**323
The sequence and
Prove by Mathematical Induction that n .
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 24082013 14:07
Last edited by Arieisit; 24082013 at 14:35. 
Felix Felicis
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 24082013 14:22
Solution 323
(Original post by Arieisit)
**323
The sequence is given by and
Prove by Mathematical Induction that .
Posted from TSR Mobile
so true for
Assume true for for some
Now we wish to prove
thereby completing the inductive step.
So the proposition is true for as true for . As it's true for it is true by the principle of mathematical induction.
This is just a standard A level question.Last edited by Felix Felicis; 24082013 at 14:38. 
Smaug123
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 24082013 14:42
(Original post by Felix Felicis)
This is just a standard A level question. 
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 24082013 15:01
Last edited by Arieisit; 24082013 at 15:03. 
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(Original post by Smaug123)
It's good to have some notrequiringstupendouslybroadanddeepmathematicalknowledge questions, though 
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 24082013 15:29
(Original post by Arieisit)
I try . I know everyone on this forum can't handle ALL those questions. I know I can't
By the way, your LaTeX is interesting… I assume you meant:

Pterodactyl
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 24082013 15:49
A neat question from BMO1 from ~10 years ago that is quite good to use to 'get back into' the maths mode.
325*
For each positive integer , define the sequence () by:
for each
Determine all the values of for which 2000 is a term of the sequence. 
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 24082013 15:51
(Original post by Smaug123)
Most of us can't :P
By the way, your LaTeX is interesting… I assume you meant:
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 24082013 17:09
(Original post by Pterodactyl)
A neat question from BMO1 from ~10 years ago that is quite good to use to 'get back into' the maths mode.
325*
For each positive integer , define the sequence () by:
for each
Determine all the values of for which 2000 is a term of the sequence.
Next, consider the sequence mod 2. If , then all terms of the sequence are odd, so 2000 isn't a term of the sequence.
So k must be odd.
The remaining 1000 cases can easily be checked, and are left as an exercise to the reader.
ETA: Suppose that a_1 = 2000. Then k = 2001.
Suppose that a_2 = 2000. Then 2k+(k1) = 2000, and k=667.
Suppose that a_3 = 2000. Then 3k(2k+(k1)) = 2000  contradiction.
Suppose that a_4 = 2000. Then 4k+1 = 2000  contradiction.
Suppose that a_5 = 2000. Then k1 = 2000, and k=2001.
Eek, I made a horrendous mistake in assuming the sequence was increasing :P hardly my finest hour…Last edited by Smaug123; 24082013 at 17:17. Reason: Oops 
metaltron
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 24082013 17:13
(Original post by Pterodactyl)
A neat question from BMO1 from ~10 years ago that is quite good to use to 'get back into' the maths mode.
325*
For each positive integer , define the sequence () by:
for each
Determine all the values of for which 2000 is a term of the sequence.
Base cases:
Inductive step  now assume a_{2k1} is (k1).
So we conclude that and .
t, k and p are integers (n,p > 1) and t> 1.
Therefore:
So for odd n, k can only be 2001.
And for even n, (4t+4)k + 1 = 2000, which has no integer solutions or:
.
k cannot = 1, and 4t + 3 cannot = 1 when t is an integer. So we just have to check pairs of these factors. 3 is three more than a multiple of 4, when t=0, which leaves k = 667. 3 x 29 = 87 = 4(21) + 3. So k = 23 is a solution. 166 x 4 + 3 = 27x29, so k=3 is a solution. Finally, 23 = 4x5 + 3, so k = 87 is a solution.
ThereforeLast edited by metaltron; 24082013 at 17:19. 
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 24082013 17:22
(Original post by Smaug123)
First, note that , because otherwise .
Next, consider the sequence mod 2. If , then all terms of the sequence are odd, so 2000 isn't a term of the sequence.
So k must be odd.
The remaining 1000 cases can easily be checked, and are left as an exercise to the reader.
(Original post by metaltron)
...
Nongeometry BMO questions are very enjoyable.
Another:
326*
Show that for every positive integer , is divisible by 2000. 
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 24082013 17:33
Solution 324*
Problem 326 looks somewhat doable, though I don't think I'd be quick enough to solve and post it. (Edit: Scratch that, it wasn't at all)Last edited by ctrls; 24082013 at 17:50. 
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 24082013 17:39
Solution 326
Note ; also . Assume . By the fact that and , we are done.
We can, of course, easily check the case when .Last edited by Mladenov; 24082013 at 17:52. 
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 24082013 17:41
(Original post by Mladenov)
Solution 326
Note ; also . Assume . By the fact that and , we are done.
We can, of course, easily check the case when . 
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 24082013 17:50
(Original post by Smaug123)
This isn't clear to me  I may be half asleep, but why is 100 mod 2000 relevant? (From this you may gather that I don't follow the argument at all :P )Last edited by Mladenov; 24082013 at 17:53. 
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 24082013 18:41
(Original post by Mladenov)
Since I did it without a paper, just for convenience, I put , and noted that for it is divisible by (hence we can ignore it from then on). Then, by and CRT, we are done. 
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 24082013 19:31
[QUOTE=ctrls;44169930]Albeit simple, it's nice to have a question that I can actually solve.
Solution 324*
Think of it in this way?

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 28082013 17:57
Question 327*/**
Find a Maclaurin expansion of .
Using the factor theorem, prove also that:
(You may assume without proof that you can factorise this particular function in this way)
By comparing these expansions, prove that
(On a side note, can people spoiler their solutions?)Last edited by DJMayes; 28082013 at 20:08. 
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 28082013 20:03
Solution 327
Last edited by 0x2a; 28082013 at 20:25. 
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 28082013 20:15
(Original post by 0x2a)
Solution 327
Spoiler:Show
The Maclaurin expansion of
Since is merely of stretch of in the yaxis, the roots of and are equal, these being , so therefore the x can be any of the nonnegative integers. And it is obvious that any matches up with one of the expansions, namely , if x were to be 0, then because of the at the start the expansion would also evaluate to 0.
We have that
If these expansions are equal then the quadratic terms of both expansions should equal one another.
Spoiler:Show
Why does x have to be a nonnegative integer? sinx has negative roots as well. This is also behind the fact that the factorised expression contains quadratics (Not linear factors, which is what the factor theorem states you can take out as your quadratic can contain extra roots you haven't proved it has). As it is your explanation of this is insufficient. Secondly, you haven't proved that the constant on the outside has to be pi, which is necessary.
On a final note (although I think this was a silly mistake) you should be matching up cubic terms, not quadratic terms which sinx doesn't have in its expansion.
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