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
Let n = 1.
? 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
What we wish to prove is
Now,
according to our assumption. So plugging this back into what we need to prove, we see that we need to show
or
So let's work a bit more on that left hand side.
and the sum of the first k numbers is
as we needed.
So the theorem is true for n = 1 and thus for all integers by induction.
-Dan
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