Grades 11-12 Video Solutions 2021-video 2021 11-12/23

Video Transcription

Problem number 23 states, the function f of x is such that f of x plus y is equal to f of x times f of y, and f of 1 is equal to 2.  What is the value of f of 2 over f of 1 plus f of 3 over f of 2, on and on, plus f of 2021 over f of 2020?  So these are the two equations that the problem gave us. Let's try and evaluate f of n. This is not possible with our current knowledge, so let's break it up into a sum.  We can say that f of n minus 1 plus 1, and now we have an x and y value to plug into our second equation.  So by the second identity, we can say that that's f of n minus 1 times f of 1. And we know what f of 1 is, we know that it is 2. So f of n is equal to 2 times f of n minus 1.  So now let's try evaluating the sum with this knowledge. So we're summing up fractions where the numerator is f of n, and the denominator would be f of n minus 1.  Using the identity we found earlier, we can see that f of n can be substituted as 2 times f of n minus 1. So the f of n minus 1's will cancel out, and we will just get 2.  This means that every one of these fractions, regardless of what the value of n is, will result in 2.  So all that this is going to be is just a massive sum of 2's, and we're going to have 2,020 of them. So let's multiply 2 by 2,020 to get our final answer of 4,040.  So the question asked us, what is the

Video Summary

The problem involves determining the value of a sum using a function \( f(x) \) defined by two equations: \( f(x+y) = f(x)f(y) \) and \( f(1) = 2 \). To solve it, we evaluate \( f(n) = 2^n \) by applying the given properties. The sum in question is \( \sum_{n=1}^{2020} \frac{f(n+1)}{f(n)} \). Substituting \( f(n) = 2^n \) results in each term simplifying to 2, making the sum equal to \( 2 \times 2020 = 4040 \). Thus, the final answer is 4040.

Keywords

functional equation

sum evaluation

exponential function

mathematical solution

series simplification