1. ## Induction Help!!

Question
Prove by induction that n in set positive

1**3 + 2**3 + 4**3 ... n**3 = n**2(n+1)**2/4

base case n = 1

n**(n+1)**2/4

1**2(1+1)**2/4

1(2)**2/4

4/4 = 1 so true

inductive step n=k

n=k+1

1**3 + 2**3 + 4**3 ... k**3 = k**2(k+1)**2/4

k**2(k+1)**2/4 + (k+1)**3 = (k+1)**2 (k+1+1)**2/4

k**2(4+1)**2/4+ 4(k+1)**3/4 = (k**2+1)(k+2)**2/4

Am i on the right track ?

2. ## Re: Induction Help!!

Your post is extremely hard to read. What I think you are trying to prove is:

$\displaystyle \sum_{i = 1}^n i^3 = \dfrac{n^2(n+1)^2}{4}$

It is clear this holds for $n = 1$.

Assuming that:

$\displaystyle \sum_{i = 1}^k i^3 = \dfrac{k^2(k+1)^2}{4}$ for a given $k$, we have:

$\displaystyle \sum_{i = 1}^{k+1} i^3 = \sum_{i = 1}^k i^3 + (k+1)^3 = \dfrac{k^2(k+1)^2}{4} + (k+1)^3$

$= \dfrac{(k+1)^2}{4}[k^2 + 4(k+1)]$

and you may continue from here. You are not supposed to just "replace" $k$ with $k+1$ and see if it works.

3. ## Re: Induction Help!!

Thanks much appreciated