已知 Iab-2I+Ib-1I=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2001)(b+2001)等于多少
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已知 Iab-2I+Ib-1I=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2001)(b+2001)等于多少
已知 Iab-2I+Ib-1I=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2001)(b+2001)等于多少
已知 Iab-2I+Ib-1I=0 求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2001)(b+2001)等于多少
因为 Iab-2I+Ib-1I=0 所以b=1 a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2001)(b+2001)
=1/1*2+1/2*3+1/3*4+1/4*5...+1/2002*2003
=1-1/2+1/2-1/3+1/3-1/4+1/4...+1/2002-1/2003
=1-1/2003
=2002/2003
已知 |ab-2|+|b-1|=0 那么 ab-2=0 b-1=0
所以b=1 a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+....+1/(a+2001)(b+2001)
=1/1×2 +1/2×3 +1/3×4.....+1/2002×2003
=1-1/2+1/2-1/3+1/3+1/4-1/4+......+1/2002-1/2003
=1-1/2003
=2002/2003
等于2002/2003
由前式可得a=2,b=1;则原式化为:
1/(2*1)1/(3*2)+1/(4*3)+……+1/(2003*2002)
=1-1/2+1/2-1/3+1/3-1/4+……+1/2002-1/2003
=1-1/2003
=2002/2003.