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Prove that:

$1^3+2^3+...+n^3=\frac{n^2(n+1)^2}{4}$ for $n \in N$

So I am thinking that I need to do a proof by mathematical induction. Here's my attempt:

Let S(n) be the statement $1^3+2^3+...+n^3=\frac{n^2(n+1)^2}{4}$ for $n \in N$. When n=1,

$$1^3=\frac{1^2(1+1)^2}{4}$$

$$1=\frac{(2)^2}{4}$$

$$1=\frac{4}{4}$$

So P(1) is true.

Suppose that k is any integer with $k \ge 1$ such that $p(k)1^3+2^3+...+k^3=\frac{k^2(k+1)^2}{4}$ for $k \in N$. Then adding $(k+1)^3$ to both sides,

$$1^3+2^3+...+k^3=\frac{k^2(k+1)^2}{4}$$

$$1^3+2^3+...+k^3+(k+1)^3=\frac{k^2(k+1)^2}{4}+(k+1)^3$$

$$1^3+2^3+...+k^3+(k+1)^3=\frac{k^2(k+1)^2+4(k+1)^3}{4}$$

$$1^3+2^3+...+k^3+(k+1)^3=\frac{(k+1)^2(k^2+4(k+1)}{4}$$

$$1^3+2^3+...+k^3+(k+1)^3=\frac{(k+1)^2(k^2+4k+4)}{4}$$

$$1^3+2^3+...+k^3+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4}$$

$$1^3+2^3+...+k^3+(k+1)^3=\frac{(k+1)^2(k+1+1)^2}{4}=P(k+1)$$

Thus, since we have proved that P(k+1) is true, we can conclude that P(k) is true for all $k \in N$.

How does it look, any edits I should make?

Martin Sleziak
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Math Major
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    Looks fine. ${}{}{}$. – copper.hat Feb 05 '15 at 19:53
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    One small error the equal sign in $=P(k+1)$. $P(k+1)$ is not a number, but the expression of the equality of the sum and the formula. I would replace it by "which is exactly the proposition $P(k+1)$" – MasB Feb 05 '15 at 20:12
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    Very minor comments: (i) for dots between the plus signs, use \cdots (for low dots use \dots); (ii) I would prefer, for no good reason, to have in the conclusion $P(n)$ is true $\dots$; there are inconsistencies of notation, first it was $S(n)$ then sometimes $P$, and once $p$. In the line after "So $P(1)$ i true there is a strange $p(k)$, and after that "for $k\in \mathbb{N}$ is unnecessary, you have already said $k$ is a positive integer. – André Nicolas Feb 05 '15 at 20:18
  • This may be one of the most duplicated questions on this site. – Daniel W. Farlow Feb 05 '15 at 20:39
  • @induktio I suppose it may be a duplicate, but I thought my answer was right but I just wanted to fix the very minor flaws (that everyone has been pointing out!) and just looking at that question wouldn't have really done that – Math Major Feb 05 '15 at 21:17
  • @MathMajor You weren't supposed to look at the question but all of the answers given. – Daniel W. Farlow Feb 05 '15 at 21:19
  • @induktio Of course I meant looking at the quesiton + answers, but for example the not putting an equal sign with the P(k+1) was never addressed in that question. So yes, it may be duplicated. But it was still beneficial and helpful to me to ask this question separately. – Math Major Feb 05 '15 at 21:24
  • @MathMajor In that case, a comment would be a lot more appropriate than an entirely new question. – Daniel W. Farlow Feb 05 '15 at 21:27

2 Answers2

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Only issue to me is that you should not be saying $\frac{(k+1)^2(k+2)^2}{4} = P(k+1)$. $P$ is a predicate, not a polynomial. In other words, it is a mathematical statement which is either true or false, but it does not possess a numerical value, unlike the arithmetic expression given on the left side. Thus you might instead write $1^3 + 2^3 + \dots + (k+1)^3 = \frac{(k+1)^2(k+2)^2}{4} \implies P(k+1)$.

Having logical connectives (e.g., $\implies$)between each of your equations is probably a good idea as well, because even though your reader likely recognizes that these statements are meant to be taken together, in theory they could just be unrelated statements.

dalastboss
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Ok, It is good.

I would do it as this. In the first step for $(k+1)$,

Left side: $$P_{k+1} - P_k = (k+1)^3$$

Right side: $$P_{k+1}-P_k={{(k+1)^2}{(k+2)^2} \over 4}-{{(k)^2}{(k+1)^2} \over 4}$$ $$P_{k+1}-P_k={1 \over 4}{(k+1)^2}({(k+2)^2-(k)^2})$$ $$P_{k+1}-P_k={1 \over 4}{(k+1)^2}({4k+4})$$ $$P_{k+1}-P_k={(k+1)^3}$$ Therefor: $L=R$

PdotWang
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