1. ## Proof by Induction

I apologize if this has been posted before. I have done many searches, but most ended up proving the Binomial Theorem. Anyhow, here goes:

Prove $\displaystyle 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2$

I'm basically looking for a shove in the right direction, as my induction step:

$\displaystyle P(k + 1) : 1^3 + 2^3 + ... + k^3 + (k + 1)^3 = ???$

I'm unsure here since the RHS can be: $\displaystyle (1 + 2 + ... + k + [k+1])^2$ or $\displaystyle (2 + 3 + ... + 2k)^2$

which is where my confusion begins. Thoughts?

2. Originally Posted by flabbergastedman
I apologize if this has been posted before. I have done many searches, but most ended up proving the Binomial Theorem. Anyhow, here goes:

Prove $\displaystyle 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2$

I'm basically looking for a shove in the right direction, as my induction step:

$\displaystyle P(k + 1) : 1^3 + 2^3 + ... + k^3 + (k + 1)^3 = ???$

I'm unsure here since the RHS can be: $\displaystyle (1 + 2 + ... + k + [k+1])^2$ or $\displaystyle (2 + 3 + ... + 2k)^2$

which is where my confusion begins. Thoughts?
hi flabbergastedman,

If $\displaystyle 1^3+2^3+3^3+....+n^3=(1+2+3+....+n)^2$

then $\displaystyle 1^3+2^3+3^3+....+n^3+(n+1)^3$ should $\displaystyle =[1+2+3+...+n+(n+1)]^2$

Hence

$\displaystyle (1+2+3+....+n+(n+1))^2$$\displaystyle =(1+2+3+...+n)^2+(n+1)[1+2+3+...+n+(n+1)]+(1+2+3+...+n)(n+1)$

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

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

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

giving

$\displaystyle [1+2+3+...+n+(n+1)]^2=(1+2+3+...+n)^2+n(n+1)^2+(n+1)^2$

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

Proven

3. Originally Posted by flabbergastedman
I apologize if this has been posted before. I have done many searches, but most ended up proving the Binomial Theorem. Anyhow, here goes:

Prove $\displaystyle 1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2$

I'm basically looking for a shove in the right direction, as my induction step:

$\displaystyle P(k + 1) : 1^3 + 2^3 + ... + k^3 + (k + 1)^3 = ???$

I'm unsure here since the RHS can be: $\displaystyle (1 + 2 + ... + k + [k+1])^2$ or $\displaystyle (2 + 3 + ... + 2k)^2$

which is where my confusion begins. Thoughts?

$\displaystyle 1^3+2^3+\ldots+k^3+(k+1)^3=\left(1+2+\ldots+k\righ t)^2+(k+1)^3$ , using the inductive hypothesis, and now use that $\displaystyle 1+2+\ldots +n=\frac{n(n+1)}{2}$

Tonio