# Prove by mathematical induction that $1^3+2^3+…n^3=1/4(n^2(n+1)^2)$ [duplicate]

In my attempt I arrived at $$1/4(k^2(k+1)^2+4(k+1)^3)$$ and don't know how to make any progress from there. I'm supposed to go from there to $$(1/4(k+1)^2(k+2)^2)$$ but have no idea how.

## marked as duplicate by José Carlos Santos, Arnaud D., Community♦Oct 5 '18 at 12:46

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• I do not see where the 4 of at 4(k+1)^3 comes from. It seems wrong. Anyways, I would try to factor out (k+1) at that point. – Cornman Oct 5 '18 at 12:30
• A good way to use induction to prove that two sequences coincide, especially when one of them is written as a sum, is to show that $a_1=b_1$ and then to show that $a_n-a_{n-1}=b_n-b_{n-1}$. – lulu Oct 5 '18 at 12:31

## 1 Answer

Factor $$(k+1)^2$$ out of $$k^2(k+1)^2+4(k+1)^3$$.

You will find that it is $$(k+1)^2(k^2+4k+4)$$ which further simplifies to $$(k+1)^2(k+2)^2$$