(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15=?
来源:学生作业帮助网 编辑:六六作业网 时间:2025/01/13 20:12:39
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15=?(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15=?(1+1/2)(1+1/2^
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15=?
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15=?
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15=?
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15
(1-1/2)(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)/(1-1/2)
=(1-1/2^16)/(1/2)=2(1-1/2^16)=2-1/2^15
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15
=2-1/2^15+1/2^15
=2
平方差公式的运用
在式子前面乘以(1-1/2)利用平方差公式可求得答案
原式=(1-1/2)(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)/(1-1/2)
=(1-1/2^16)/(1/2)=2(1-1/2^16)
=2-1/2^15(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15
=2-1/2^15+1/2^15
=2
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