作业帮[1+(-1/2)]+[1/2+(-1/3)]+[1/3+(-1/4)]+…+[1/1999+(-1/2000)]=?等于多少啊?A.2000/1999 B.1999/2000 C.1/2000 D.1/1999
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![[1+(-1/2)]+[1/2+(-1/3)]+[1/3+(-1/4)]+…+[1/1999+(-1/2000)]=?等于多少啊?A.2000/1999 B.1999/2000 C.1/2000 D.1/1999](/uploads/image/z/1743060-12-0.jpg?t=%5B1%2B%28-1%2F2%29%5D%2B%5B1%2F2%2B%28-1%2F3%29%5D%2B%5B1%2F3%2B%28-1%2F4%29%5D%2B%E2%80%A6%2B%5B1%2F1999%2B%28-1%2F2000%29%5D%3D%3F%E7%AD%89%E4%BA%8E%E5%A4%9A%E5%B0%91%E5%95%8A%3FA.2000%2F1999+B.1999%2F2000+C.1%2F2000+D.1%2F1999)
[1+(-1/2)]+[1/2+(-1/3)]+[1/3+(-1/4)]+…+[1/1999+(-1/2000)]=?等于多少啊?A.2000/1999 B.1999/2000 C.1/2000 D.1/1999
[1+(-1/2)]+[1/2+(-1/3)]+[1/3+(-1/4)]+…+[1/1999+(-1/2000)]=?等于多少啊?
A.2000/1999 B.1999/2000 C.1/2000 D.1/1999
[1+(-1/2)]+[1/2+(-1/3)]+[1/3+(-1/4)]+…+[1/1999+(-1/2000)]=?等于多少啊?A.2000/1999 B.1999/2000 C.1/2000 D.1/1999
B
你把中括号全部去掉就成了下面的式子,(-1/2)+1/2=0,(-1/3)+1/3=0,以此类推,就得到
1+(-1/2)+1/2+(-1/3)+1/3+(-1/4)+…+1/1999+(-1/2000)=1-（1/2000）=1999/2000
1 2 1 2 () () ()()
1+1/2^2
1 1 1 1 1 1 1 1 1 1 1 -+-2 2
1+2+1+2+1+2+1+2+1+2+1+2+1+2 =( )*( ) =()
(1-1/2^2)(1-1/3^2)(1-1/4^2).(1-1/2009^2),
（1+1/2+1/3+...
计算行列式 2 1 1 1 ,1 2 1 1 ,1 1 2 1,1 1 1 2,
1/2-1/(n+1)
证明 1+1/1+1/1*2+1/1*2*3+.+1/1*2*3*...*n
[(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(1+2^-(1/2)]
(1-1/2^2)*(1-1/3^2)*(1-1/4^2)*.*(1-1/2002^2)*(1-1/2003^2)
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16),
1*1/2=?
1+1+2+56
5,2,1,1,
1/2+56/1
1.1/2.
2x-1-1