题目应该是1-1/2-1/4-1/8-1/16-1/32-1/64吧,此题须用到以下知识点,即
1/2=1-1/2
1/4=1/2-1/4
1/8=1/4-1/8
1/16=1/8-1/16
1/32=1/16-1/32
1/64=1/32-1/64
知道了这个知识点,解题就比较方便了
解:
1-1/2-1/4-1/8-1/16-1/32-1/64
=1-(1-1/2)-(1/2-1/4)-(1/4-1/8)-(1/8-1/16)-(1/16-1/32)-(1/32-1/64)
=1-1+1/2-1/2+1/4-1/4+1/8-1/8+1/16-1/16+1/32-1/32+1/64
=1/64
1-1/2-1/4-1/8-1/16-1/32-1/64
=1-1/2-1/4-1/8-1/16-1/32-1/64-1/64+1/64
=1/64.
通过分析可以看出,减数的和等于一个等比数列,公比是1/2,可以直接求出等比数列的和。然后再计算减法。当然,1/1.2就等于10/12。等比数列的和为:
1/4*(1-1/2的5次方)╱1-1/2
=1/4*(63/64)*2
=1/4*63/32
=63/128
10/12-63/128
=320/384-189/384
=131/384
1/1.2-1/4-1/8+1/6-1/32-1/64
=5/6+1/6-3/8-3/64
=1-27/64
=37/64
1/2-1/4-1/8-1/16-1/32-1/64=32/64-16/64-8/64-4/64-2/64-1/64=32/64-(16/64+8/64+4/64+2/64+1/64)=32/64-31/64=1/64