# Finding the sum of 1^3 + 2^3 + + n^3 by induction

1. Jul 6, 2014

### jonroberts74

$$1^3+2^3+...+n^3 = \left[ \frac{n(n+1)}{2}\right]^2; n\ge 1$$

$$P(1) = 1^3 = \frac{8}{8} = 1$$

$$P(k) = 1^3+...+k^3 = \left[ \frac{k(k+1)}{2}\right]^2$$ (induction hypothesis)

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

I start getting stuck here
I foiled it out then let m = P(k)

$$\left[ m + \frac{2(k+1)}{2}\right]^2$$

Last edited: Jul 6, 2014
2. Jul 6, 2014

You probably meant $m = \sqrt{P(k)}$. Now expand the square and use the induction hypothesis.

3. Jul 6, 2014

### ehild

You assume that $$P(k) = 1^3+...+k^3 = \left[ \frac{k(k+1)}{2}\right]^2$$ To get P(k+1), you have to add (k+1)^3 to P(k). $$P(k+1)=P(k)+(k+1)^3=\left[ \frac{k(k+1)}{2}\right]^2+(k+1)^3$$ Write it out, and show that it is equal to $$P(k+1) = \left[ \frac{(k+1)(k+2)}{2}\right]^2$$
pull out (k+1)2

ehild

4. Jul 6, 2014

### jonroberts74

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

$$(k+1)^2 \left[\frac{(k+2)}{2}\right]^2=(k+1)^2\left[\frac{k^2}{2^2} + (k+1)\right]$$

$$(k+1)^2 \left[\frac{(k+2)}{2}\right]^2=(k+1)^2\left[\frac{k^2+4k+4}{2^2}\right]$$

$$(k+1)^2 \left[\frac{(k+2)}{2}\right]^2=(k+1)^2\left[\frac{(k+2)^2}{2^2}\right]$$

???

5. Jul 6, 2014

### ehild

Apply that a2b2=(ab)2

$$(k+1)^2 \left[\frac{(k+2)}{2}\right]^2=\left[(k+1)\frac{(k+2)}{2}\right]^2= P(k+1)$$

is it familiar? Is it what you wanted to arrive at?

ehild