Grades 11-12 Video Solutions 2011-11&12 Video Solutions 2011 problem23

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Question number 23. Functions given by the sequence f1 of x, f2 of x, f3 of x, and so on, satisfy the following conditions.  f1 of x is equal to x, and then recursively for positive integers n, we have fn plus 1 of x equal to 1 over 1 minus fn of x. What is the value of f2011 evaluated at 2011?  We can write out more terms of this sequence and maybe we will find a pattern. So let's begin with n is equal to 1, we have f of 2 then is 1 over 1 minus f1 of x, and  that is given as x by definition, so f2 of x is equal to 1 over 1 minus x. Next we have with n equal to 2, f3 is given by 1 over 1 minus f2 of x.  We copy that from the previous step and simplify, first with a common denominator, and that gives us 1 divided by negative x divided by 1 minus x, which is x minus 1 divided by x.  So that's our function f3. Let's do one more. When n is equal to 3, we have f4. We copy the definition of f3 from the previous step, and simplifying first in the denominator,  we have 1 over 1 over x, which is x. And then this is exactly the same as the function f1, so we note that f1 is equal to f4, and  that will be equal to f7, which will be equal to f10, and so on. There will be some f sub m here.  All of these are equal to x, and that number m is divisible by 3 with remainder 1, so that's 3 times some integer plus 1, where k here is 1, 2, 3, and so on.  So now we have to find f of 2011, the function f2011 evaluated at 2011. That is going to be equal to 2011 because f2011 of x is equal to x.  Since 2011 is equal to 2010 plus 1, and 2010 is a number that's divisible by 3, the sum of its digits is divisible by 3. So our answer here is a, 2011.

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