Combinatorics - Mathematical Induction?

nintendo424
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Hello, I am having trouble solving this problem. Maybe I'm just overreacting to it. In my two semesters in discrete math/combinatorics, I've never seen a problem like this (with two summations) and been asked to prove it. Can some one help?

\sum^{n}_{i=1} i^3 = \frac{n^2(n+1)^2}{4} = (\sum^{n}_{i=1} i)^2

I mean, I know the whole S(n), S(1), S(k), S(k+1) steps, but I'm just unsure of how to write it. The solutions manual for the book skip that problem.

Book: Discrete And Combinatorial Mathematics: An Applied Introduction by Ralph P. Grimaldi, 5th Edition.
 
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First of all, this is a textbook problem, so it belongs in the homework forums. I moved it for you :smile:

Second, you actually need to show two things:

\sum_{i=1}^n i^3=\frac{n^2(n+1)^2}{4}

and

\sum_{i=1}^n i = \frac{n(n+1)}{2}

(and square both sides)

Can you do that?
 
Thank you very much! That helped a lot, I just finished my proof. :D That makes sense why you'd have to break it up. I didn't put the relationship between \sum^{n}_{i=1}i = \frac{n(n+1)}{2} and (\sum^{n}_{i=1}i)^2 = \frac{n^2(n+1)^2}{4} together. lol
 
nintendo424 said:
Hello, I am having trouble solving this problem. Maybe I'm just overreacting to it. In my two semesters in discrete math/combinatorics, I've never seen a problem like this (with two summations) and been asked to prove it. Can some one help?

\sum^{n}_{i=1} i^3 = \frac{n^2(n+1)^2}{4} = (\sum^{n}_{i=1} i)^2

I mean, I know the whole S(n), S(1), S(k), S(k+1) steps, but I'm just unsure of how to write it. The solutions manual for the book skip that problem.

Book: Discrete And Combinatorial Mathematics: An Applied Introduction by Ralph P. Grimaldi, 5th Edition.


These are two separate problems. ∑i³ is one and ∑i is the other. Have you tried either?

The question belongs in mathematics, not computer science.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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