Online Course Discussion Forum
MCIII NT Chap3 3.23-3.27
I'm stuck on all 5 questions.
For 3.23, I solved that n>1, but this doesn't seem like a likely answer...
For the rest I have no clue where to start. Any hints/solutions would be appreciated!
3.23: You can set variables $n=\lfloor x\rfloor$ and $a=\{x\}$, and try to factor.
3.24 and 3.25 are about the sum of a series. There are usually two techniques that may be applied: telescoping, and pairing. Telescoping is good for 3.24, and pairing is good for 3.25. Well, 3.24 is significantly harder, so we probably should talk about it more in details in class.
3.26 is actually not that hard. It requires some careful estimation on the terms. You can notice that the first bunch of terms are all 0. Then at some point it becomes 1, and then 2, and so on; near the end, the values don't repeat any more. Each of these places where the value changes can be calculated.
3.27 Trying a few numbers will also help, but it does require some more thoughts. This will be reviewed in class.
Thank you for your reply.
3.23: that is in fact exactly what I had done. I factored out (n-1)(a-1)<0. Since a<1, then n>1, therefore x>1. However, this doesn't seem likely, but I do not know what I did wrong
3.24: Could you explain what telescoping is?
3.25: When we pair the ends together, we can get the nth and the (2016-n)th adds together to 100? (since the numbers inside the "brackets" add to 101) We would then have 100800 as our sum. But I am not quite sure of this and don't quite know how to prove it.
3.26: I've been calculating this like forever, and I now think there must be a way to obtain it in a way that is not bashing and just trying random numbers. Could you give me a hint?
3.23: Well, if an integer $n > 1$, it actually means $n \ge 2$, right?
3.24: Telescoping means you split each term into two terms, and then adjacent terms cancel out. More details in class.
3.25: You actually said it pretty well. Just give it a little bit more thought, you will know why the paired terms add up to 100 (hint: you already know that, without the floor function, the terms add up to 101. Then what happens if you take the floor of each? Remember the floor function results in integers.).
3.26: For smaller numbers, each value of obtained multiple times, so the values 0, 1, 2, ..., are all taken, up to a certain value. Up to this point, we know each of the consecutive numbers have appeared, so we know how many they are. Once this certain value is reached, the values after it will not be consecutive, but each value only appears once, so we can count by counting the terms. Combining these two steps, you can get the total number of possible values.
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