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I'm currently going through Calculus by Spivak by myself, and came across a proof by induction requiring to prove $1^3+...+n^3 = (1+...+n)^2$

Naturally, to prove this, I need to somehow convert $(1+...+n)^2+(n+1)^3$ to $(2+...+2n)^2$. After quite a bit of thinking, I'm still not sure how to do this. I think i may be forgetting about some property of squares that we're supposed to be using.

Note: Please only provide a hint, not the complete answer. Edit: I was mistakenly taking $(1+...+n)^2*(n+1)^3$ rather than $(1+...+n)^2+(n+1)^3$. Corrected.

Martin Sleziak
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Aayush Agrawal
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3 Answers3

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Previous Answer

Hint: $$1 + 2 + \ldots + n = \frac{n(n+1)}{2}$$

Can you take it from here?

Revised Answer

My apologies for not bothering to check more closely earlier, but I think your original equation $$\left(1 + 2 + \ldots + n\right)^2\left(n + 1\right)^3 = \left(2 + 4 + \ldots + 2n\right)^2$$ is not an identity.

To see why, via a quick inspection, the highest power of $n$ on the LHS is $n^7$, while the highest power of $n$ on the RHS is $n^4$, a contradiction.

Therefore, your proposed equation is not an identity.

Jose Arnaldo Bebita Dris
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  • I'm already familiar with that identity, but i'm afraid not. After converting the notation inside and multiplying it by (n+1)^3, where do i go? I.e, how do i convert it into the statement required? – Aayush Agrawal Jan 14 '16 at 07:23
  • The left side is actually a polynomial with degree $7$ not $5$ and the right side is a polynomial of degree $4$ not $2$. Each side as a summation with a number of terms that depends on $n$. – JimmyK4542 Jan 14 '16 at 07:41
  • I made a mistake where i, instead of adding $(n+1)^3$, was multiplying it. Also, i did not imply that $(1+2+...+n)^2(n+1)^2 = (2+4+...+2n)^2$ but rather $(1+2+...+n)^2(n+1)^2 = (2+2+3+4+...+2n)^2$. I tried to simplify it by integrating the n + 1 into the existing equation, something that i now realize was a very stupid mistake. Your original hint was enough to get me to the right answer after i realized my mistake. – Aayush Agrawal Jan 14 '16 at 07:44
  • Thanks Jimmy! I've made the appropriate edit. – Jose Arnaldo Bebita Dris Jan 14 '16 at 07:44
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Naturally, to prove this, i need to somehow convert $(1+...+n)^2*(n+1)^3$ to $(2+...+2n)^2$.

Unfortunately, "$(1+...+n)^2*(n+1)^3 = (2+...+2n)^2$" is not true for all $n$. Once you simplify the summations, the left side will be a polynomial with degree $2 \cdot 2 + 3 = 7$ while the right side will be a polynomial with degree $2 \cdot 2 = 4$.

The induction step is to assume that $$1^3+2^3+\cdots+n^3 = (1+2+\cdots+n)^2$$ for some integer $n$, and then prove that $$1^3+2^3+\cdots+n^3+(n+1)^3 = (1+2+\cdots+n+(n+1))^2.$$

As Arnie Dris suggested, the identity $1+2+\cdots+n = \dfrac{n(n+1)}{2}$ is useful.

Note that \begin{align*}& (1+2+\cdots+(n+1))^2 - (1+2+\cdots+n)^2 \\ &= \left[\dfrac{(n+1)(n+2)}{2}\right]^2-\left[\dfrac{n(n+1)}{2}\right]^2 \\ &=\dfrac{(n+1)^2}{4}\left[(n+2)^2-n^2\right] \\ &= \dfrac{(n+1)^2}{4}[4n+4] \\ &= (n+1)^3\end{align*}

Do you see how to complete the proof using this?

JimmyK4542
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No i dont think so . (Of above equation,induction one) see RHS can be simplied to $\frac{(n)^2(n+1)^2}{4}$ and we all know that summation $\sum n^3=\frac{(n^2)(n+1)^2}{4}$ and what you are proving doesnt seem to be true the second equation put $n=2$ and you get $243=36$ please see what you have interpreted wrongly in the proof

Archis Welankar
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    The second equation is for the step two of induction, i.e where you prove that if k is true, k+1 is true. The second equation is asking how to convert $1^3 + ... + n^3 + (n+1)^3 = (1+...+n)^2*(n+1)^3$ to the same statement as what you get when you substitute n as n + 1. – Aayush Agrawal Jan 14 '16 at 07:35
  • Update: My mistake. I was multiply by $(n+1)^3$ rather than adding it. – Aayush Agrawal Jan 14 '16 at 07:37
  • Okay please include it in body as i misinterpreted it – Archis Welankar Jan 14 '16 at 07:38