SOLUTION: 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

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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(52793)   (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.



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