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This is what I've been able to do:

Base case: $n = 1$

$L.H.S: 1^3 = 1$

$R.H.S: (1)^2 = 1$

Therefore it's true for $n = 1$.

I.H.: Assume that, for some $k \in \Bbb N$, $1^3 + 2^3 + ... + k^3 = (1 + 2 +...+ k)^2$.

Want to show that $1^3 + 2^3 + ... + (k+1)^3 = (1 + 2 +...+ (k+1))^2$

$1^3 + 2^3 + ... + (k+1)^3$

$ = 1^3 + 2^3 + ... + k^3 + (k+1)^3$

$ = (1+2+...+k)^2 + (k+1)^3$ by I.H.

Annnnd I'm stuck. Not sure how to proceed from here on.

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    $\begingroup$ One approach is to use $1+2+\cdots+n=\dfrac{n(n+1)}{2}$. $\endgroup$ – Jonas Meyer Feb 4 '13 at 3:56
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    $\begingroup$ Have you seen/proved "Gauss's formula"? $$1 + \ldots + n = \frac{n(n+1)}{2}$$ $\endgroup$ – Andrew Maurer Feb 4 '13 at 3:56
  • $\begingroup$ @JonasMeyer Jinx! $\endgroup$ – Andrew Maurer Feb 4 '13 at 3:56
  • $\begingroup$ @JonasMeyer Ahhhh, thanks for pointing out Gauss's formula. I think I can handle it from here now. $\endgroup$ – Tejen Shrestha Feb 4 '13 at 4:00
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    $\begingroup$ Here is an interesting visual "proof" that doesn't require using that formula: users.tru.eastlink.ca/~brsears/math/oldprob.htm#s32 $\endgroup$ – Jonas Meyer Feb 4 '13 at 4:00
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HINT: You want that last expression to turn out to be $\big(1+2+\ldots+k+(k+1)\big)^2$, so you want $(k+1)^3$ to be equal to the difference

$$\big(1+2+\ldots+k+(k+1)\big)^2-(1+2+\ldots+k)^2\;.$$

That’s a difference of two squares, so you can factor it as

$$(k+1)\Big(2(1+2+\ldots+k)+(k+1)\Big)\;.\tag{1}$$

To show that $(1)$ is just a fancy way of writing $(k+1)^3$, you need to show that

$$2(1+2+\ldots+k)+(k+1)=(k+1)^2\;.$$

Do you know a simpler expression for $1+2+\ldots+k$?

(Once you get the computational details worked out, you can arrange them more neatly than this; I wrote this specifically to suggest a way to proceed from where you got stuck.)

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Consider the case where $n = 1$. We have $1^3 = 1^2$. Now suppose $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for some $n \in \mathbb N$. Recall first that $\displaystyle (1 + 2 + 3 + \cdots + n) = \frac{n(n+1)}{2}$ so we know $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 = \bigg(\frac{n(n+1)}{2}\bigg)^2$. Now consider $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 + (n + 1)^3 = \bigg(\frac{n(n+1)}{2}\bigg)^2 + (n+1)^3 = \frac{n^2 (n+1)^2 + 4(n+1)^3}{4} = \bigg( \frac{(n+1)(n+2)}{2} \bigg)^2$. Hence, the statement holds for the $n + 1$ case. Thus by the principle of mathematical induction $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for each $n \in \mathbb N$.

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  • $\begingroup$ What are the steps to get from the pre last expression (...)/4 to the last one (...)/2 ? I mess up here a lot $\endgroup$ – LurioTabasco Oct 18 '17 at 13:10
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IMHO, this fact is a coincidence; a better approach is to prove the closed-form formula for both. As we know

$$ 1 + 2 + \cdots + k = \frac{k(k+1)}{2} $$

the corresponding claim to prove is

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

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See http://www.youtube.com/watch?v=-usK5CMpUPo <-- this video . The case you want is at about time stamp 5:53.

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