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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> | <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> | ||
| + | ---- | ||
| + | ==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. | ||
| + | |||
| + | ---- | ||
| + | [[2011_Fall_MA_181_Bell|Back to MA 181, Prof. Bell]] | ||
[[Category:MA181Fall2011Bell]] | [[Category:MA181Fall2011Bell]] | ||
Latest revision as of 04:17, 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.
