# Simple Sequences

1. Oct 14, 2008

### ila

If there is a sequence of numbers 1,4,9,16...... what is the 12th term in the sequence (where 1 is the first term). The textbook says that it is 144 since each term of the sequence is the square of the term value. I find (and have always found) this to be confusing, why can't an alternate algorithm generate those few numbers, namely add 3 for the first term, five to the second term, 7 to the second term. Now there there appears to me an obvious pattern, the amount that one is adding to each term is increasing by 2 in the number of terms. I still haven't figured out an expression for the nth term for this alternative though. The general point though is this, with these sequences if we have given only a few numbers why can't we propose many different algorithms and then suggest that that is the correct generating process?

2. Oct 14, 2008

### CRGreathouse

1. There are uncountably many integer sequences that continue the partial sequence you were given.
2. Your sequence happens to be the same as your teacher's: the squares are the sum of consecutive odd numbers!

3. Oct 14, 2008

### d_leet

You are certainly correct that there are other algorithms that would generate this same sequence for the first few terms. So strictly speaking the next number could be anything. Ironically though if you continue you the sequence with your pattern you will arrive at exactly the same sequence the book does, namely the perfect squares. And it is not too terribly difficult to prove this fact, that the sum of the first n odd numbers is n^2.

4. Oct 14, 2008

### ila

I had a sinking suspicion that that may have been the case.

5. Oct 14, 2008

### robert Ihnot

All you have to do is see that n^2 = ((n-1) +1)^2 = (n-1)^2+2(n-1)+1 = (n-1)^2 + (2n-1). So that, 1, 1+3=4, 4+5=9, 9+7= 16. So that each term increases by the next odd number, and is a square.

Thus the nth term is $$\sum_1^n (2j-1)=n^2.$$

Last edited: Oct 15, 2008