Very few people can be as smart as Gauss, but it is important to know how to do the sum of n, n^2, n^3, etc.

namely, 1+2+…+n = n(n+1)/2

1^2+2^2+…+n^2 = n(n+1)(2n+1)/6

1^3+2^3+…+n^3 = n^2(n+1)^2/4

Those are very important sums in mathematics.

You are right about the n*2^n thing. Yes, I forget to put the factor n there. Nice work on spotting that!

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The expansion in question 9 is simply Taylor’s expansion of the function (1-4x)^(-1/2). You should have learned this if you have learn calculus.

The step in question 7 is even easier. W_1 is equal to 1 as we have calculated at the very beginning.

[(n+1) + … + 2] is a well-known summation. Gauss knew how to calculate this type of sum when he was only 7 years old.

You may have heard of the story of how he calculated 1+2+…+100.   

[(n+1) + … + 2] is done in the same way.

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Great, I have just uploaded the answers to your other questions. Please check them.

I will be going to sleep very soon, so if you have additional questions, please post them and I will take a look at them tomorrow when I wake up.

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Proof of the useful result. Suppose we want to decompose integer n into k nonnegative smaller integers.

We assume we have n+k-1 empty places in a row. We put k-1 obstacles to separate the row into k different regions. There is a total of “Choose k-1 from n+k-1” ways for placing those obstacles. For each way of placing obstacles, it corresponds to a  way of decomposing the integer n, by simply counting how many empty places are there in each region.

For instance, if n = 7, k=3, meaning we want to decompose 7 into 3 non-negative integers. We now have 7+3-1= 9 empty spaces, I will use ⚪ to represent, namely:

⚪⚪⚪⚪⚪⚪⚪⚪⚪

Now we place 2 obstacles in those places, indicated by X. If I place them in this way:

⚪⚪⚪X⚪⚪⚪X⚪

It means  3 + 3 + 1 since there are 3 empty places in region 1 and 2, and 1 empty place in region 3

If in this way

X X⚪⚪⚪⚪⚪⚪⚪

It means 0 + 0 + 7 because there is no empty places in region 1 and 2, while 7 empty places in region 3.

Do you get it now?

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Not really, the theorem is talking about a different thing. We are focused on how to partition an integer into smaller integers, not combinations.

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This is how you decompose the number 34 into 4 nonnegative integers. It is not equivalent to “choose 4 objects from 34 objects”.

It is a very useful result that if you want to decompose an integer n into k nonnegative integers, the total number of possible decomposition is given by “Choose k-1 object from n+k-1 objects “.

I recommend you think about how to prove this useful result. Also, I have taken the other questions you posted and am working on them. I will explain to you if you cannot figure out why to prove this result. (Hint: suppose there are n+k-1 places in a row, and you put k-1 obstacles to separate the n+k-1 places into four regions).

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Hi, you may try to post them on this site, but I cannot guarantee that I will be the one to pick them up. They could be picked up by other people as well, but I think other people can also give you satisfactory answers.

Hi, after you pick 1 and 2 (or 39 and 40), you claim that you can pick another 2 books in the rest of 37 books.

Have you considered the possibility that the other two books you pick can be adjacent and form another pair, thus breaking the assumption that there is only one pair?

The best strategy to avoid introducing new pairs, is to fixed the pair at the very beginning, and separate the pair with the other two books by inserting the rest of 36 books, as I have done in my “insertion” method.

Fixed some typos.

ans

Feel free to ask if still confused.