Lesson Upper league problem on evaluating the sum

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Upper league problem on evaluating the sum


Problem 1

If   f(n) = log%28%28n%29%29%2Flog%28%282006n-n%5E2%29%29,    find   f(1) + f(2) + f(3) + . . . + f(2005).

Solution 

Let m = 2006 - n, and consider the sum of two symmetric terms

    f(n) + f(m) = log%28%28n%29%29%2Flog%28%282006n-n%5E2%29%29 + log%28%28m%29%29%2Flog%28%282006m-m%5E2%29%29.


Notice that  2006m-m^2 = 2006*(2006-n) - (2006-n)^2 = 2006^2 - 2006n - 2006^2 + 2*2006n - n^2 = 2006n-n^2.


So, f(n) and f(m) have THE SAME DENOMINATOR value of log(2006n-n^2).   


    +---------------------------------------------------------+
    |  This remarkable observation is a KEY to the solution.  |
    +---------------------------------------------------------+


Thus we can write these two addends and their sum with the common denominator

    f(n) + f(m) = log%28%28n%29%29%2Flog%28%282006n-n%5E2%29%29 + log%28%28m%29%29%2Flog%28%282006n-n%5E2%29%29 = %28log%28%28n%29%29+%2B+log%28%28m%29%29%29%2Flog%28%282006n-n%5E2%29%29


which we can continue

    f(n) + f(m) = log%28%28nm%29%29%2Flog%28%282006n-n%5E2%29%29    (1)


Next,  nm = n*(2006-n) = 2006n-n^2,

so, in (1)  the numerator is the same as the denominator.


It gives us  f(n) + f(m) = 1 for each and every pair of positive integers (n,m) such that  m = 2006-n.


Thus the symmetric pairs give the sum of 1 taken 1002 times, i.e. the sum of 1002.


The central (unpaired) term at n= 1003 gives f(n) = f(1003) = log%28%281003%29%29%2Flog%28%282006%2A1003-1003%5E2%29%29 = log%28%281003%29%29%2Flog%28%281003%5E2%29%29 = log%28%281003%29%29%2F%282%2Alog%28%281003%29%29%29 = 1%2F2.


Therefore, the total sum and the  highlight%28highlight%28ANSWER%29%29  is 10021%2F2 = 1002.5.


ANSWER.  The total sum is 10021%2F2 = 1002.5.


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