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Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!
ankhtahr:
Yay, you're giving me hope.
Tht's exactly the way I solved these problems. It took me literally the whole night (as you can see in the university thread), but I finished everything.
ankhtahr:
I seriously hope that the exercise sheets will get easier, when I'm more used to doing stuff like this. This is one of our current exercises:
We have to prove these:
The first one was easy. And seeing that all these are best proven by induction is pretty obvious. I'm currently working on the second one. I'm absolutely sure the solution involves the binomial theorem, which we've learnt as .
The toughest exercise though is supposed to be this one:
Nobody I know has found a solution for it. Somebody found a solution for it online though.
Edit: I'm currently kinda stuck at (ii) at . I don't know how to get the +1 out of the binomial coefficient. I know that , but I don't know if that helps me any further.
Wait. I got that wrong. It's , and now I can see that it might help me.
Barmymoo:
I find this thread completely fascinating and completely baffling.
ankhtahr:
Damnit, I just noticed that is of course only valid for . So it does not help me, as in the sum there is k=0.
snalin:
Do you have to pay to use the site that makes the latex maths symbols?
Anyways, (iii):
I'll use (n k) for ("n choose k")
(n k) is the number of distinct subsets of size k of a set of size n
The sum from k=(0 to n) of (n k) is thus the number of subsets of any size of a set of size n*
This sum is now pretty easy to figure out: for every element of the original (size n) set, you make a distinct subset by either selecting, or not selecting each element of the original set. So, for every element you have two choices, and there are n elements - thus you can create a subset in 2^n different ways. The sum is equal to 2^n.
* the same as the number of elements in the power set of a set of size n, which is generally known to be 2^n, but I doubt you can use that fact.
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