Hello everyone, I am in desperate need of help on a mathematical induction question. I realize that this forum may not be the right place to post "simple" math questions, but what can I say, I am desperate. So, please bear with me. I will greatly appreciate any help on the following question: Prove by mathematical industion that, (2n+1)+(2n+3)+(2n+5)+...+(4n-1)= 3n^2 Thank you very much, and I sincerely hope I get a detailed answer. Thanks again. Pete
Spleenie | (165.228.129.12) | Tuesday, 24 April 2001 8:44:39 PM
Pete,
Well my detailed answer is that by any method this expression cannot be true.
First of all, the leftmost part of the expression makes no sense. Perhaps let t = 1,2,3..etc for each bracketed part of the leftmost expression, it can be resolved to:
(2n+1)+(2n+3)+...+(2n+[2t-1]), which is an arithmetic series with limit infinity
Even with this, simply placing a number for n (for instance "1") should show that there is no way the expression on the left can equal 3n^2 for all integers n
letting n=1 gives:
3 + 5 + 7 + infinity = 3
infinity does not equal 3, not even for very large values of 3.
I suggest either the question is flawed, or it has been typed wrong
Cheers
Spleenie | (165.228.129.12) | Tuesday, 24 April 2001 8:47:47 PM
Oops
Sorry about the lack of formatting. Trust me, I was pressing enter.
What a silly
Mitch | (213.122.71.97) | Thursday, 26 April 2001 1:19:09 AM
Hi,
The formula makes sense if interpretted as follows:
When n = 1 use just the first term, so that:
(2n+1) = 3n^2 which is true.
When n = 2, use the first two terms, so that:
(2n+1)+(2n+3) = 3n^2 which is true. And so on.
This is what I beleive is intended. I've had a go at an induction proof but have not yet been able to push it to the end result.
The trouble is that it's not as simple as the usual formulae then easily succumb to induction. For example:
1 + 2 + 3 + 4 + ..... n = n(n+1) / 2 is a simple infinite formula involving one variable. Using S to denote sum we can re-write this simple formula as:
S(n = 1 to k) of n.
As n variens from 1 to k we generate 1 + 2 + 3 + ..... + k
I have been unable to write the formula given in such a manner. I have written it as:
S(from k = 1 to n) of 2n + 2k - 1
as k caries from 1 to n then we generate the terms of the equation given:
2n + 1 + 2n + 3 + 2n + 5 etc etc
the trouble is that when we come to add the next term on to this equation every term changes so that we generate:
2(n+1) + 1 + 2(n+1) + 3 and so on.
These key differences between the formulae make the given one harder to crack with induction.
I have no doubt it is possible and need to keep trying - I just thought I'd write these thoughts in case they helped.
Mitch | (213.122.71.97) | Thursday, 26 April 2001 1:40:48 AM
In case it helps (and I guess it probably won't because you specified induction).
As Spleenie pointed out this series is an arithmetic series and it's easy enough to prove your formula using the properties of arithmetic series.
An arithmetic series (AP) is defined as follows:
a + (a + d) + (a + 2d) + (a + 3d) + .........
now the property which prooves your formula is:
The sum of the first k terms of an AP = k/2(2a + (k - 1)d)
Now in your formula a = 2n + 1 (The first term) and d = 2 (The common difference)
so the sum of the first k terms is easily seen to be:
2kn + k^2
Now looking at this formula you can trivially see that whenever n = k (as is always the case in your formula) then the sum of the series is 2k^2 + k^2 = 3k^2
So your formula is porven.
Spleenie | (203.35.170.199) | Friday, 27 April 2001 11:23:52 PM
Looking at the expression in the way described above does yield it seems the correct interpretation. It wasn't immediately obvious that the n was both an indexer and a variable. Still...should've seen it.
Due to the ever dynamic comparisons to the kth and (k+1)th component of this sum I see difficulties in solving this through induction. My main problem is the dual role of n. I don't think I have come across this before, and am unsure of how to deal with it when it is acting as both an index (defining where we cease the addition according to its number) and as a value _in_ the addition, when relating it to the general nth term (the (4n-1) and the result (3n^2), both of which are using n as the index. Maybe I am still not reading the question correctly. It's been a while.
I will introduce a new variable "i" which I will use as an indexer and then see if I can solve it using induction. As shown by Mitch, the end result is seen using the rules of arithmetic progression.
Cheers.
netwerp | (212.126.144.12) | Friday, 19 April 2002 4:18:34 AM
This looks tricky you are using n to mean two things are you sure question is correct, if so i will get back to you
netwerp | (212.126.144.12) | Thursday, 25 April 2002 9:37:47 PM
Hi i think i have the soloution u r after but i wrotr it in an editor then saved as gif iwill try and paste it into the chatroom otherwise i need an email adress
mine is
gundaddy@freeuk.com
netwerp@freeuk.com
L.F.P | (62.119.58.64) | Wednesday, 22 May 2002 10:26:57 AM
Spleenie is right.
L.F.P | (62.119.58.64) | Wednesday, 22 May 2002 10:27:19 AM
Spleenie is right.
Miffy | (141.213.46.181) | Friday, 4 October 2002 4:05:12 PM
The summation is *bounded* by the rightmost term and have a finite number of terms, and definitely does not approach infinity. Notice, the number of terms is equal to n. i.e. When n=1, it is (2(1)+1)= 3(1)^2. When n=2, it is (2(2)+1) + 2(2)+3 = 12 = 3(2)^2.
Let pn be the proposition "(2n+1)+(2n+3)+(2n+5)+...(4n-1)= 3n^2"
Base case: p_1: 4(1)-1 =3
2(1)+1 =3. Obviously we only need to 'add' the term 2(1)+1 =3 = 3(1)^2 = 3.
(In other words, the number of terms, k, in the equation is given by
Inductive: Assume pn is true, i.e. (2n+1)+(2n+3)+(2n+5)+ ...(4n-1)= 3n^2
(we want to prove it implies (2(n+1)+1) + (2(n+1)+3)+...(4(n+1)-1) = 3(n+1)^2.
In this case, we have n+1 terms in the eqn.
(2(n+1)+1) + (2(n+1)+3)+...(4(n+1)-1)
=(2n+1)+(2n+3)+(2n+5)+ ...(4n-1)+ 2n + 2(n+1)+ 2(n+1)-1
= 3n^2 + 2n + 2n +2n +2+2-1.
= 3n^2 + 6n + 3
=3(n+1)^2
Thus pn implies pn+1.
By the principle of mathematical induction, the proposition is true.
| (141.219.34.108) | Monday, 8 September 2003 6:32:52 AM
| (141.219.34.108) | Monday, 8 September 2003 6:32:56 AM