已知字母ab满足已知:字母a、b满足√(a-1)+√(b-2)=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2011)(b+2001)
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![已知字母ab满足已知:字母a、b满足√(a-1)+√(b-2)=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2011)(b+2001)](/uploads/image/z/8392893-69-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AD%97%E6%AF%8Dab%E6%BB%A1%E8%B6%B3%E5%B7%B2%E7%9F%A5%EF%BC%9A%E5%AD%97%E6%AF%8Da%E3%80%81b%E6%BB%A1%E8%B6%B3%E2%88%9A%EF%BC%88a-1%29%2B%E2%88%9A%28b-2%29%3D0%2C%E6%B1%821%EF%BC%8Fab%2B1%2F%28a%2B1%29%28b%2B1%29%2B1%2F%28a%2B2%29%28b%2B2%29%2B%E2%80%A6%E2%80%A6%2B1%2F%28a%2B2011%29%28b%2B2001%29)
已知字母ab满足已知:字母a、b满足√(a-1)+√(b-2)=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2011)(b+2001)
已知字母ab满足
已知:字母a、b满足√(a-1)+√(b-2)=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2011)(b+2001)
已知字母ab满足已知:字母a、b满足√(a-1)+√(b-2)=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2011)(b+2001)
a、b满足√(a-1)+√(b-2)=0,
则
a-1=0,b-2=0
a=1,b=2
所以
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2011)(b+2001)
=1/1×2+1/2×3+.+1/2002×2003
=1-1/2+1/2-1/3+.+1/2002-1/2003
=1-1/2003
=2002/2003