# Thread: [SOLVED] Need help with this mathematical induction

1. ## [SOLVED] Need help with this mathematical induction

This is what I have to prove:

$\displaystyle 2(1^3 + 2^3 + ... n^3)$= $\displaystyle \left[\frac{n(n+1)}{2}\right]^2$
I get the first part. I plugged in 1 and it works fine.

Now it's the second step that I have trouble with.

This is what I did:
$\displaystyle 1^3 + 2^3 ... + ^3 + (k+1)^3$ = $\displaystyle \left[\frac{k(k+1)}{2}\right]^2$

Then I added $\displaystyle \frac{2(k+1)^3}{2}$.

So now I have:
$\displaystyle \left[\frac{k(k+1)}{2}\right]^2$ + $\displaystyle \frac{2(k+1)^3}{2}$

How do I go about to simplifying this.

Do I square the denominator on the first equation and pull out a $\displaystyle (k+1)^2$ ? Then put the second equation over 4?

Thank you !

2. I'm not quite sure what you're doing...? It almost looks like you've assumed the k+1 step, and then did something to it, like working backwards...?

First, you need to show that the equation is true for n = 1.

Once you've done that, assume that the equation is true for some n = k:

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

You then do the k+1-th step for the left-hand side, and see where that leads:

$\displaystyle 2(1^3\, +\, 2^3\, +\, ...\, +\, k^3\, +\, (k\, +\, 1)^3)$

$\displaystyle 2(1^3\, +\, 2^3\, +\, ...\, \, k^3)\, +\, 2(k\, +\, 1)^3$

Now substitute from the assumption step:

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

Convert to a common denominator, combine the terms, and see what you get. It might help to note that you're wanting eventually to end up with:

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

3. Originally Posted by stapel
I'm not quite sure what you're doing...? It almost looks like you've assumed the k+1 step, and then did something to it, like working backwards...?

First, you need to show that the equation is true for n = 1.

Once you've done that, assume that the equation is true for some n = k:

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

You then do the k+1-th step for the left-hand side, and see where that leads:

$\displaystyle 2(1^3\, +\, 2^3\, +\, ...\, +\, k^3\, +\, (k\, +\, 1)^3)$

$\displaystyle 2(1^3\, +\, 2^3\, +\, ...\, \, k^3)\, +\, 2(k\, +\, 1)^3$

Now substitute from the assumption step:

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

Convert to a common denominator, combine the terms, and see what you get. It might help to note that you're wanting eventually to end up with:

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

I'm not quite sure what I'm doing myself. This subject was not covered in class, but I'm doing an extra assignment and this is a question.

I'm trying to follow the instructions in the book but they are not very clear.

So now if I put both terms on same denominator I end up with

$\displaystyle \left[\frac{k(k+1)}{2}\right]^2$ + $\displaystyle \frac{8(k+1)^3}{4}$ ?

Should it be: $\displaystyle \frac{(k+1)^2 (k) + 8(k+1)}{4}$ ?

I appreciate your help greatly. I have spent my whole weekend researching these and being stressed out so getting some help is appreciated.

4. Ok so I reworked the problem.

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

I got this

$\displaystyle \frac{(k^2+k)^2}{4}$ + $\displaystyle (k^3+3K^2+3k+1)$

Then I got:

$\displaystyle \frac{(k^4+2k^3 +k^2)}{4}$ + $\displaystyle \frac{4(k^3+3K^2+3k+1)}{4}$

Now when I combine like terms I got

$\displaystyle \frac{k^4+6k^3+13k^2+12k+4}{4}$

And now I'm stuck again. I want to try to factor this out. I'll work some more on it...

5. I got it ! Thanks for the help.