某同学在计算3（4+1）（4눀+1）时，把3写成4-1后发现可以连续运用平方差公式计算

2^2代表2的平方，2^4代表2的4次方……
(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)+1/2^15
=[(1-1/2^2)(1+1/2^2)(1+1/2^4)(1+1/2^8)]/(1-1/2)+1/2^15
=[(1-1/2^4)(1+1/2^4)(1+1/2^8)]/(1-1/2)+1/2^15
=[(1-1/2^8)(1+1/2^8)]/(1-1/2)+1/2^15
=(1-1/2^16)/(1-1/2)+1/2^15
=2-1/2^15+1/2^15
=2

原式=(1-1/2)(1+1/2)(1+1/4)(1+1/16)(1+1/256)×2+1/2的15次方
=(1-1/4)(1+1/4)(1+1/16)(1+1/256)×2+1/2的15次方
=(1-1/16)(1+1/16)(1+1/256)×2+1/2的15次方
=(1-1/256)(1+1/256)×2+1/2的15次方
=(1-1/2的16次方)×2+1/2的15次方
=2-1/2的15次方+1/2的15次方
=2
同学你初几?