I am writing the exact problem.(adsbygoogle = window.adsbygoogle || []).push({});

1. Prove that for all natural numbers n the following equality is satisfied:

[tex]1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n+1)^2}{4}[/tex]

I am doing the exact method they used in the book, which I don't entirely agree on.

My answer is different than the one in the back of the book, but they come to the same conclusion.

Let n = k + 1,

[tex]\frac{(k+1)^2(k+1+1)^2}{4} = \frac{k^4+6k^3+13k^3+12k+4}{4} = \frac{k^2(k+1)^2}{4} + (k+1)^3[/tex]

Since, [itex]\frac{k^2(k+1)^2}{4}[/itex] is true by our assumption then adding [itex](k+1)^3[/itex] does not change the solution when n = k + 1.

I left some of the calculations out so I don't have to waste an hour doing tex, but I don't think there is any mistakes in it.

Obviously, if you test it, it is right.

Does that constitute as a proof?

I don't agree with the way they do it because they are assuming that there assumption is true. I really don't know.

Any comments?

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# Proof of a Sequence

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