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#1 2007-10-06 08:12:00

qwertyuiopsadhfghgrdghd
Guest

arithmetic sequence (very hard)

We have an arithmetic sequence:


And we are given:

First term is 1; the second term is 2.
Every successive term is 3 times the preceding term plus 5 times the second preceding term.
We are to prove that
does not divide
for n≥2
A term does not divide the product of the next two terms
How can I check that?

Fixed LaTeX - Ricky

#2 2007-10-06 20:32:36

Jai Ganesh
Administrator
Registered: 2005-06-28
Posts: 45,956

Re: arithmetic sequence (very hard)

Let a(n-1)=x, a(n)=y.
By definition, a(n+1)=5x+3y
a(n+2)=5y+[3(5x+3y)]=5y+15x+9y=15x+10y
a(n+1)*a(n+2)=(5x+3y)(15x+10y)=45x² +30y² +95xy
It can be seen that a(n+1)*a(n+2) is not divisible by a(n), i.e.
y since two of the three terms forming the sum a(n+1)*a(n+2) are divisible by y, whereas the third term, that is, 45x² is independent of y.
Maybe my logic is wrong at some step, I tried making a start!


It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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#3 2007-10-07 01:26:45

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: arithmetic sequence (very hard)

You’ve made a start but it’s not enough. 45x[sup]2[/sup] is not independent of y. If x is a[sub]n−1[/sub] and y is a[sub]n[/sub], then y = 5a[sub]n−2[/sub]+3x – so they are related.

Last edited by JaneFairfax (2007-10-07 03:16:36)

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#4 2007-10-07 03:12:36

bossk171
Member
Registered: 2007-07-16
Posts: 305

Re: arithmetic sequence (very hard)

How does one do subscript without LaTeX?


There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

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#5 2007-10-07 06:53:35

HallsofIvy
Guest

Re: arithmetic sequence (very hard)

The standard ASCII method is an underline: a_n.

#6 2007-10-08 06:35:47

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: arithmetic sequence (very hard)

After banging on my head on this brick wall for two or three days, I think I’ve finally found a solution. I’m post the whole thing when I get home (I’m now at a friend’s house).

First you establish two preliminary results:

P1: a[sub]n[/sub] is not divisible by 3 or 5 for any n.

P2: gcd(a[sub]n[/sub],a[sub]n+1[/sub]) = 1.

P1 is easily proved by induction, so I’ll skip the proof here. P2 is also proved by induction; I’ll post my proof later.

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#7 2007-10-08 10:05:32

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: arithmetic sequence (very hard)

So we’re all right for n = 1.

Hence, by induction, P2 is proved.

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#8 2007-10-08 10:24:29

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: arithmetic sequence (very hard)

Now we’re set. Suppose there is some n ≥ 2 for which a[sub]n[/sub] divides a[sub]n+1[/sub]a[sub]n+2[/sub]. Then

But gcd(a[sub]n[/sub],a[sub]n+1[/sub]) = 1. So a[sub]n[/sub] can’t possibly divide a[sub]n+1[/sub][sup]2[/sup] – unless a[sub]n[/sub] = 1 (which doesn’t hold for n ≥ 2).

Contradiction! Quod erat demonstrandum.

Last edited by JaneFairfax (2007-10-08 10:35:34)

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