the image you have posted is not visible probably due to some site problem. If you can tell me in which section of the book you have this then i can help.

Results 1 to 9 of 9

- Apr 16th 2011, 11:27 PM #1

- Joined
- Apr 2011
- From
- Recife, Brazil
- Posts
- 18

- Apr 17th 2011, 01:52 AM #2

- Apr 17th 2011, 02:38 AM #3
We'll call this sum S.

We can write S = 1 + 2 + 3 + ... + (n - 2) + (n - 1) + n.

We can also write it as S = n + (n - 1) + (n - 2) + ... + 3 + 2 + 1.

If we add the sums written in these two forms together we get

2S = (n + 1) + (n - 1 + 2) + (n - 2 + 3) + ... + (3 + n - 2) + (2 + n - 1) + (1 + n)

2S = (n + 1) + (n + 1) + (n + 1) + ... + (n + 1) + (n + 1) + (n + 1)

2S = n(n + 1)

S = n(n + 1)/2.

- Apr 17th 2011, 04:03 AM #4

- Joined
- Apr 2011
- From
- Recife, Brazil
- Posts
- 18

- Apr 17th 2011, 04:46 AM #5

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 12,028
- Thanks
- 846

- Apr 17th 2011, 04:54 AM #6

- Apr 17th 2011, 05:31 AM #7

- Apr 17th 2011, 07:34 AM #8

- Joined
- Apr 2011
- From
- Recife, Brazil
- Posts
- 18

What sounds strange to me is that he adds a (r + 1) to both sides of the equation, then it became r(r + 1) + 2(r+1) and so on.

I don't understand this process.

Prove it already tried to explain me but i still can't see why to do a Sum S.

This page is a part of the book where the author tries to explain Proof by induction.

- Apr 17th 2011, 07:40 AM #9

i think it must have been this way in the book

define:

f(r) = 1+2+3+...+r

to prove that f(r)= r*(r+1)/2

induction:

f(1) is true

assume f(r) is true.

then f(r+1) = f(r) +(r+1). do you see why?

so f(r+1)= r(r+1)/2 + (r+1)

so f(r+1)= (r+1)[r/2 + 1]

so f(r+1)= (r+1)(r+2)/2

so f(r+1) is also true.