| Line 10: | Line 10: | ||
<math>\vdots</math> <math>\vdots</math> | <math>\vdots</math> <math>\vdots</math> | ||
| + | |||
| + | From the observation, we can assume the following formula is true: | ||
<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> | ||
[[Category:MA181Fall2011Bell]] | [[Category:MA181Fall2011Bell]] | ||
Revision as of 15:16, 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 $
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} $
