0

I need to prove that for each natural n:

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

How do I do that? how do I know whether I should choose strong induction or simple induction?

FigureItOut
  • 1,835
  • Start by checking it for, say, $n=1,2,3.$ Not difficult – Will Jagy Nov 08 '14 at 19:37
  • Write $(n(n+1)/2)^2$ for the RHS, then prove it with induction. It's easy. – GDumphart Nov 08 '14 at 19:45
  • 5
    Maybe it's not so helpful to say a problem is easy when someone appears to have trouble with it. It may make some people more confident to try it out, but it also makes a lot of people feel stupid. – Zavosh Nov 08 '14 at 19:46

3 Answers3

4

Use induction on the right hand side :

$(1+2+...+n+(n+1))^2 = (1+2+...+n)^2 + 2(1+2+...+n)(n+1) + (n+1)^2$

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

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

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

Ojas
  • 2,146
  • can you tell me how you splitted (1 + 2 + ... + n + (n+1)) ^2 to (1 + 2 + ... + n) ^2 + 2(1 + 2 + .... + n)(n+1)+ (n+1)^2 ? – FigureItOut Nov 08 '14 at 20:15
  • $(a+b)^2 = a^2 + 2ab + b^2$. In this case, $a=(1+2+...+n)$ and $b=(n+1)$. – Ojas Nov 08 '14 at 20:17
1

You can check the base case by plugging in $n=1$. The induction step is simplified by first noting that $(1+2+\cdots+n)=n(n+1)/2$. From this, you can proceed as follows $$ (1+2+\cdots+n)^2+(n+1)^3=\frac{(n+1)^2}{4}[n^2+4(n+1)]\\ =\frac{(n+1)^2}{4}[n^2+4n+4]=\left[\frac{(n+1)(n+2)}{2}\right]^2=[1+\cdots+n+(n+1)]^2. $$

Edit: to see $1+\cdots+n=n(n+1)/2$, let $S_n=1+\cdots+n$. Then $$ 2S_n=S_n+S_n=(1+\cdots+n)+(n+\cdots+1)\\ =(1+n)+(2+(n-1))+\cdots+(n+1)=\underbrace{(1+n)+\cdots+(1+n)}_{n\text{ times}}=n(n+1). $$

Kim Jong Un
  • 14,794
  • 1
  • 24
  • 49
  • Can u tell me why (1+2 +... +n) = [n(n+1)]/2 ? – FigureItOut Nov 08 '14 at 19:57
  • The formula $1+2+\dots + n = \frac{n(n+1)}{2}$ is very well-known. It can also be proved using induction. If you're learning about induction in a course and you're given the problem you posted as execise, chances are that formula is either an example from the course, or an earlier exercise. – Zavosh Nov 08 '14 at 20:01
1

I propose neither of them.

Since both the LHS and the RHS are polynomials in $n$ having degree $4$, it is sufficient to check that the identity holds for $n\in\{0,1,2,3,4\}$ to be sure that it holds for any $n$ - since two polynomials with degree $d$ assuming the same values over $d+1$ points are the same polynomial.

Jack D'Aurizio
  • 353,855
  • How do you prove that the LHS is a polynomial of $n$ of degree $4$? –  Nov 08 '14 at 20:19
  • @Rafflesiaarnoldii: In virtue of the identity $\sum_{k=r}^{n}\binom{k}{r}=\binom{n+1}{r+1}$. It is sufficient to write $n^p$ as $p!\binom{n}{p}+\ldots$ and sum over $n$ to be sure that $\sum_{n=1}^{N}n^p$ is a polynomial in $N$ with degree $p+1$. – Jack D'Aurizio Nov 08 '14 at 20:21
  • 1
    So that's two more identities to be proved (by induction or otherwise), when one begins at the level of OP's question. –  Nov 08 '14 at 20:23
  • @Rafflesiaarnoldii: ok, I get your point. Anyway, I think that the "interpolation" technique is worth mentioning. – Jack D'Aurizio Nov 08 '14 at 20:30