[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
[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