- #1
otto
- 4
- 0
So look at what I've done:
[tex]
{n+1 \choose k} = \frac {(n+1)!} {(n+1-k)! \cdot k!} = \frac {(n+1)\cdot n!}{(n-(k-1))!\cdot k \cdot (k-1)!}
[/tex][tex]
=
\frac {(n+1)}{k} \cdot
\frac { n!}{(n-(k-1))!\cdot (k-1)!} =
\frac {(n+1)}{k}
\cdot {n \choose k-1}
[/tex]
oops I accidentally posted this before I finished my calculations, please ignore this until I've finished it. Thanks
[edit] So Yea, that's it. Thing is, somethings wrong (try plugging numbers into it). Anyways, I need to figure out what I did wrong. This is just part of a massive can of worms. I am trying to figure out the proof for: [tex]\sum\limits_{i=1}^n {n \choose k} = 2^n
[/tex]
[tex]
{n+1 \choose k} = \frac {(n+1)!} {(n+1-k)! \cdot k!} = \frac {(n+1)\cdot n!}{(n-(k-1))!\cdot k \cdot (k-1)!}
[/tex][tex]
=
\frac {(n+1)}{k} \cdot
\frac { n!}{(n-(k-1))!\cdot (k-1)!} =
\frac {(n+1)}{k}
\cdot {n \choose k-1}
[/tex]
oops I accidentally posted this before I finished my calculations, please ignore this until I've finished it. Thanks
[edit] So Yea, that's it. Thing is, somethings wrong (try plugging numbers into it). Anyways, I need to figure out what I did wrong. This is just part of a massive can of worms. I am trying to figure out the proof for: [tex]\sum\limits_{i=1}^n {n \choose k} = 2^n
[/tex]
Last edited: