2
$\begingroup$

Prove, by algebraic manipulation, that:

\[ {{2n} \choose {n}} + {{2n} \choose {n+1}}={1\over2} {{2n+2} \choose {n+1}} \]

  • 0
    I don't really like the notation used, and its not typed using math tex, if you could refine your question perhaps with binomial coeifficents, typed correctly, I could help you2012-12-09
  • 2
    This just looks like a special case of pascals identity, Jillian you should slow down on the question posting and try to research some of these yourself, you could have found an algebraic proof on wikipedia, which I found after several searchs2012-12-09
  • 0
    If you're going to go on a Math forum just to bring people down, perhaps rethink why you're here.2012-12-09
  • 0
    There's a button called "Edit" @Ethan2012-12-11

3 Answers 3