Question number 29. The function f taking the integers to the integers satisfies the conditions that the value of f at 4 is 6 and then recursively we have the definition x times f of x is the product of the factor x minus 3 with the value of f at x plus 1. The question is what is the value of this product here where each factor starting with the first one is f evaluated at 3 more than the previous factor. So let's see if there is a pattern to the function f when we evaluate it at consecutive integers. So the first value is given 6 and we're going to use the recursive formula here f of x plus 1 is going to be equal to x times f of x divided by x minus 3. So when x is equal to 3 we have f of 4 is equal to 6 when x is equal to 4 we're going to have f of 5 and now we use our formula 4 times f of 4 divided by 1 so that's 4 times 6 and we can just write 24. Now when x is equal to 5 we have f of 6 is 5 f evaluated at 5 divided by now we have 5 minus 3 so 2 so we have 5 times 24 divided by 2 and that gives us 5 times 12 or 60. When x is equal to 6 we have f evaluated at 7 that is going to be 6 f of 6 divided by 3 so that is 2 times f of 6 and that is 2 times 60 so 120 is the value here and let's see if we can notice a pattern now. Well the first couple of numbers seems to follow the factorials 6 is 3 factorial 3 times 2 times 1 24 is 4 factorial that's 4 times 3 times 2 times 1 but 60 is not 5 factorial 5 factorial is a 120 in fact 60 we can write as 5 times 4 times 3 20 times 3 and 120 we can write as 6 times 5 times 4 so 30 times 4 and so that is our pattern instead of following the factorial we have a product always of three consecutive positive integers so we see a pattern and that pattern is that f evaluated at n for n greater than or equal to 4 is the product if we plug in 7 we have n minus 1 times n minus 2 times n minus 3 for example so that that is our product and that is equal to we can say this is n minus 1 factorial and then we have to divide by n minus 4 factorial so we can do a proof by induction but we don't have time or space here to prove that this formula is actually the pattern that we have seen and it holds in general but we can look at the suggested solutions so C suggested solutions for a proof for a proof by induction so we're going to trust our instincts that this pattern holds and then what do we have well f of 4 would be then given by 3 factorial over 0 factorial f of 7 is next that would be 6 factorial over 3 factorial f of 10 is next and that would be 9 factorial divided by 6 factorial and then if we we see that if we were to multiply all of these together like that then the denominators always cancel and 0 factorial is 1 so we end up looking at always the last factorial here in the numerator so if we keep multiplying them like that we see that f of 2011 is 2000 factorial divided by 1997 factorial and the last factor over here would be when we are looking at 2014 I will just write 2013 factorial divided by 2000 factorial and these cancel again the 1997 will cancel with the previous factor and so the answer here is what remains so we are looking at 2013 factorial in this sort of telescoping product where factors cancel but the answer is here d 2013 factorial

recursive function

telescoping sequence

factorial

product simplification

integer pattern