上下同乘以(1-/2),反复应用平方差公式求解
(1+1/2)×(1+1/2ˆ2)×(1+1/2ˆ4)×(1+1/2ˆ8)+1/2ˆ15
=(1-1/2)*(1+1/2)×(1+1/2ˆ2)×(1+1/2ˆ4)×(1+1/2ˆ8)/[(1-1/2)]+1/2ˆ15
=(1-1/2^2)(1+1/2^2)(1+1/2ˆ4)×(1+1/2ˆ8)/[(1-1/2)]+1/2ˆ15
=(1-1/2^4)(1+1/2ˆ4)×(1+1/2ˆ8)/[(1-1/2)]+1/2ˆ15
=(1-1/2ˆ8)×(1+1/2ˆ8)/[(1-1/2)]+1/2ˆ15
=(1-1/2^16)/[(1-1/2)]+1/2ˆ15
=1/2-1/2^15+1/2^15
=1/2
为什么(1-1/2^16)/[(1-1/2)]+1/2ˆ15=1/2-1/2^15+1/2^15?
=(1-1/2^16)/(1/2)+1/2^15=(1-1/2^16)*2+1/2^15=2-1/2^15+1/2^15=2,我打错了,