document.write( "Question 1203303: Hi, I was wondering if I could get some help on this problem?
\n" ); document.write( "If f(n) = $\"log%28n%29%2Flog%282006n-n%5E2%29\"$ , find f(1) + f(2) + f(3)+...+ f(2005)
\n" ); document.write( "All I can figure out is the split the logarithm on the bottom into $\"log%28n%29+%2B+log%282006+-+n%29\"$
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Algebra.Com's Answer #838750 by ikleyn(52803) $\"\"$ $\"About$

You can put this solution on YOUR website!
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\n" ); document.write( "If f(n) = $\"log%28%28n%29%29%2Flog%28%282006n-n%5E2%29%29\"$ , find f(1) + f(2) + f(3) + . . . + f(2005).
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document.write( "Let m = 2006 - n, and consider the sum of two symmetric terms\r\n" );
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document.write( "    f(n) + f(m) =  + .\r\n" );
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document.write( "Notice that  2006m-m^2 = 2006*(2006-n) - (2006-n)^2 = 2006^2 - 2006n - 2006^2 + 2*2006n - n^2 = 2006n-n^2.\r\n" );
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document.write( "So, f(n) and f(m) have THE SAME DENOMINATOR log(2006n-n^2).   \r\n" );
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document.write( "    |  This remarkable observation is a KEY to the solution.  |\r\n" );
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document.write( "Thus we can write these two addends and their sum with the common denominator\r\n" );
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document.write( "    f(n) + f(m) =  +  = \r\n" );
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document.write( "which we can continue\r\n" );
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document.write( "    f(n) + f(m) =     (1)\r\n" );
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document.write( "Next,  nm = n*(2006-n) = 2006n-n^2,\r\n" );
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document.write( "so, in (1)  the numerator is the same as the denominator.\r\n" );
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document.write( "It gives us  f(n) + f(m) = 1 for each and every pair of positive integers (n,m) such that  m = 2006-n.\r\n" );
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document.write( "Thus the symmetric pairs give the sum of 1 taken 1002 times, i.e. the sum of 1002.\r\n" );
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document.write( "The central (unpaired) term at n= 1003 gives f(n) = f(1003) =  =  =  = .\r\n" );
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document.write( "Therefore, the total sum and the    is 1002 = 1002.5.\r\n" );
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document.write( "ANSWER.  The total sum is 1002 = 1002.5.\r\n" );
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