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         <math>\vdots</math>                  <math>\vdots</math>
 
         <math>\vdots</math>                  <math>\vdots</math>
  
<math>\sum_{n=1}^N \dfrac{\left(n+k\right)!}{\left(n-1\right)!} = \dfrac1{k+2}\cdot\dfrac{\left(N+k+1\right)!}{\left(N-1\right)!}\quad for k\in\mathbb{N}</math>
+
<math>\sum_{n=1}^N \dfrac{\left(n+k\right)!}{\left(n-1\right)!} = \dfrac1{k+2}\cdot\dfrac{\left(N+k+1\right)!}{\left(N-1\right)!}\quad \mathrm{for} k\in\mathbb{N}</math>
  
  
 
[[Category:MA181Fall2011Bell]]
 
[[Category:MA181Fall2011Bell]]

Revision as of 06:04, 4 September 2011

Homework 2 collaboration area

Here's some interesting stuff:

$ \sum_{n=1}^N 1 = \dfrac11N $

$ \sum_{n=1}^N n = \dfrac12N\left(N+1\right) $

$ \sum_{n=1}^N n\left(n+1\right) = \dfrac13N\left(N+1\right)\left(N+2\right) $

       $ \vdots $                  $ \vdots $

$ \sum_{n=1}^N \dfrac{\left(n+k\right)!}{\left(n-1\right)!} = \dfrac1{k+2}\cdot\dfrac{\left(N+k+1\right)!}{\left(N-1\right)!}\quad \mathrm{for} k\in\mathbb{N} $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn