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This is a question from my textbook, it goes like this:

Prove (by induction) that the series $1^3+2^3+...+n^3=(1+2+...+n)^2$

Here is my attempt at a solution:

The base case would be: $n = 1 \implies 1^3=1^2$

Then we suppose that the statement works for a certain: $n=k$.

This gives us: $1^3+2^3+...+k^3=(1+2+...+k)^2$ (the inductive assumption) and this can be written as: $1^3+2^3+...+k^3=(\frac{k(k+1)}{2})^2$ by using the sum formula.

Further: $n=(k+1) \implies LHS = 1^3+2^3+...+k^3+(k+1)^3 =$ (by using the inductive assumption) $=(\frac{k(k+1)}{2})^2 +(k+1)^3$ and the $RHS = (1+2+...+k+(k+1))^2$

The proof would be complete if I could show that the $LHS = RHS$ but how do I do that?

Thank you for your help!

Lukas Arvidsson
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  • Not sure about the downvote... The question is clear and OP shows work. Still a duplicate. – The Chaz 2.0 Jul 19 '13 at 15:26
  • Sorry for that! If someone would like to please help me with the last part of this question I would still appreciate that very much! – Lukas Arvidsson Jul 19 '13 at 15:27
  • Write $1 + 2+...+k+(k+1)$ as $(1+2+...+k)+(k+1)$. Then when you square it, you'll get something of the form $a^2+2ab+ b^2$, where $a^2$ is the RHS of the inductive hypothesis. Not sure if that makes sense...! – The Chaz 2.0 Jul 19 '13 at 15:39
  • Thank you! The question is solved, I am sorry for making a duplicate, should have been more thorough in my search before posting! – Lukas Arvidsson Jul 19 '13 at 15:50
  • Duplicates are hard to avoid, and nothing to apologize for (usually)! – The Chaz 2.0 Jul 19 '13 at 16:08

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