+6 If \(f(n) = \frac{log (n)}{log(2006n-n^2)}\), find the sum of \(f(1) + f(2) + f(3) + ... + f(2005).\)
I want some help on this, I haven't been able to deduct anything from it except for making the function a bit simpler:
\(f(n) = \frac{log (n)}{log (n) + log (2006-n)}\)
Thank you!
+129883 I like your thinking !!!
Note that we can first add f(1) + f(2005) and we get
log (1) / ( log (1) + log (2005) ) + log (2005) / ( log (2005) + log (1) ) =
[ log (1) + log (2005) ] / [ log(1) +Log (2005)] = 1
Next we pair f(2) + f(2004) and we have
log (2) / ( log 2 + log 2004) + log (2004) /( log (2004) + log (2) ) = (log 2 + log 2004) /(log 2 + log 2004) = 1
Continuing this pattern....We end up with (2005 - 1 ) / 2 = 1002 pairs that each evalute to 1
And we have one evaluation which is n = (2005 + 1) / 2 = 1003
The evaluation of this is log (1003) /( log 1003 + log (2006 -1003)) =
log (1003) / ( log 1003 + log 1003) = 1* log (1003) / 2 *log ( 1003 ) = 1/2 = .5
So.....the sum is 1002 + 1/2 = 1002.5
