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# The Proof is Trivial! watch

1. **323

The sequence and
Prove by Mathematical Induction that n .

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2. 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.
3. (Original post by Felix Felicis)
This is just a standard A level question.
It's good to have some not-requiring-stupendously-broad-and-deep-mathematical-knowledge questions, though
4. 324*
Evaluate

sin 2r + cos 2r )

Where n is a positive integer.
5. (Original post by Smaug123)
It's good to have some not-requiring-stupendously-broad-and-deep-mathematical-knowledge questions, though
I try . I know everyone on this forum can't handle ALL those questions. I know I can't
6. (Original post by Arieisit)
I try . I know everyone on this forum can't handle ALL those questions. I know I can't
Most of us can't :P
By the way, your LaTeX is interesting… I assume you meant:
7. 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.
8. (Original post by Smaug123)
Most of us can't :P
By the way, your LaTeX is interesting… I assume you meant:
Yep, that's what I meant... I'm new to this Latex stuff

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9. (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.
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.

ETA: Suppose that a_1 = 2000. Then k = 2001.
Suppose that a_2 = 2000. Then 2k+(k-1) = 2000, and k=667.
Suppose that a_3 = 2000. Then 3k-(2k+(k-1)) = 2000 - contradiction.
Suppose that a_4 = 2000. Then 4k+1 = 2000 - contradiction.
Suppose that a_5 = 2000. Then k-1 = 2000, and k=2001.

Eek, I made a horrendous mistake in assuming the sequence was increasing :P hardly my finest hour…
10. (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.
We shall show by induction that all odd terms alternate between having values of 1 and (k-1).

Base cases:

Inductive step - now assume a_{2k-1} is (k-1).

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.

Therefore
11. (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.
You brought it down from an infinite number of solutions, that's as good as solved. Much like Graham's number as an upper bound.

(Original post by metaltron)
...

Non-geometry BMO questions are very enjoyable.

Another:

326*
Show that for every positive integer , is divisible by 2000.
12. (Original post by Arieisit)
324*
Evaluate

Where n is a positive integer.
Albeit simple, it's nice to have a question that I can actually solve.

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)
13. Solution 326

Note ; also . Assume . By the fact that and , we are done.
We can, of course, easily check the case when .
Solution 326

Note ; also . Assume . By the fact that and , we are done.
We can, of course, easily check the case when .
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 )
15. (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 )
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.
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.
Aah, it didn't help that I thought you were answering question 325 :P it is now perfectly clear!
17. [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?

18. 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?)
19. Solution 327

Spoiler:
Show

(Original post by DJMayes)
Question 327*/**

Find a Maclaurin expansion of .
The Maclaurin expansion of

(Original post by DJMayes)
By comparing these expansions, prove that
We have that

If these expansions are equal then the quadratic terms of both expansions should equal one another.

20. (Original post by 0x2a)
Solution 327

Spoiler:
Show

The Maclaurin expansion of

Since is merely of stretch of in the y-axis, the roots of and are equal, these being , so therefore the x can be any of the non-negative 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 non-negative 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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