# Prove that for each n in N, 1^3+2^3+ +n^3=[n(n+1)/2]^2

1. Aug 29, 2011

### A93

prove that for each n in N, 1^3+2^3+...+n^3=[n(n+1)/2]^2

2. Aug 29, 2011

### micromass

Staff Emeritus
What did you try?? If you let us know where you're stuck, then we'll know how to help...

3. Aug 29, 2011

### A93

induction.
my dumb butt got "lost" in the wanting to prove p(k+1) is true (if that even makes sense, lol)

4. Aug 29, 2011

### micromass

Staff Emeritus
and where are you stuck in induction??

5. Aug 29, 2011

### A93

the proving of p(k+1) is true.
basically the area where for each k>=1, if P(k) is true, then p(k+1) is true.....basically the induction part, lol

6. Aug 29, 2011

### micromass

Staff Emeritus
Yes, so to show that p(k+1) is true, you need to prove that

$$1^3+2^3+...+k^3+(k+1)^3=\left(\frac{(k+1)(k+2)}{2}\right)^2$$

Now, what happens if you appy "p(k) is true" on that??

7. Aug 29, 2011

### lineintegral1

Induction has a few steps. Let's see if this clarifies them a bit,

1) Base Case: Show that your summation formula works for k = 1 case (which is probably easiest here lol)

2) Induction Case: Create an induction hypothesis. For this case, you assume that the kth case holds. In other words,

$\sum_{k=1}^{n}k^3=\left (\frac{n(n+1)}{2} \right )^2$

is true. Now, show that the kth case implies the (k+1)th case. How do you think you can do this?

Last edited: Aug 29, 2011