Geometric progression

**weasley74** · March 25th 2008, 11:47 AM

Just a quick question;

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

**CaptainBlack** · March 26th 2008, 04:07 AM

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 $n$ (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: $\left(\frac{n(n+1)}{2}\right)^2$ .

Expand and show the required equality.

RonL

**CaptainBlack** · March 26th 2008, 04:36 AM

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

**mr fantastic** · March 26th 2008, 04:41 AM

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

**weasley74** · March 26th 2008, 04:47 AM

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

**topsquark** · March 26th 2008, 05:05 AM

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

**weasley74** · March 26th 2008, 05:17 AM

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? =)

**CaptainBlack** · March 26th 2008, 05:52 AM

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

**topsquark** · March 26th 2008, 05:56 AM

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.
$(1)^2 = 1^3$ ? 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
$(1 + 2 + ~...~ + k)^2 = 1^3 + 2^3 + ~...~ + k^3$

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

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

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

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

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

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

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

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

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

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

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

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

-Dan

**topsquark** · March 26th 2008, 05:57 AM

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

**CaptainBlack** · March 26th 2008, 05:58 AM

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

**CaptainBlack** · March 26th 2008, 06:00 AM

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

**CaptainBlack** · March 26th 2008, 06:02 AM

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

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