Question 1203303
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Claim:
f(n) + f(2006-n) = 1 where n is an integer in the set {1,2,3,...,2005}


Here's a proof:
{{{f(n) = log((n))/log((2006n-n^2))}}}


{{{f(n) = log((n))/log((n(2006-n)))}}}


{{{f(2006-n) = log((2006-n))/log(((2006-n)(2006-(2006-n))))}}}


{{{f(2006-n) = log((2006-n))/log(((2006-n)(2006-2006+n)))}}}


{{{f(2006-n) = log((2006-n))/log(((2006-n)n))}}}


{{{f(2006-n) = log((2006-n))/log((n(2006-n)))}}}
Both f(n) and f(2006-n) have the same denominator.


It allows us to add the numerators and use the log rule log(A)+log(B) = log(AB)
{{{f(n)+f(2006-n) = log((n))/log((n(2006-n)))+log((2006-n))/log((n(2006-n)))}}}


{{{f(n)+f(2006-n) = (log((n))+log((2006-n)))/log((n(2006-n)))}}}


{{{f(n)+f(2006-n) = (log((n(2006-n))))/log((n(2006-n)))}}}


{{{f(n)+f(2006-n) = 1}}}


Then using that formula we can write
f(1) + f(2005) = 1
f(2) + f(2004) = 1
f(3) + f(2003) = 1
f(4) + f(2002) = 1
...
...
f(2002) + f(4) = 1
f(2003) + f(3) = 1
f(2004) + f(2) = 1
f(2005) + f(1) = 1
We have 2005 pairs of values. Each pair adds to 1.
We have a sum of 1+1+1+...+1 = 2005*1 = 2005


This appears it could be the final answer. However, it is not. Pay close attention that double-counting has taken place here.
For instance, f(1)+f(2005) would be treated the same as f(2005)+f(1).


To correct this error, divide by 2.
2005/2 = <font color=red>1002.5</font>


Put another way, if we add straight down through all 2005 equations and do a bit of algebraic housekeeping, the left hand sides combine to:
2*( f(1)+f(2)+f(3)+...+f(2005) )
while the right hand sides combine to 1+1+1+..+1, aka 2005 copies of '1' being added.
Therefore, we have the equation
2*( f(1)+f(2)+f(3)+...+f(2005) ) = 2005
which rearranges to
f(1)+f(2)+f(3)+...+f(2005) = 2005/2 = <font color=red>1002.5</font>




<font color=black size=4>Answer:</font>
f(1) + f(2) + f(3)+...+ f(2005) = <font color=red size=4>1002.5</font>
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