SOLUTION: The function f satisfies f(a + b) = f(a) + f(b) - ab for all nonnegative integers a and b, and f(1) = 7. Compute f(123).

Question 1209928: The function f satisfies
f(a + b) = f(a) + f(b) - ab
for all nonnegative integers a and b, and f(1) = 7. Compute f(123).

Found 2 solutions by CPhill, greenestamps:
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let the given equation be
$$f(a+b) = f(a) + f(b) - ab$$
for all nonnegative integers $a$ and $b$, and $f(1) = 7$.
First, we find $f(2)$:
$$f(2) = f(1+1) = f(1) + f(1) - (1)(1) = 7 + 7 - 1 = 13$$
Next, we find $f(3)$:
$$f(3) = f(2+1) = f(2) + f(1) - (2)(1) = 13 + 7 - 2 = 18$$
Next, we find $f(4)$:
$$f(4) = f(3+1) = f(3) + f(1) - (3)(1) = 18 + 7 - 3 = 22$$
Let's try to find a pattern.
$f(1) = 7 = 1^2 + 6(1)$
$f(2) = 13 = 2^2 + 6(2) + 1$
$f(3) = 18 = 3^2 + 6(3) - 9 = 9+18 = 27$
$f(3) = 18 = 3^2 + 6(3) = 9 + 9 = 18$
$f(4) = 22 = 4^2 + 6(4) - 18 = 16+24-18=22$
We can guess that $f(x) = x^2 + 6x$.
Let's check if this satisfies the given equation.
$$f(a+b) = (a+b)^2 + 6(a+b) = a^2 + 2ab + b^2 + 6a + 6b$$
$$f(a) + f(b) - ab = (a^2 + 6a) + (b^2 + 6b) - ab = a^2 + 6a + b^2 + 6b - ab$$
These are not equal. Let's try to find a pattern.
Let $f(x) = Ax^2 + Bx$.
$f(1) = A+B = 7$
$f(2) = 4A + 2B = 13$
Multiply the first equation by 2:
$2A + 2B = 14$
Subtract this from the second equation:
$2A = -1$
$A = -1/2$
$B = 7 - A = 7 + 1/2 = 15/2$
Thus $f(x) = -\frac{1}{2} x^2 + \frac{15}{2} x$.
Check:
$f(a+b) = -\frac{1}{2} (a+b)^2 + \frac{15}{2} (a+b) = -\frac{1}{2} (a^2 + 2ab + b^2) + \frac{15}{2} a + \frac{15}{2} b$
$f(a) + f(b) - ab = (-\frac{1}{2} a^2 + \frac{15}{2} a) + (-\frac{1}{2} b^2 + \frac{15}{2} b) - ab = -\frac{1}{2} a^2 - \frac{1}{2} b^2 + \frac{15}{2} a + \frac{15}{2} b - ab$
$f(a+b) = f(a) + f(b) - ab$
So $f(x) = -\frac{1}{2} x^2 + \frac{15}{2} x$.
Now, compute $f(123)$:
$f(123) = -\frac{1}{2} (123)^2 + \frac{15}{2} (123) = \frac{1}{2} (15(123) - 123^2) = \frac{123}{2} (15 - 123) = \frac{123}{2} (-108) = 123(-54) = -6642$
Final Answer: The final answer is $\boxed{-6642}$

Answer by greenestamps(13206) (Show Source): You can put this solution on YOUR website!

Given:
f(1)=7
f(a+b)=f(a)+f(b)-ab

To find: f(123)

Find f(2):
f(2)=f(1+1)=f(1)+f(1)-1*1=7+7-1=13

Find f(3):
f(3)=f(1+2)=f(1)+f(2)-2=7+13-2=18

Find f(4):
f(4)=f(1+3)=f(1)+f(3)-3=7+18-3=22

Find f(4) in a different way to make sure the recursive definition is valid:
f(4)=f(2+2)=f(2)+f(2)-4=13+13-4=22

Find f(5) in two different ways:
f(5)=f(1+4)=f(1)+f(4)-4=7+22-4=25
f(5)=f(2+3)=f(2)+f(3)-6=13+18-6=25

The recursive definition appears to be valid.

The values of f(1) to f(5) form a sequence with a clear pattern:

7, 13, 18, 22, 25, ...

The differences between successive terms are decreasing by 1.

To find the value of f(123), we want to have an explicit formula for the n-th term. One way we can find that formula is using the method of finite differences.

Here is a display of the first few terms of the sequence and the first and second differences:
7 13 18 22 26 6 5 4 3 -1 -1 -1

The constant difference of 1- means the sequence can be produced with a polynomial of degree 2 with leading coefficient -1/(2!) = -1/2. So the sequence can be formed with a polynomial of the form

To find the coefficients a and b, we can compare the given sequence to the sequence formed by the polynomial .
t(n) (-1/2)n^2 an+b ---------------------- 7 -1/2 15/2 13 -2 15 = 30/2 18 -9/2 45/2 22 -8 30 = 60/2 ...

We can see that the differences are just (15/2)n, so the explicit formula for the n-th term is

So

ANSWER:

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