数学题（1+1/2）（1+1/22）（1+1/24）（1+1/28）+1/215（1+1/2）（1+1/22）（1+1/24）（1+1/28）+1/215怎么计算 用乘法公式简化计算

(1+1/2）（1+1/22）（1+1/24）（1+1/28）+1/215
(乘以1-1/2，再除以1-1/2)
=(1-1/2)(1+1/2）（1+1/22）（1+1/24）（1+1/28）/(1-1/2)+1/215
=(1-1/22)(1+1/22)(1+1/24）（1+1/28）/(1/2)+1/215
=(1-1/24)(1+1/24)(1+1/28)/(1/2)+1/215
=(1-1/28)(1+1/28)/(1/2)+1/215=2×1-2×1/216+1/215
=2-1/215+1/215
=2

（1+1/2）（1+1/22）（1+1/24）（1+1/28）+1/215
=(1-1/2)（1+1/2）（1+1/22）（1+1/24）（1+1/28）*2+1/215
一直用平方差公式,有:
原式=(1-1/2^16)*2+1/2^15
=2-1/2^15+1/2^15
=2

(1+1/2）（1+1/22）（1+1/24）（1+1/28）+1/215乘以1-1/2,再除以1-1/2=(1-1/2)(1+1/2）（1+1/22）（1+1/24）（1+1/28）/(1-1/2)+1/215平方差=(1-1/22)(1+1/22)(1+1/24）（1+1/28）/(1/2)+1/215=(1-1/24)(1+1/24)(1+1...

(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15
=2*(1-1/2)(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15
=2*(1-1/2^2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15
=2*(1-1/2^4)(1+1/2^4)(1+1/2^8)+1/2^15
=2*(1-1/2^8)(1+1/2^8)+1/2^15
=2*(1-1/2^16)+1/2^15
=2-1/2^15+1/2^15
=2