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=Homework 2 collaboration area=
 
=Homework 2 collaboration area=
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Here's some interesting stuff:
  
 
<math>\sum_{n=1}^N 1 = \dfrac11N</math>
 
<math>\sum_{n=1}^N 1 = \dfrac11N</math>
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<math>\sum_{n=1}^N n\left(n+1\right) = \dfrac13N\left(N+1\right)\left(N+2\right)</math>
 
<math>\sum_{n=1}^N n\left(n+1\right) = \dfrac13N\left(N+1\right)\left(N+2\right)</math>
  
      <math>\vdots</math>                     <math>\vdots</math>
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        <math>\vdots</math>                 <math>\vdots</math>
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From the observation, we can assume the following formula is true:
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<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>
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----
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==Discussion==
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*Would somebody care to add these to the [[Collective_Table_of_Formulas]]? Perhaps one should create be a new page dedicated to summation formulas.
  
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[[2011_Fall_MA_181_Bell|Back to MA 181, Prof. Bell]]
  
 
[[Category:MA181Fall2011Bell]]
 
[[Category:MA181Fall2011Bell]]

Latest revision as of 03:17, 6 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 $

From the observation, we can assume the following formula is true:

$ \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} $


Discussion

  • Would somebody care to add these to the Collective_Table_of_Formulas? Perhaps one should create be a new page dedicated to summation formulas.

Back to MA 181, Prof. Bell

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn