|
Question 1196482: A function f is defined for integers m and n as given: f(mn)=f(m)f(n)-f(m+n)+ 1001, where either m or n is equal to 1, and f(1)=2.
a) Prove that f(x)=f(x-1)+1001.
b) Find the value of f(9999).
Answer by ikleyn(52788)   (Show Source):
You can put this solution on YOUR website! .
A function f is defined for integers m and n as given: f(mn)=f(m)f(n)-f(m+n)+ 1001, where either m or n is equal to 1, and f(1)=2.
a) Prove that f(x)=f(x-1)+1001.
b) Find the value of f(9999).
~~~~~~~~~~~~~~~
In the given formula, take x = m, n = 1. You will get
f(x*1) = f(x)*f(1) - f(x+1) + 1001, or
f(x) = 2*f(x) - f(x+1) + 1001, which is the same as
0 = f(x) - f(x+1) + 1001,
f(x+1) = f(x) + 1001. (1)
Thus part (a) is just solved and completed.
From the formula (1), we conclude that the sequense (1) is an arithmetic progression
with the first term f(1) = 2 and the common difference of 1001. So
f(9999) = f(1) + (9999-1)*1001 = 2 + (9999-1)*1001 = 10,008,000.
Thus part (b) is completed, too.
Solved.
|
|
|
| |
