(1/3-7/12+9/20-11/30+13/42-15/56)*2的3次方+21=?
问题描述：
(1/3-7/12+9/20-11/30+13/42-15/56)*2的3次方+21=?
答
(1/3-7/12+9/20-11/30+13/42-15/56)*2的3次方+21
=（1/3-1/3-1/4+1/4+1/5-1/5-1/6+1/6+1/7-1/7-1/8）*8+21
=-1/8*8-21
=-1+21
=20
答
1/3-7/12+9/20-11/30+13/42-15/56
=1/3-1/3-1/4+1/4+1/5-1/5-1/6+1/6+1/7-1/7-1/8
=-1/8
(1/3-7/12+9/20-11/30+13/42-15/56)*2的3次方+21
=(-1/8)*2^3+21
=(-1/8)*8+21
=-1+21
=20
答
1/3-7/12+9/20-11/30+13/42-15/56
=1/3-(1/3+1/4)+(1/4+1/5)-(1/5+1/6)+(1/6+1/7)-(1/7+1/8)
=1/3-1/3-1/4+1/4+1/5-1/5-1/6+1/6+1/7-1/7-1/8
=-1/8
所以
(1/3-7/12+9/20-11/30+13/42-15/56)乘2^3+21
=(-1/8)*8+21
=-1+21
=20