如果/ab-2/+/b-1/=0,试求下式的值:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a+2000)(b+2000)
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如果/ab-2/+/b-1/=0,试求下式的值:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a+2000)(b+2000)
如果/ab-2/+/b-1/=0,试求下式的值:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a+2000)(b+2000)
如果/ab-2/+/b-1/=0,试求下式的值:1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a+2000)(b+2000)
/ab-2/+/b-1/=0
得到ab-2=0,b-1=0
解得a=2,b=1
所以要求的式子变成了
1/2+1/(2)(3)+1/(3)(4)…+1/(2001)(2002)
=1-1/2+1/2-1/3+……+1/2001-1/2002
=2001/2002
这个是列项相消法
w45ey
由题意知:
ab -2 =0;
b -2 =0.
由此解得:a =1, b =2.
因为: 1/(1 +n)*(2 +n) = 1/(1 +n) -1/(2 +n).
所以
原式=1/1*2 + 1/2*3 + 1/3*4 + .... + 1/2000*2000
=1 -1/2 + 1/2 - 1/3 + ... + 1/2000 -1/2000
=1 -1/2005
=1999/2000.
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