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# Sum of a Logarithm Function

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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!

Aug 17, 2023

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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   Aug 17, 2023