Question

数学题：已知2*4分之2=2分之1减4分之1，求2*4分之1+4*6分之1+……+48*50分之1

Answer 1

将后面式子全部相加后乘以2，再用定义新运算后可知：2分之1－50分之1＝50分之24
化简得25分之12
然后前面乘了2，所以用25分之12除于2
所以答案为25分之6

=1/2x(1/2-1/4+1/4-1/6+....1/48-1/50)
=1/2x(1/2-1/50)
=1/2x24/50
=6/25
希望采纳

Answer 2

根据2/(2×4)=1/2-1/4，我们同理可以构造2/(4×6)=1/4-1/6,…,2/[2n×2(n+1)]=1/n-1/(n+1)
则1/(2×4)+1/(4×6)+…+1/(48×50)
=1/2×2×[1/(2×4)+1/(4×6)+…+1/(48×50)]
=1/2×[2/(2×4)+2/(4×6)+…+2/(48×50)]
=1/2×(1/2-1/4+1/4-1/6+1/6-1/8…+1/48-1/50)
上述式子中，从第2项开始，到尾二项为止。双数项都可以跟后面的单数项相消，即相加为0.
例如-1/4+1/4=0,-1/6+1/6=0.
原式=1/2×(1/2-1/50)=1/2×24/50=6/25

Answer 3

2/(2*4)=1/2+1/4
则1/(2*4)=(1/2-1/4)*(1/2)
同理:1/(6*4)=(1/4-1/6)*(1/2)
…………………………
1/(48*50)=(1/48-1/50)*(1/2)
所以
原式=（1/2+1/4)*(1/2)+(1/6+1/4)*(1/2)+………………+(1/48+1/50)*(1/2)
=1/2(1/2-1/4+1/4-1/6+…………+1/48-1/50)
=1/2(1/2-1/50)
=6/25

Answer 4

2/(2*4)=1/2+1/4
1/(2*4)=(1/2-1/4)*(1/2)
1/(6*4)=(1/4-1/6)*(1/2)
…………………………
1/(48*50)=(1/48-1/50)*(1/2)

原式=（1/2+1/4)*(1/2)+(1/6+1/4)*(1/2)+………………+(1/48+1/50)*(1/2)
=1/2(1/2-1/4+1/4-1/6+…………+1/48-1/50)
=1/2(1/2-1/50)
=6/25