Question 1209928
<br>
Given:
f(1)=7
f(a+b)=f(a)+f(b)-ab<br>
To find: f(123)<br>
Find f(2):
f(2)=f(1+1)=f(1)+f(1)-1*1=7+7-1=13<br>
Find f(3):
f(3)=f(1+2)=f(1)+f(2)-2=7+13-2=18<br>
Find f(4):
f(4)=f(1+3)=f(1)+f(3)-3=7+18-3=22<br>
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<br>
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<br>
The recursive definition appears to be valid.<br>
The values of f(1) to f(5) form a sequence with a clear pattern:<br>
7, 13, 18, 22, 25, ...<br>
The differences between successive terms are decreasing by 1.<br>
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.<br>
Here is a display of the first few terms of the sequence and the first and second differences:<br><pre>

   7   13   18   22   26
     6    5    4    3
       -1   -1   -1</pre>
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<br>
{{{t(n)=(-1/2)n^2+an+b}}}<br>
To find the coefficients a and b, we can compare the given sequence to the sequence formed by the polynomial {{{(-1/2)n^2}}}.<br><pre>

   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
   ...</pre>
We can see that the differences are just (15/2)n, so the explicit formula for the n-th term is<br>
{{{t(n)=(-1/2)n^2+(15/2)n}}}
{{{t(n)=(15n-n^2)/2}}}
{{{t(n)=(n(15-n))/2}}}<br>
So<br>
ANSWER: {{{t(123)=(123(15-123))/2=((123)(-108))/2=-6642}}}<br>