# Thread: Proof By Induction #2

1. ## Proof By Induction #2

I'm using sigma notation and this is what it breaks down to:

p(k) is k^3= (k^2(k+1)^2)/4

2. Originally Posted by Edbaseball17
I'm using sigma notation and this is what it breaks down to:

p(k) is k^3= (k^2(k+1)^2)/4
What would i do to get p(k+1)

3. Originally Posted by Edbaseball17
I'm using sigma notation and this is what it breaks down to:

p(k) is k^3= (k^2(k+1)^2)/4
Are you asking how to show that
$\sum_{i = 1}^n i^3 = \frac{1}{4} \cdot n^2(n+ 1)^2$?

The base case is n = 1:
$\sum_{i = 1}^1 1^3 = 1$
and
$\frac{1}{4} \cdot 1^2(1 + 1)^2 = \frac{1}{4} \cdot 4 = 1$
so this checks out.

Now we assume the theorem is true for some n = k. That is:
$\sum_{i = 1}^k i^3 = \frac{1}{4} \cdot k^2(k + 1)^2$

So we want to show this for n = k + 1.
$\sum_{i = 1}^{k + 1} i^3 = \sum_{i = 1}^k i^3 + (k + 1)^3 = \frac{1}{4} \cdot k^2(k + 1)^2 + (k + 1)^3$

You need to show that this is the same thing as:
$\frac{1}{4} \cdot (k + 1)^2((k + 1) + 1)^2 = \frac{1}{4} \cdot (k + 1)^2(k + 2)^2$

-Dan

4. kinda, but the right side is all over 4, not just the 1

5. Originally Posted by Edbaseball17
kinda, but the right side is all over 4, not just the 1
sorry, just have to learn how to make all the fancy characters, lol.