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Use mathematical induction to prove that $\forall n\in \mathbb{N}$, $$\sum ^{n}_{i=1}i^{3}=\dfrac {n^{2}(n+1)^{2}}{4}$$

$$\begin{align*} \sum_{k=1}^{n+1} k^3 &= \sum_{k=1}^{n} k^3 + (n+1)^2 \stackrel{\rm(IH)}{=} \dfrac {n^{2}(n+1)^{2}}{4} + (n+1)^2 \\ &= \dfrac {n^{2}(n+1)^{2}+4(n+1)^2}{4} \end{align*}$$

Is it true? What to do next?

amWhy
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Zhenya
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  • first you need to correct the error in the second line – David Holden Nov 02 '14 at 14:38
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    I think closing as a duplicate is misguided with this question. The OP is asking about his/her work, which isn't present in the linked post, and with a little direction (pointing out the error in adding $(n+1)^2$ instead of $(n+1)^3$. When questions are tagged "proof writing" and "proof verification", the point is not to point the OP to others' proofs! – amWhy Nov 03 '14 at 14:51

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First, don't forget the base case, $n = 1$: It holds.

Your inductive hypothesis is fine, but your inductive step is off.

You want: $$\begin{align*} \sum_{k=1}^{n+1} k^3 &= \left(\sum_{k=1}^{n} k^3\right) + (n+1)^{\color{blue}{\bf 3}} \stackrel{\rm(IH)}{=} \dfrac {n^{2}(n+1)^{2}}{4} + (n+1)^\color{blue}{\bf 3} \\ &= \dfrac {n^{2}(n+1)^{2}+4(n+1)^3}{4}\\ &= \frac{(n+1)^2(n^2 + 4(n+1))}{4}\\ &=\frac{(n+1)^2 (n^2 + 4n + 4)}{4}\\ & = \frac{(n+1)^2(n+2)^2}{4} \end{align*}$$

amWhy
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  • What to do after this? How to write description? – Zhenya Nov 02 '14 at 14:48
  • $IH(1)\land(\forall n\ge1:IH(n)\implies IH(n+1))\implies\forall n\ge1:IH(n)$ –  Nov 02 '14 at 14:51
  • @Tom What we have shown is that the proposition $P(1)$ is true, and most important, that the inductive hypothesis $P(n)$ implies $P(n+1)$. Therefore, by the principle of induction, the proposition is true for all $n \geq 1$. Note that the last expression is $\dfrac{(n+1)^2((n+1)+1)^2}4$ which means that $P(n+1)$ does indeed follow from the inductive hypothesis. – amWhy Nov 02 '14 at 16:50