Question

1/4( )1/4( )1/4( )1/4( )1/4=1/8 括号内填运算符号或括号， 使等式成立

Answer 1

1〉解：因为对于正整数n，1/[n*(n+1)*(n+2)]=[1/(2n)]-[1/(n+1)]+
原式=(1/4）-(1/3)+(1/8)
+(1/6)-(1/4)+(1/10)
+(1/8)-(1/5)+(1/12)
+(1/10)-(1/6)+(1/14)
+(1/12)-(1/7)+(1/16)
+(1/14)-(1/8)+(1/18)
+(1/16)-(1/9)+(1/20)
=-(1/3)+(1/8)+(1/5)+(1/6)+(1/7)+(1/8)+(1/18)+(1/20)-(1/9)-(1/7)-(1/5)
=-(1/3)+(1/4)+(1/6)+(1/18)+(1/20)-(1/9)
=(1/12)+(1/20)-(1/18)
=7/90

2〉解：原式=1/3+4/3*5+6/3*5*7+8/3*5*7*9+（10/3*5*7*9*11+1/3*5*7*9*11)
=(1/3)+(4/3*5)+(6/3*5*7)+(8/3*5*7*9)+1/(3*5*7*9)
=(1/3)+(4/3*5)+(6/3*5*7)+((8+1)/(3*5*7*9)]
=(1/3)+(4/3*5)+[(6+1)/(3*5*7)]
=(1/3)+[(4+1）/(3*5)]
=2/3

Answer 2

1/4+(1/4+1/4)×1/4-1/4=1/8

Answer 3

【1/4(+ )1/4】(× )1/4(+ )1/4(- )1/4=1/8

Answer 4

(1/4+1/4)*1/4+1/4-1/4=1/8