Sum of n^3 = (sum of n)^2 induction proof

cnwilson2010
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Homework Statement


Prove:
1^3 + 2^3 +...+n^3=(1+2+...+n)^2
n=Natural number

Homework Equations





The Attempt at a Solution


Using induction -
n=1 obvious

Assume for n=k equation is true.

Show for k+1.

I have that the right side prior to k+1 is (k^2(k+1)^2)/4
After k+1 I have (k^4+6k^3+13k^2+12k+4)/4,
but I can't figure out the trick to get the form above equal to the left +(k+1)^3.

Any ideas?
Thanks.
 
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hi cnwilson2010! welcome to pf! :smile:

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erm :redface: … a2 - b2 = … ? :wink:
 
cnwilson2010 said:

Homework Statement


Prove:
1^3 + 2^3 +...+n^3=(1+2+...+n)^2
n=Natural number

Homework Equations





The Attempt at a Solution


Using induction -
n=1 obvious

Assume for n=k equation is true.

Show for k+1.

I have that the right side prior to k+1 is (k^2(k+1)^2)/4
After k+1 I have (k^4+6k^3+13k^2+12k+4)/4,
but I can't figure out the trick to get the form above equal to the left +(k+1)^3.

Any ideas?
Thanks.

On the left side you've added (k+1)^3. On the right side you've added the results after k+1 minus the results after k. The thing you want to prove equal to (k+1)^3 is (k+1)^2*(k+2)^2/4-k^2*(k+1)^2/4.
 
Thank you for your replies they were very helpful. Now I'm working on n2<=2n.

For what values of Natural numbers is this statement true and prove by induction. Obviously, 1 and 2 are true, now I'm thinking proof by contradiction and the idea of the existence of a least member of the set. Does this sound reasonable or way too much work for the results?
 
Seems begging for proof by induction, you will have to show that 2k+1&lt;k^{2} at some point.

Or failing that, you could note that the two sequences a_{n}=n^{2},b_{n}=2^{n} are both monotonically increasing sequences and that one increases faster that the other.
 
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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