# Series property

1. Mar 23, 2008

### ramsey2879

Given $$a_{0} = 0$$, $$a_{1} = 1$$, and $$a_n = Ba_{n-1} - a_{n-2}$$; prove that
$$\sum_{i=1}^{m} a_{2i-1} = (a_{m})^2$$

2. Mar 23, 2008

### al-mahed

$$a_0 \ = 0$$
$$a_1 \ = 1$$
$$a_2 \ = B$$
$$a_3 \ = B^2 - 1$$
$$a_4 \ = B^3 - 2B$$
$$a_5 \ = B^4 - 3B^2 + \ 1$$
$$a_6 \ = B^5 - 4B^3 + \ 3B$$
$$a_7 \ = B^6 - 5B^4 + \ 6B^2 - 1$$
$$a_8 \ = B^7 - 6B^5 + 10B^3 - 4B$$
$$a_9 \ = B^8 - 7B^6 + 15B^4 - 10B^2 + 1$$
$$a_1_0 = B^9 - 8B^7 + 21B^5 - 20B^3 + 5B...$$

You made an interesting conjecture, unfortunally I have tried for some minutes and nothing... the proof seems to be not too complicate but long, I think usign the binomial will be helpfull

As you see above, the terms and the powers seems to have a relations with the paschal triangle. The powers form a simple arithmetical progression, and the terms forms a progression like the paschal triangle

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
...​

3. Mar 24, 2008

### tiny-tim

Hi ramsey

$$\sum_{i=1}^{m} a_{2i-1}\,=\,\sum_{i=1}^{m} Ba_{2i-2}\,-\,\sum_{i=0}^{m-1} a_{2i-1}$$​

Does that help?

4. Mar 24, 2008

### ramsey2879

al-mahed l

That seems to be on the right tract towards a short proof. I arrived at this via a related conjecture which I think I did prove. This gives another approach to the proof.
It can be proven by direct substitution, the math is not too complicated but is tedious, and I do not think I made an error in my proof.

Lema 1
Given $$a,b$$, and $$B$$

define $$c = Bb-a$$

then $$a^2 - Bab + b^2 = b^2 -Bbc + c^2$$

Lema 2
Given $$n = a^2 - Bab + b^2$$
define $$d = b^2-a^2 \, e = Bb^2-2ab$$
Then$$d^2 - Bde + e^2 = n^2$$

Proof
Lema 1 and 2 each can be applied to get next terms in the series for $$n^2$$Via Lema 1 [tex] f = Be - d = B(Bb^2 - 2ab) -b^2 + a^2$$ Via Lema 2 $$f = c^2-b^2 = (B^2-1)b^2 - B2ab + a^2 Inspection shows that they are equivalent Also the following relation forms [tex] g = Bf - e = Bc^2 - 2cb = (B^3-2B)b^2 -a((2B62+2)b + Ba)$$ Since a and b can take any value this formula can be applied recursively to show that the series for n^2 has every other term as the difference in the squares fo the sucessive terms for in the series for n. But when n = 1 and you start with a = 0, b = 1 the series for n^2 becomes the same as the series for n. So every other term $$a_{n}^2 - a_{n-1}^2. I ran out of time but this will be completed later Last edited: Mar 25, 2008 5. Mar 25, 2008 ### ramsey2879 Good point. At least it helps in the summing of the other half of the series. I will edit my other post to complete my proof now so I will not need this relation at this moment. 6. Mar 25, 2008 ### ramsey2879 al-mahed Sorry that I have to redo this but in my haste to post i left a mess and then found that it could not be edited. I arrived at this via a related conjecture which I also proved. There is probably a better proof using induction and properties of Pascel's triangle as you have shown is so involved here. My proof follows and also used induction. The following Lema can be proven by direct substitution, the math is not too complicated but is tedious, and I did double check the math. Lema 1 Given [tex] a,b$$, and $$B$$ define $$c = Bb-a$$ then $$a^2 - Bab + b^2 = b^2 -Bbc + c^2$$ Lema 2 Given $$n = a^2 - Bab + b^2$$ define $$d = b^2-a^2 \, e = Bb^2-2ab$$ Then $$d^2 - Bde + e^2 = n^2$$ Proof: Lema 1 and 2 each can be applied to get next terms in the series for $$n^2[tex] Via Lema 1 [tex] f = Be - d = B(Bb^2 - 2ab) -b^2 + a^2$$ Via Lema 2 $$f = c^2-b^2 = (B^2-1)b^2 - B2ab + a^2 Inspection shows that they are equivalent Also the following relation forms [tex] g = Bf - e = Bc^2 - 2cb = (B^3-2B)b^2 -a((2B62+2)b + Ba)$$ Since a and b can take any value this formula can be applied recursively to show that the series for n^2 has every other term as the difference in the squares fo the sucessive terms for in the series for n. But when n = 1 and you start with a = 0, b = 1 the series for n^2 becomes the same as the series for n. So every other term $$a_{n}^2 - a_{n-1}^2. I ran out of time but this will be completed later[/QUOTE] 7. Mar 25, 2008 ### ramsey2879 al-mahed Once again I have to redo this. I inadvertently submitted this before finishing. Mabe I made an inadvertant mouse click when I was trying to bring up more of the text to edit or something. Again what was posted can not be edited for some reason. I arrived at this via a related conjecture which I also proved. There is probably a better proof using induction and properties of Pascel's triangle as you have shown is so involved here. My proof follows and also used induction. The following Lema can be proven by direct substitution, the math is not too complicated but is tedious, and I did double check the math but will not reproduce it below. Lema 1 Given [tex] a,b$$, and $$B$$ define $$c = Bb-a$$ then $$a^2 - Bab + b^2 = b^2 -Bbc + c^2$$ Lema 2 Given $$n = a^2 - Bab + b^2$$ define $$d = b^2-a^2 \, e = Bb^2-2ab$$ Then $$d^2 - Bde + e^2 = n^2$$ Lema 3 By Lema 1, $$a,b,c$$ can be the first terms of an infinite series where adjacent terms can be substuted for the $$a$$ and $$b$$ to give the value $$n$$ while $$d,e$$ can be the beginnings of an infinite series where each pair of adjacent terms can be substituted to give the value $$n^2$$. Let $$f$$ and $$g$$ are the next terms in the series for $$n^2$$ through the recursive formula of Lema 1. Then the following can also be established by direct substitution of terms using Lema 2 applied to the second pair $$b$$ and $$c$$: $$f = c^2 - b^2$$ $$g = Bc^2 - 2bc$$ Proof Since a and b can take any value the above formula can be applied recursively such that every other term of the second series, $$d,f,h,\dots$$ is equal to the difference of squares, $$b^2-a^2$$, $$c^2-b^2$$, \quad \dots$$.

Also if a = 0 and b = 1 then $$d=b$$, and $$e=c$$ and since each series have the same recursive formula they are the same series. Thus the proof of the relation of my first post is established.