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Topic review (newest first)
- jd
- 2005-10-27 08:22:51
Thank you I get how to do it what I was confused about is "Prove that 2n natural numbers is n(2n+1)" I thought it meant prove 2+4+6+8+...+2n is n(2n+1).
- mathsyperson
- 2005-10-25 05:48:18
Well, fair enough, then. Don't worry, we all make mistkaes.
- kylekatarn
- 2005-10-25 05:30:19
Sorry the whole mess but THAT was my REPLY! I clicked REPORT instead of QUOTE.
Again, it was NOT my intention to report that post.
- mathsyperson
- 2005-10-25 05:23:30
kylekatarn: You should only report abusive or spammy posts, things you think should be deleted because they are inappropriate. If you think ganesh's explanation isn't full enough, just say so within the topic.
ganesh: kylekatarn "would not accept that as a full demonstration".
- ganesh
- 2005-10-24 14:09:46
jd wrote:
I only started a level is there a easier way? I think I saw "S=n/2[2a + (n-1)d]" when the teacher did it.
You can think of a sum of n natural numbers as an Arithmetic Progression and use this formula.
S = n/2[2a + (n-1)d]
Here, n = n, a = 1, d = 1
Therefore, S = n/2[2 + (n-1)1]
S = n(n+1)/2
- kylekatarn
- 2005-10-24 02:49:47
If you read the demonstration carefully, you will see right at the beginning the ◊ 'operator' as I define it:
◊n=1+2+3+...+n
I just reduced your problem to a simpler version. Instead of trying to build an expression for the sum of first 2n naturals, I got the expression for the sum of first N naturals, ◊n. After that you just plug 2n: ◊(2n) and get the expression.
Think '◊n' as ◊(n) or f(n). There is nothing special about ◊.. It's just a symbol I usually associate with the 'function' that sums the 1st N naturals.
Now read my post 2 or 3 more times and try to catch up with a sheet of paper, line by line.
- jd
- 2005-10-23 21:35:40
I'm really confused what does
◊0=0
◊1=1
mean?
I only started a level is there a easier way? I think I saw "S=n/2[2a + (n-1)d]" when the teacher did it.
thanks, JD
- kylekatarn
- 2005-10-23 10:00:55
Code:
n
let ◊n = 1+2+3+...+n = ∑ k
k=1
◊0=0
◊1=1
//finding a formula is easy
◊n=◊(n-1)+n
◊n=◊(n-2)+(n-1)+n
◊n=◊(n-3)+(n-2)+(n-1)+n
◊n=◊(n-4)+(n-3)+(n-2)+(n-1)+n
.
.
.
◊n=(n-n+0)+(n-n+1)+...+(n-3)+(n-2)+(n-1)+n
◊n=0+1+2+...+n-3+n-2+n-1+n
`----- n+1 terms -----´
◊n=[n+n+n+....+n+n]+(-n+0)+(-n+1)+(-n+2)+...+(-3)+(-2)+(-1)
◊n=[n²+n]+(-n+0)+(-n+1)+...+(-3)+(-2)+(-1)
◊n=[n²+n]-n+0-n+1-n+2+...-3-2-1
◊n=[n²+n]-(n-0+n-1+n-2+...+3+2+1)
◊n=[n²+n]-(1+2+3+...+(n-2)+(n-1)+(n-0))
◊n=[n²+n]-(1+2+3+...+(n-2)+(n-1)+n)
◊n=[n²+n]-◊n
2◊n=n²+n
n²+n
◊n = ----
2
//great! now we have a formula for ◊n.
//but the sum of the first 2n natural numbers is
2.n
∑ k
k=1
//wait..don't we have a formula for that? Yes!
λ
∑ k = ◊(λ)
k=1
2.n (2n)²+2n (2n)(2n+1) n(2n+1)
∑ k = ◊(2n) = -------- = ---------- = ------- >>>>> Done!
k=1 2 2 2- jd
- 2005-10-23 07:52:52
Could someone help I'm not sure how to do this.
"Show that the sum of the first 2n natural numbers is n(2n+1)"
It's a question from the heinemann c1 book.
I would ask the teacher again but it's half term and I don't know his phone number.
good day