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109 bytes removed ,  09:55, 10 December 2015
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Temos:
<math>\sum_{k=1}^{n} k = \frac {n(n-1)}{2}</math>
<pre>Incompleto
</pre>
Separando o somatório:
<math>\sum_{k=1}^{n} k\cdot k! =\sum_{k=1}^{n} k\cdot \sum_{k=1}^{n} k! </math> Temos: <math>\sum_{k=1}^{n} k = \frac{n(n+1)}{2}</math>
e teremos que descobrir o
<math>\sum_{k=1}^{n} k!</math>