Hi Mike.
You're thinking along the right lines. Let's look at that sequence {4, 12, 24, 40...} more closely.
Those are all products of 4, so if we divide by 4, we get this {1, 3, 6, 10, ... } which is just the
sum of all the previous numbers { 1, 1+2, 1+2+3, 1+2+3+4, ... }. How do we express the sum of integers starting at 1 and ending at n?
Here's a short proof using "mathematical induction":
Theorem: The sum of all positive integers up to n equals n(n+1)/2.
Proof:
When n = 1, n(n+1)/2 = 1(1+1)/2 = 1, which is the first element in the {1, 3, 6, 10...} sequence, and that's what we want. Now let's assume that this works for all values of n up to k, that is for n from {1, 2, 3, ..., k} for some integer k. In particular, we know that the k(k+1)/2 is the kth element in the list. Now we will try to show that it works for the next element in the list as well.
1+2+3+...+k+(k+1)
= [ k(k+1)/2 ] + (k+1)
= (k*k + k + 2k + 2) / 2
= (k*k +3k + 2) / 2
and we can factor this as:
= (k + 1) (k + 2) / 2.
And that's the same as replacing the "k" in k(k+1)/2 with "k+1". So if it works for one value of k (and we showed that it does already; k=1), then it works for the next value of k (k=2), and the next and the next forever. So we have proved, using "mathematical induction" that the numbers in the sequence {1, 3, 6, 10, ... } can be generated using the formula n(n+1)/2.
To get the original sequence {4, 12, 24, 40, ...} we just multiply by 4.
Now you know how to express the nth term of this series too.
Hope this helps,
Stephen La Rocque>
