (k+1)!</math>