# Thread: Geometric progression

1. ## Geometric progression

Just a quick question;

How do I show that (1+2+...+n)^2 = 1^3+2^3+...+n^3 ?

2. Originally Posted by weasley74
Just a quick question;

How do I show that (1+2+...+n)^2 = 1^3+2^3+...+n^3 ?
The sum on the right can be shoen to be a quartic in $\displaystyle n$ (construct the
difference table and it becomes obviouse).

Which quartic can be established by fitting the first few terms to the
quartic.

The sum on the left is: $\displaystyle \left(\frac{n(n+1)}{2}\right)^2$.

Expand and show the required equality.

RonL

3. Originally Posted by weasley74
Just a quick question;

How do I show that (1+2+...+n)^2 = 1^3+2^3+...+n^3 ?
Second method from the department of incredibly elegant but indirect proofs.

The expression on the left is obviously a quartic in n, since the table of
second differences of what is inside the square is constant it is a quadratic in
n and so its square is a quartic in n.

As explained before what is on the right is also a quartic.

To show that two quartics are identical you need only show that they are
equal at five distinct points. So compute the first 5 values of what is on the
left and the corresponding values for what is on the right and if the
corresponding terms are equal the identity is proven.

RonL

4. Originally Posted by CaptainBlack
Second method from the department of incredibly elegant but indirect proofs.

The expression on the left is obviously a quatic in n, since the table of
second differences of what is inside the square is constant it is a quadratic in
n and so its square is a quartic in n.

As explained before what is on the right is also a quartic.

To show that two quartics are identical you need only show that they are
equal at five distinct points. So comopute the first 5 values of what is on the
left and the corresponding values for what is on the right and if the
corresponding terms are equal the identity is proven.

RonL

5. So I just have to show that the five first values are equal? it sounds a bit too easy..

6. Originally Posted by weasley74
So I just have to show that the five first values are equal? it sounds a bit too easy..
Well, if you want something more "logic" oriented, you could always do a proof using Mathematical Induction.

-Dan

7. Originally Posted by topsquark
Well, if you want something more "logic" oriented, you could always do a proof using Mathematical Induction.

-Dan
and how do I do that? =)

8. Originally Posted by topsquark
Well, if you want something more "logic" oriented, you could always do a proof using Mathematical Induction.

-Dan

I was just about to post method 3, Mathematical Induction (which I can
confirm does work quite nicely, but is not as elegant as method 2).

RonL

9. Originally Posted by weasley74
Just a quick question;

How do I show that (1+2+...+n)^2 = 1^3+2^3+...+n^3 ?
Let n = 1.
$\displaystyle (1)^2 = 1^3$? True.

So assume the theorem is true for some n = k. Then we need to prove that it is true for n = k + 1.

Our assumption is that
$\displaystyle (1 + 2 + ~...~ + k)^2 = 1^3 + 2^3 + ~...~ + k^3$

What we wish to prove is
$\displaystyle (1 + 2 + ~...~ + k + (k + 1))^2 = 1^3 + 2^3 +~...~+k^3 + (k + 1)^3$

Now,
$\displaystyle (1 + 2 + ~...~ + k + (k + 1))^2 = (1 + 2 + ~...~ + k)^2 + (k + 1)^2 + 2(1 \cdot (k + 1) + 2 \cdot (k + 1) + ~...~ k(k + 1))$

$\displaystyle = (1^3 + 2^3 + ~...~ + k^3) + (k + 1)^2 + 2(1 \cdot (k + 1) + 2 \cdot (k + 1) + ~...~ k(k + 1))$
according to our assumption. So plugging this back into what we need to prove, we see that we need to show
$\displaystyle = (1^3 + 2^3 + ~...~ + k^3) + (k + 1)^2 + 2(1 \cdot (k + 1) + 2 \cdot (k + 1) + ~...~ k(k + 1))$$\displaystyle = 1^3 + 2^3 +~...~+k^3 + (k + 1)^3$

or
$\displaystyle (k + 1)^2 + 2(1 \cdot (k + 1) + 2 \cdot (k + 1) + ~...~ k(k + 1)) = (k + 1)^3$

So let's work a bit more on that left hand side.
$\displaystyle (k + 1)^2 + 2(1 \cdot (k + 1) + 2 \cdot (k + 1) + ~...~ k(k + 1))$

$\displaystyle = (k^2 + 2k + 1) + 2(k + 1)(1 + 2 + ~...~ + k)$

and the sum of the first k numbers is
$\displaystyle = (k^2 + 2k + 1) + 2(k + 1) \frac{k(k + 1)}{2}$

$\displaystyle = (k^2 + 2k + 1) + k(k + 1)^2$

$\displaystyle = k^2 + 2k + 1 + k^3 + 2k^2 + k$

$\displaystyle = k^3 + 3k^2 + 3k + 1$

$\displaystyle = (k + 1)^3$
as we needed.

So the theorem is true for n = 1 and thus for all integers by induction.

-Dan

10. Originally Posted by CaptainBlack
I was just about to post method 3, Mathematical Induction (which I can
confirm does work quite nicely, but is not as elegant as method 2).

RonL
Agreed. And Mathematical Induction here is a bit tedious.

-Dan

11. Originally Posted by weasley74
and how do I do that? =)
First you should know how induction works:

You have a proposition about a netural number n, such as your identity.

Show it holds for a base case, typical n=1 or n=0.

The if you assume that it hold for some k, show this implies it holds for k+1

Then you are done the general proposition is proven for all n greater than or
equal to the base case.

Checking the base case is trivial in this case as for n=1 both sides of the
equality are 1.

I will let you have a go at proving the induction step its not difficult.

RonL

12. Originally Posted by weasley74
So I just have to show that the five first values are equal? it sounds a bit too easy..
That part is easy, the key part of the argument is identifying that both
sides are quartics.

RonL

13. Originally Posted by topsquark
Agreed. And Mathematical Induction here is a bit tedious.

-Dan
On my a5 note pad its half a side of paper!

RonL