2013 AMC 12A Problems/Problem 21
Let and , and from the problem description,
We can reason out an approximation, by ignoring the :
And a better approximation, by plugging in our first approximation for in our original definition for :
And an even better approximation:
Continuing this pattern, obviously, will eventually terminate at our original definition of .
However, at , going further than our second approximation will not distinguish between our answer choices.
So we take our second approximation and plug in.
Since , we know . This gives us our answer range:
We see that , and that the quantity within "" is between and , i.e. , therefore , or .