已知|ab-2|与|b-1|互为相反数,试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2002)(b+2002)的值
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![已知|ab-2|与|b-1|互为相反数,试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2002)(b+2002)的值](/uploads/image/z/2711006-62-6.jpg?t=%E5%B7%B2%E7%9F%A5%7Cab-2%7C%E4%B8%8E%7Cb-1%7C%E4%BA%92%E4%B8%BA%E7%9B%B8%E5%8F%8D%E6%95%B0%2C%E8%AF%95%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8F1%2Fab%2B1%2F%EF%BC%88a%2B1%29%EF%BC%88b%2B1%29%2B1%2F%EF%BC%88a%2B2%29%28b%2B2%29%2B%E2%80%A6%2B1%2F%28a%2B2002%29%28b%2B2002%29%E7%9A%84%E5%80%BC)
已知|ab-2|与|b-1|互为相反数,试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2002)(b+2002)的值
已知|ab-2|与|b-1|互为相反数,试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2002)(b+2002)的值
已知|ab-2|与|b-1|互为相反数,试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2002)(b+2002)的值
因|ab-2|与|b-1|互为相反数,即ab-2=0,b-1=0 解得a=2,b=1
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2002)(b+2002)
=1/2 + 1/2x3 +1/3x4 .+1/(2003x2004)
因为1/a(a+1) = 1/a - 1/(a+1); 例如 1/2x3=1/2 - 1/3,1/3x4 = 1/3-1/4
所以代数式 1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2002)(b+2002)
=1/2 + 1/2x3 +1/3x4 .+1/(2003x2004)
=1/2+1/2-1/3 +1/3-1/4 +.+1/2003-1/2004
=1/2 + 1/2 -1/2004
=2003/2004