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**323

The sequence
(edited 10 years ago)
Solution 323

Original post by Arieisit
**323

The sequence {un}\{ u_n \} is given by u1=1u_1 = 1 and un+1=(n+1)un, n1u_{n+1} = (n+1) \cdot u_n, \ n \geq 1
Prove by Mathematical Induction that un=n! nZ+u_n = n! \ \forall n \in \mathbb{Z^{+}}.

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*** :eyeball:

u1=1=1!u_{1} = 1 = 1! so true for n=1n=1

Assume true for n=k:uk=k!n=k: u_{k} = k! for some kZ+k \in \mathbb{Z^{+}}

Now we wish to prove uk+1=(k+1)!u_{k+1}=(k+1)!

uk+1=(k+1)uk=(k+1)k!=(k+1)!u_{k+1} = (k+1) \cdot u_{k} = (k+1) \cdot k! = (k+1)!

thereby completing the inductive step.

So the proposition is true for n=k+1n = k + 1 as true for n=kn = k. As it's true for n=1n = 1 it is true nZ+\forall n \in \mathbb{Z^{+}} by the principle of mathematical induction.

This is just a standard A level question. :redface:
(edited 10 years ago)
Original post by Felix Felicis

This is just a standard A level question. :redface:

It's good to have some not-requiring-stupendously-broad-and-deep-mathematical-knowledge questions, though :smile:
324*
Evaluate

[latex]\displaystyle\sum_{r=1}^n (tan r θ\theta sin 2r θ\theta + cos 2r θ\theta)

Where n is a positive integer.
(edited 10 years ago)
Original post by Smaug123
It's good to have some not-requiring-stupendously-broad-and-deep-mathematical-knowledge questions, though :smile:


I try :redface: . I know everyone on this forum can't handle ALL those questions. I know I can't :getmecoat:
Original post by Arieisit
I try :redface: . I know everyone on this forum can't handle ALL those questions. I know I can't :getmecoat:


Most of us can't :P
By the way, your LaTeX is interesting… I assume you meant:
r=1n(tan(rθ)sin(2rθ)+cos(2rθ))\displaystyle\sum_{r=1}^n (tan (r \theta) sin (2r \theta) + cos (2r \theta))
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 k>1k>1, define the sequence (ana_n) by:
a0=1, an=kn+(1)nan1a_0=1,\ a_n = kn+(-1)^na_{n-1} for each n1n\geq 1
Determine all the values of kk for which 2000 is a term of the sequence.
Original post by Smaug123
Most of us can't :P
By the way, your LaTeX is interesting… I assume you meant:
r=1n(tan(rθ)sin(2rθ)+cos(2rθ))\displaystyle\sum_{r=1}^n (tan (r \theta) sin (2r \theta) + cos (2r \theta))


Yep, that's what I meant... I'm new to this Latex stuff :erm:

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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 k>1k>1, define the sequence (ana_n) by:
a0=1, an=kn+(1)nan1a_0=1,\ a_n = kn+(-1)^na_{n-1} for each n1n\geq 1
Determine all the values of kk for which 2000 is a term of the sequence.

First, note that k2001k \leq 2001, because otherwise a1>2000a_1 > 2000.
Next, consider the sequence mod 2. If k0(mod2)k \equiv 0 \pmod{2}, 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…
(edited 10 years ago)
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 k>1k>1, define the sequence (ana_n) by:
a0=1, an=kn+(1)nan1a_0=1,\ a_n = kn+(-1)^na_{n-1} for each n1n\geq 1
Determine all the values of kk 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:

a1=ka0=k1 a_1 = k - a_0 = k-1
a2=2k+(k1)=3k1 a_2 = 2k + (k-1) = 3k - 1
a3=3k(3k1)=1 a_3 = 3k - (3k-1) = 1

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

a2p=2pk+(k1)=(2p+1)k1 a_{2p} = 2pk + (k-1) = (2p+1)k -1
a2p+1=(2p+1)k(2p+1)k+1=1 a_{2p+1} = (2p+1)k - (2p+1)k + 1 = 1
a2p+2=(2p+2)k+1 a_{2p+2} = (2p+2)k + 1
a2p+3=(2p+3)k(2p+2)k1=k1 a_{2p+3} = (2p+3)k - (2p+2)k - 1 = k - 1

So we conclude that a4t+1=k1 a_{4t+1} = k - 1 and a4t+3=1 a_{4t+3} = 1 .

t, k and p are integers (n,p > 1) and t> -1.

Therefore:

a4t+2=(4t+2)k+(k1)=(4t+3)k1 a_{4t+2} = (4t+2)k + (k-1) = (4t+3)k - 1
a4t+4=(4t+4)k+1 a_{4t+4} = (4t+4)k + 1

So for odd n, k can only be 2001.
And for even n, (4t+4)k + 1 = 2000, which has no integer solutions or:

(4t+3)k1=2000 , (4t+3)k=2001=3×23×29 (4t+3)k -1 = 2000 \ , \ (4t+3)k = 2001 = 3 \times 23 \times 29 .

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 k=3 , 23 , 87 , 667 or 2001 k = 3 \ , \ 23 \ , \ 87 \ , \ 667 \ or \ 2001
(edited 10 years ago)
Original post by Smaug123
First, note that k2001k \leq 2001, because otherwise a1>2000a_1 > 2000.
Next, consider the sequence mod 2. If k0(mod2)k \equiv 0 \pmod{2}, 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
...

:clap2:

Non-geometry BMO questions are very enjoyable.

Another:

326*
Show that for every positive integer nn, 121n25n+1900n(4)n121^n-25^n+1900^n-(-4)^n is divisible by 2000.
Reply 2171
Original post by Arieisit
324*
Evaluate

r=1n(tan(rθ)sin(2rθ)+cos(2rθ))\displaystyle\sum_{r=1}^n (\tan {(r\theta)}\sin{(2r\theta)}+\cos{(2r\theta)})

Where n is a positive integer.

Albeit simple, it's nice to have a question that I can actually solve. :tongue:

Solution 324*

r=1n(tan(rθ)sin(2rθ)+cos(2rθ))\displaystyle\sum_{r=1}^n (\tan {(r\theta)}\sin{(2r\theta)}+\cos{(2r\theta)})

=r=1nsin(rθ)cos(rθ)(2sin(rθ)cos(rθ))+(12sin2(rθ))\displaystyle=\sum_{r=1}^n \frac{\sin{(r\theta)}}{\cos{(r \theta)}}\left(2\sin{(r\theta)}\cos{(r\theta)}\right)+\left(1-2\sin^2{(r\theta)}\right)

=r=1n2sin2(rθ)+12sin2(rθ)\displaystyle=\sum_{r=1}^n 2\sin^2{(r\theta)}+1-2\sin^2{(r\theta)}

=r=1n1=n\displaystyle=\sum_{r=1}^n 1 = n

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)
(edited 10 years ago)
Solution 326

Note 2000=24532000=2^{4}5^{3}; also 1900100(mod2000)1900 \equiv -100 \pmod {2000}. Assume n2n \ge 2. By the fact that 121n9n25n(mod24)121^{n} \equiv 9^{n} \equiv 25^{n} \pmod {2^{4}} and 121n(4)n(mod53)121^{n} \equiv (-4)^{n} \pmod {5^{3}}, we are done.
We can, of course, easily check the case when n=1n = 1.
(edited 10 years ago)
Original post by Mladenov
Solution 326

Note 2000=24532000=2^{4}5^{3}; also 1900100(mod2000)1900 \equiv -100 \pmod {2000}. Assume n4n \ge 4. By the fact that 121n9n25n(mod24)121^{n} \equiv 9^{n} \equiv 25^{n} \pmod {2^{4}} and 121n(4)n(mod53)121^{n} \equiv (-4)^{n} \pmod {5^{3}}, we are done.
We can, of course, easily check the case when n3n \le 3.


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 )
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 1900100(mod2000)1900 \equiv -100 \pmod {2000}, and noted that for n2n \ge 2 it is divisible by 20002000 (hence we can ignore it from then on). Then, by gcd(2,5)=1\gcd(2,5)=1 and CRT, we are done.
(edited 10 years ago)
Original post by Mladenov
Since I did it without a paper, just for convenience, I put 1900100(mod2000)1900 \equiv -100 \pmod {2000}, and noted that for n2n \ge 2 it is divisible by 20002000 (hence we can ignore it from then on). Then, by gcd(2,5)=1\gcd(2,5)=1 and CRT, we are done.

Aah, it didn't help that I thought you were answering question 325 :P it is now perfectly clear!
Original post by ctrls
Albeit simple, it's nice to have a question that I can actually solve. :tongue:

Solution 324*

r=1n(tan(rθ)sin(2rθ)+cos(2rθ))\displaystyle\sum_{r=1}^n (\tan {(r\theta)}\sin{(2r\theta)}+\cos{(2r\theta)})

=r=1nsin(rθ)cos(rθ)(2sin(rθ)cos(rθ))+(12sin2(rθ))\displaystyle=\sum_{r=1}^n \frac{\sin{(r\theta)}}{\cos{(r \theta)}}\left(2\sin{(r\theta)}\cos{(r\theta)}\right)+\left(1-2\sin^2{(r\theta)}\right)

=r=1n2sin2(rθ)+12sin2(rθ)\displaystyle=\sum_{r=1}^n 2\sin^2{(r\theta)}+1-2\sin^2{(r\theta)}

=r=1n1=n\displaystyle=\sum_{r=1}^n 1 = n

Think of it in this way?

1cos(2θ)sin(2θ)=tanθ\frac{1-\cos{(2\theta)}}{\sin{(2 \theta)}} = \tan \theta

tan2θ=?\tan 2\theta = ?

tan3θ=?\tan 3\theta = ?
Question 327*/**

Find a Maclaurin expansion of sin(πx) sin( \pi x) .

Using the factor theorem, prove also that:

sin(πx)=πx(1x212)(1x222)(1x232)... sin ( \pi x ) = \pi x (1-\dfrac{x^2}{1^2})(1-\dfrac{x^2}{2^2})(1-\dfrac{x^2}{3^2})...

(You may assume without proof that you can factorise this particular function in this way)

By comparing these expansions, prove that n=11n2=π26 \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2} = \dfrac{\pi ^2}{6}

(On a side note, can people spoiler their solutions?)
(edited 10 years ago)
Reply 2178
Solution 327

Spoiler

(edited 10 years ago)
Original post by 0x2a
Solution 327

Spoiler



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