Mock AIME 1 2006-2007 Problems/Problem 13
Let , , and be geometric sequences with different common ratios and let for all integers . If , , , , , and , find .
The problem tempts us to simply introduce six variables: the first term and the ratio for each of the three geometric sequences. The problem statement then gives us six independent (non-linear) equations in these variables, which does uniquely determine these variables - but there is no obvious way of computing them. We will show a different solution.
Consider the polynomial with , and distinct.
Let , , and . Then we can write .
Now consider the recurrence . ( is called the characteristic polynomial of this recurrence.)
We can easily verify that each of the three sequences , and satisfies the recurrence. Moreover, we know the following facts:
- If a sequence satisfies the recurrence, then for all the sequence does, too.
- If sequences and satisfy the recurrence, then the sequence does, too.
Thus each linear combination of the three sequences we have satisfies the recurrence. In other words, each sequence of the form solves the recurrence. And vice versa, each solution of the recurrence can be written in this form, because given the values , and one can always uniquely determine the coefficients .
Solving the task
We just proved the following observation: If we have three geometric sequences with distinct ratios , , and , their sum always satisfies the recurrence for the defined above, and some determined by the initial values of these sequences.
In our situation, we do not need to know the ratios , , and . All we need to compute are the coefficients , , and .
But this is easy. We know that:
This is a set of three linear equations. In our case, it has a unique solution , hence .