# Thread: 1^3 +....+n^3 =... (induction)

1. ## 1^3 +....+n^3 =... (induction)

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

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

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

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

from here I foiled and let m = p(k) for

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

$\left[ m+k+1\right]^2$

2. ## Re: 1^3 +....+n^3 =... (induction)

Originally Posted by Jonroberts74
$1^3+2^3+...+n^3 = \left[ \frac{n(n+1)}{2}\right]^2; n\ge 1$

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

$P(k) = 1^3+...+k^3 = \left[ \frac{k(k+1)}{2}\right]^2$
Okay, you want to prove that if P(k) is true then so is P(k+1).

$P(k+1) = 1^3+...+k^3+(k+1)^3 = \left[\frac{(k+1)(k+2)}{2}\right]^2$
No, you can't say this. You don't know it is true, this what you want to prove
What you can say is that $1^3+ ...+ k^3+ (k+ 1)^3= \left(1^3+ ...+ k^3\right)+ (k+1)^3= \left[\frac{k(k+1)}{2}\right]^2+ (k+1)^3$

Now try to show that the right side is the same as $\left[\frac{(k+1)(k+2)}{2}\right]^2$. For example you can factor $(k+1)^2$ out of both terms.

from here I foiled and let m = p(k) for

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

$\left[ m+k+1\right]^2$

4. ## Re: 1^3 +....+n^3 =... (induction)

I thought you assume P(k) to be true then show P(k+1) is true therefore P(k) is also true

I also can't see what the image is, even opening it it is too small for me. I can only read the title of it, I googled it but couldn't find the same image

5. ## Re: 1^3 +....+n^3 =... (induction)

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

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

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

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

??