## Ch 1 Natural Numbers proof |

04 Aug 2013, 03:21
David McKenna (1 post) |
I’m a bit confused - during the inductive part of the proof, where does the right side of the equation come from? (0+1+2+3+…+n+n+1) = |

07 Aug 2013, 17:54
Joe Legris (1 post) |
A well-known formula for the sum of consecutive positive integers up to n is: S = n(n+1)/2 You can see this by writing, S = 1+2+3+…+(n-1)+n Now reverse the order and write, S’ = n+(n-1)+…+3+2+1 Clearly, S = S’, so S + S’ = 2S. Now add S + S’ term by term: S + S’ = 2S = (n+1)+(n-1+2)+(n-2+3)+…+(n-1+2)+(n+1) = (n+1)+(n+1)+(n+1)+…+(n+1)+(n+1) There are n terms, so: 2S = n(n+1) and, S = n(n+1)/2 In the text, the series has 1 more term, i.e. n+1, so: S = (n+1)(n+2)/2 |

You must be logged in to comment |