Ah, this exam felt so good!
After the STA257 exam, and an unexpectedly brutal CSC207 exam, this exam was like a blessing.
I really like the way this course is set-up, with all the required problem sets and other things, just because it really forces you into the material.
I'm glad I had this exam as my last.
Friday, December 12, 2008
Monday, December 1, 2008
Polya Goes to Work
Not quite sure where we were supposed to be pulling the problem from and no mathematical problem immediately jumps to mind, so I looked back at old exams for one.
Question 1 of the December 2006 exam seems intriguing.
Q: Prove that 4^(n+1) + 5^(2n-1) is a multiple of 21 for all natural numbers n >= 1.
Understanding the problem:
An incoherent and unintuitive mixture of 4 and 5 to some weird combination of exponents gives a multiple of 21.
Plan:
Try to see a pattern. Or just mess around with some induction and hope something pops out. I do like it when a seemingly random mess generates order. Chaos into something intelligible. Scrambled... Ah, off topic.
Carrying it out:
Well, we must begin where the darkness meets the light, where total balance exists, because nothing exists at all - zero.
So, popping in n = 0:
4^(0+1) + 5^(2(0)-1) = 4^1 + 5^(-1) = 4/5.
Eh? Oh wait, yes, that is right, the question says n >= 1... Phew.
n = 1:
4^(1+1) + 5^(2(1)-1) = 4^2 + 5^1 = 21.
Yay, we have our foundation. 21 is a multiple of 21. At least in this universe. And I do exist here. I assumed that when I created this blog.
Now, for some induction.
We assume that for n it's, true, and prove it is true for n+1.
4^(n+1+1) + 5^(2(n+1)-1) = 4^(n+2) + 5^(2n+1)
Now how can I make this so that what we assumed appears within...
So I guess the obvious step is to pull apart the exponents.
Hm...
= 4*4^(n+1) + 5^2*5^(2n-1) Doesn't look good.
We can see what we assumed inside this, but I don't see a way to get at it.
= 4^2*4^n + 5*5^2n Neither does that.
We have 21 from 4^2 + 5 but I don't see a way to work with that either.
It's a exam Q, so it shouldn't be that hard.
Speaking of exams, I need to study for them... Hopefully will finish this later.
Question 1 of the December 2006 exam seems intriguing.
Q: Prove that 4^(n+1) + 5^(2n-1) is a multiple of 21 for all natural numbers n >= 1.
Understanding the problem:
An incoherent and unintuitive mixture of 4 and 5 to some weird combination of exponents gives a multiple of 21.
Plan:
Try to see a pattern. Or just mess around with some induction and hope something pops out. I do like it when a seemingly random mess generates order. Chaos into something intelligible. Scrambled... Ah, off topic.
Carrying it out:
Well, we must begin where the darkness meets the light, where total balance exists, because nothing exists at all - zero.
So, popping in n = 0:
4^(0+1) + 5^(2(0)-1) = 4^1 + 5^(-1) = 4/5.
Eh? Oh wait, yes, that is right, the question says n >= 1... Phew.
n = 1:
4^(1+1) + 5^(2(1)-1) = 4^2 + 5^1 = 21.
Yay, we have our foundation. 21 is a multiple of 21. At least in this universe. And I do exist here. I assumed that when I created this blog.
Now, for some induction.
We assume that for n it's, true, and prove it is true for n+1.
4^(n+1+1) + 5^(2(n+1)-1) = 4^(n+2) + 5^(2n+1)
Now how can I make this so that what we assumed appears within...
So I guess the obvious step is to pull apart the exponents.
Hm...
= 4*4^(n+1) + 5^2*5^(2n-1) Doesn't look good.
We can see what we assumed inside this, but I don't see a way to get at it.
= 4^2*4^n + 5*5^2n Neither does that.
We have 21 from 4^2 + 5 but I don't see a way to work with that either.
It's a exam Q, so it shouldn't be that hard.
Speaking of exams, I need to study for them... Hopefully will finish this later.
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