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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> | ||
+ | ----------------------------- | ||
+ | S5.2_45 | ||
+ | \left( \right) | ||
+ | <math>f\left( x \right)=2x^3</math> | ||
+ | <math> | ||
+ | |||
+ | aaa | ||
+ | |||
+ | </math> | ||
[[Category:MA181Fall2011Bell]] | [[Category:MA181Fall2011Bell]] |
Revision as of 14:55, 5 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} $
S5.2_45 \left( \right) $ f\left( x \right)=2x^3 $
$ aaa $