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I have to use the axiom of induction to prove the summation of k^3 from 0 to n is $(n(n+1)/2)^2$. Here's what I have so far:

Let P(n) be the assertion that $0^3+1^3+⋯+n^3=(n(n+1)/2)^2$

Base Case P(0): $0^3=(0(0+1)/2)^2=0$

The base case of P(0) is true using P(n) and laws of arithmetic.

Inductive Step: Assume P(k) is true for some k greater than or equal to 0.

$P(k): 0^3+1^3+⋯+k^3=(k(k+1)/2)^2$

We wish to prove P(k+1): $0^3+1^3+⋯+k^3+(k+1)^3=((k+1)((k+1)+1)/2)^2$

Simplifying the right hand side, we get

$0^3+1^3+⋯+k^3+(k+1)^3=(((k+1)(k+2))/2)^2\\S=0^3+1^3+⋯+k^3+(k+1)^3\\ +(k+1)^3+k^3+⋯+1^3+0^3\\2S=(k+1)^3+k^3+1^3+(k+1)^3+k^3+1^3\\2S=2(k+1)^3+2k^3+2\\S=(k+1)^3+k^3+1=k^3+3k^2+3k+1+k^3+1=2k^3+3k^2+3k+2$

From this, I'm stuck on how to get the answer to prove the statement.

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1 Answer 1

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Hint:

Write $0^3 + 1^3 + ... + k^3 + (k+1)^3 = \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3$ using your induction hypothesis. Then check if $\left(\frac{(k+1)(k+2)}{2}\right)^2=\left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3$.

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  • $\begingroup$ At this point, all you need to do is expand both sides of that last equation and see that they are in fact the same. $\endgroup$
    – TravisJ
    Apr 10, 2015 at 21:23

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