document.write( "Question 1165362: A function f is defined for integers m and n as given:
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Algebra.Com's Answer #789857 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
f(mn) = f(m)∙f(n) - f(m+n) + 1001
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document.write( "Let m=1, then\r\n" );
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document.write( "        f(1n) = f(1)∙f(n) - f(1+n) + 1001\r\n" );
document.write( "         f(n) = 2∙f(n) - f(n+1) + 1001\r\n" );
document.write( "       f(n+1) = f(n) + 1001\r\n" );
document.write( "f(n+1) - f(n) = 1001 \r\n" );
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document.write( "This tells us that after the first term the common difference between\r\n" );
document.write( "a term and its preceding term equals 1001.\r\n" );
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document.write( "So it is an arithmetic sequence with first term f(1) = 2 and common\r\n" );
document.write( "difference 1001.\r\n" );
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document.write( "     The formula is an = a1 + (n-1)∙d\r\n" );
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document.write( "                  f(n) = f(1) + (n-1)∙1001\r\n" );
document.write( "                  f(n) = 2 + (n-1)∙1001\r\n" );
document.write( "                  f(n) = 2 + 1001n - 1001\r\n" );
document.write( "                  f(n) = 1001n - 999\r\n" );
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document.write( "So when n=12\r\n" );
document.write( "               f(9999) = 1001(9999) - 999\r\n" );
document.write( "               f(9999) = 10008999 - 999\r\n" );
document.write( "               f(9999) = 10008000\r\n" );
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document.write( "Edwin

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