Proving Induction and Divisibility - Two Simple Homework Problems

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Homework Statement


Prove 1^3 + 2^3 + ... + n^3 = (1 + 2 + ... + n)^2 for all natural numbers n.


Homework Equations





The Attempt at a Solution



Well, this seems like the typical induction proof, so I start by testing the hypothesis at 1: 1^3 = 1^2 = 1. Then I assume that the equation is true for n, and try to prove it true for n + 1.

1 ^3 + 2^3 + ... + n^3 + (n + 1)^3 = (1 + 2 + ... + n)^2 + (n + 1)^3

I have this, but how do I get it to be (1 + 2 + ... + n + n + 1)^2? Is it some crazy factorization that I'm overlooking? It seems pretty easy.


Homework Statement



Prove that 7^n - 6n - 1 is divisible by 36 for all positive integers n.

Homework Equations





The Attempt at a Solution



Well, the statement seems to be saying the following: 7^n - 6n - 1 = 36k, where k is a positive integer. As with the first problem, testing it with 1 works. Assuming it is true for n + 1 would mean the following, if I'm correct:

7^(n + 1) - 6(n + 1) - 1 = 36j, where j is a positive integer. At this point, I'm not sure how to proceed.


I know I'm probably overlooking some obvious stuff here, so a little hint is probably all I'll need to finish these problems off. Thanks! :)
 
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steelphantom said:

Homework Statement


Prove 1^3 + 2^3 + ... + n^3 = (1 + 2 + ... + n)^2 for all natural numbers n.


Homework Equations





The Attempt at a Solution



Well, this seems like the typical induction proof, so I start by testing the hypothesis at 1: 1^3 = 1^2 = 1. Then I assume that the equation is true for n, and try to prove it true for n + 1.

1 ^3 + 2^3 + ... + n^3 + (n + 1)^3 = (1 + 2 + ... + n)^2 + (n + 1)^3

I have this, but how do I get it to be (1 + 2 + ... + n + n + 1)^2? Is it some crazy factorization that I'm overlooking? It seems pretty easy.
Try to substitute something else for (1+2+...+n). The formula for it is not that hard to derive.
steelphantom said:

Homework Statement



Prove that 7^n - 6n - 1 is divisible by 36 for all positive integers n.

Well, the statement seems to be saying the following: 7^n - 6n - 1 = 36k, where k is a positive integer. As with the first problem, testing it with 1 works. Assuming it is true for n + 1 would mean the following, if I'm correct:

7^(n + 1) - 6(n + 1) - 1 = 36j, where j is a positive integer. At this point, I'm not sure how to proceed.
7^(n+1) = 7^n*7. You are assuming that 7^n = 36k + 6n + 1. See what happens.
 
steelphantom said:

Homework Statement


Prove 1^3 + 2^3 + ... + n^3 = (1 + 2 + ... + n)^2 for all natural numbers n.

1^3 + 2^3 + ... + n^3=\sum_{n=0} ^{N} n^3

(1 + 2 + ... + n)^2 =(\sum_{n=0} ^{N} n)^2


Do you know what \sum_{n=0} ^{N} n represents?
 
Thanks guys for your help! I solved the second problem. The first problem is still puzzling me a bit. Here's what I have:

1^3 + 2^3 + ... + n^3 = \sum_{n=0} ^{N} n)^2 + (n + 1)^3

Still not sure how to add those two terms together. I'm sure it's very easy, but I'm just not seeing it.
 
The same question, again, can you write
\sum_{n=0}^N n
in a "closed" form and then square?
 
HallsofIvy said:
The same question, again, can you write
\sum_{n=0}^N n
in a "closed" form and then square?

Got it! Thank you. :)
 
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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