Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
One approach is to use $1+2+\cdots+n=\dfrac{n(n+1)}{2}$. –  Jonas Meyer Feb 4 '13 at 3:56
    
Have you seen/proved "Gauss's formula"? $$1 + \ldots + n = \frac{n(n+1)}{2}$$ –  andybenji Feb 4 '13 at 3:56
    
@JonasMeyer Jinx! –  andybenji Feb 4 '13 at 3:56
    
@JonasMeyer Ahhhh, thanks for pointing out Gauss's formula. I think I can handle it from here now. –  icanc Feb 4 '13 at 4:00
1  
Here is an interesting visual "proof" that doesn't require using that formula: users.tru.eastlink.ca/~brsears/math/oldprob.htm#s32 –  Jonas Meyer Feb 4 '13 at 4:00

4 Answers 4

up vote 5 down vote accepted

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.)

share|improve this answer

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

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

share|improve this answer
1  
It is not a coincidence. It is a geometric phenomenon. See <a href="southalabama.edu/mathstat/personal_pages/carter/…; this illustration,</a> for example. –  Scott Carter Feb 4 '13 at 4:03
1  

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 $n \in \mathbb N$.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.