1/2x1+1/(2+1)(1+1)+1/(2+2)(1+2)+1/(2+3)(1+3)+.1/(2+2003)(1+2

gister2022-10-04 11:39:543条回答

1/2x1+1/(2+1)(1+1)+1/(2+2)(1+2)+1/(2+3)(1+3)+.1/(2+2003)(1+2003)=?
求详细计算结果我已经知道结果是2004/2005 就是想知道计算的步骤 如果有理数a ,b满足丨ab-2丨+(1-b)²=0 试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2003)(b+2003)的值

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raykee 共回答了27个问题 | 采纳率92.6%
1/2x1+1/(2+1)(1+1)+1/(2+2)(1+2)+1/(2+3)(1+3)+.1/(2+2003)(1+2003)=
=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+……+(1/2004-1/2005)
=1/1-1/2+1/2-1/3+1/3-1/4……+1/2004-1/2005
=1/1-1/2005
=2004/2005
1年前
therefor 共回答了505个问题 | 采纳率

有理数a,b满足|ab-2|+(1-b)^2=0
ab-2=0,1-b=0
a=2,b=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a+2003)(b+2003)
=1/1*2+1/2*3+1/3*4+...+1/2004*2005
=1-1/2+1/2-1/3+1/3-1/4+...+1/2004-1/2005
1年前
wsxrfhn3895 共回答了10个问题 | 采纳率
1/2 = 1- 1/2,1/(2+1)(1+1) = 1/(1+2) - 1/(2+1).....,如此分解,所求算式化解后只剩第一项和最后一项,记得答案
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=
(1+2x−1)x
2=x2
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点评:
本题考点: 数列的求和.

考点点评: 本题考查了数列求和的两个常用方法:公式法、裂项求和,裂项求和时注意公式[1n(n+k)=1/k•(1n −1n+k)中的1k]不要漏掉.

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