# 2014 AIME II Problems/Problem 7

## Problem

Let . Find the sum of all positive integers for which

## Solution 1

First, let's simplify that big ugly sigma notation:

Now we write out the notation and simplify:

Converting to exponential form we have the much nicer equation:

OKAY. Now let's look at the function f. Well we have the base which factors nicely into . And then there's the exponent. Hmm well there's a pi inside. That must count for something. Well, if x is odd, then the exponent will be -1 because the cosine of an odd multiple of pi is always -1. However, if it's an even multiple of pi, the cosine is 1. Remember raising to an exponent of -1 just gives the reciprocal. So we have fractions and then anti-fractions and we're multiplying them? Let's plug in the values without simplifying:

Aha! MASS CANCELATION...however, notice we can't really end because we don't know if the value of n is going to be odd or even. We can prove this mass cancelation happens by simply looking at consecutive functions of f:

Therefore this does indeed cancel and was not a clever trap set by AIME committee. However, we still don't know where to end. So we branch off into 2 cases here:

**Case 1: n is odd**

Okk so if n is odd, then the exponent of f(n) is -1 and we have

Now we simply solve for n in both situations and see which one gives us an integer n:

**Case 2 : n is even**

Okk so if n is even, then the exponent of f(n) is 1 and we have:

Now we simply solve for n in both situations and see which one gives us an integer n:

OKKK FINALLY BACK TO THE SOLUTION:

We've got n=18,3. So the sum is clearly

## Solution2

Note that is when is odd and when is even. Also note that for all . Therefore Because of this, is a telescoping series of logs, and we have Setting each of the above quantities to and and solving for , we get possible values of and so our desired answer is