代数题解答
已知a=14,b=2.求1/ab+1/(a+1)(b+1)+...+1/(a+2004)(b+2004)的值...
已知a=14,b=2.求1/ab+1/(a+1)(b+1)+...+1/(a+2004)(b+2004)的值
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解：1/ab+1/(a+1)(b+1)+...+1/(a+2004)(b+2004)
=(1/12)(1/2-1/14+1/3-1/15+.....+1/2006-1/2018)
=(1/12)(1/2-1/14+1/3-1/15+.....+1/2006-1/2018)
=(1/12){(1/2+1/3+...+1/2004}-1/14-1/15-...-1/2018)
=(1/12)(1/2+1/3+...+1/13-1/2007-1/2008-...-1/2018)
=(1/12)(1/2+1/3+...+1/13-12/2012)
=0.188
=(1/12)(1/2-1/14+1/3-1/15+.....+1/2006-1/2018)
=(1/12)(1/2-1/14+1/3-1/15+.....+1/2006-1/2018)
=(1/12){(1/2+1/3+...+1/2004}-1/14-1/15-...-1/2018)
=(1/12)(1/2+1/3+...+1/13-1/2007-1/2008-...-1/2018)
=(1/12)(1/2+1/3+...+1/13-12/2012)
=0.188
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1/ab+1/(a+1)(b+1)+...+1/(a+2004)(b+2004)
=(1/12)(1/2-1/14+1/3-1/15+.....+1/2006-1/2018)
=(1/12)(1/2-1/14+1/3-1/15+.....+1/2006-1/2018)
=(1/12)(1/2+1/3+...+1/13-1/2007-1/2008-...-1/2018)
=(1/12)(1/2-1/14+1/3-1/15+.....+1/2006-1/2018)
=(1/12)(1/2-1/14+1/3-1/15+.....+1/2006-1/2018)
=(1/12)(1/2+1/3+...+1/13-1/2007-1/2008-...-1/2018)
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