1. ## Mathematical Induction Proof

My question is to prove this by Induction, 1^3 + 2^3 + … + n^3 = n^2(n+1)^2/4

I got the base step, but I am struggling on the Inductive Step. Here is what I have:
Assume Pk is true
Consider Pk+1
Σ i^3 (from i=1 to k+1) = Σ i^3 (from i=1 to k) + (k+1)
= k^2(k+1)/4 + (k+1)
= (k+1)[k(k+1)/4 + 1]
= (k+1)[k(k+1)+4/4]
Therefore, by Induction Pn is true for all n≥1
I was told that my answer is incorrect. Any help on where I went wrong?

2. Originally Posted by page929
My question is to prove this by Induction, 1^3 + 2^3 + … + n^3 = n^2(n+1)^2/4

I got the base step, but I am struggling on the Inductive Step. Here is what I have:
Assume Pk is true
Consider Pk+1
Σ i^3 (from i=1 to k+1) = Σ i^3 (from i=1 to k) + (k+1) incorrect
= k^2(k+1)/4 + (k+1)
= (k+1)[k(k+1)/4 + 1]
= (k+1)[k(k+1)+4/4]
Therefore, by Induction Pn is true for all n≥1
I was told that my answer is incorrect. Any help on where I went wrong?
Hi page929,

As you are summing cubes,

$\displaystyle \displaystyle\sum_{i=1}^{ k+1}i^3=1^3+2^3+...+k^3+(k+1)^3=\sum_{i=1}^ki^3+(k +1)^3$

and you should be attempting to show that this is $\displaystyle \displaystyle\frac{(k+1)^2(k+1+1)^2}{4}$

3. I think I have it now. Thanks,