搜搜考试网[(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(
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[(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(1+2^-(1/2)]
![[(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(](/uploads/image/z/19946482-34-2.jpg?t=%5B%281%2B2%5E-%281%2F32%29%5D%2A%5B%281%2B2%5E-%281%2F16%29%5D%2A%5B%281%2B2%5E-%281%2F8%29%5D%2A%5B%281%2B2%5E-%281%2F4%29%5D%2A%5B%28)
[(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(1+2^-(1/2)]
=[(1-2^-(1/32)]*[(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(1+2^-(1/2)]/[(1-2^-(1/32)]
=[1-2^(-1)]/[(1-2^-(1/32)]
=1/[(2-2^(31/32)]
再问： 为什么这样做，有什么方法？
=[(1-2^-(1/32)]*[(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(1+2^-(1/2)]/[(1-2^-(1/32)]
=[1-2^(-1)]/[(1-2^-(1/32)]
=1/[(2-2^(31/32)]
再问： 为什么这样做，有什么方法？
1/2+1/4+1/8+1/16+1/32+.
1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256
1/2-1/4-1/8-1/16-1/32-1/64-1/128-1/256
1/2-1/4-1/8-16/1-32/1-64/1-128/1-256/1
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1计算方法是什么?
(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1=
计算:(1+1/2^32)(1+1/2^16)(1+1/6^8)(1+1/2^4)(1+1/2^2)(1+1/2) .
简便计算1/2+1/4+1/8+1/16+1/32+1/64+1/128
1/2+1/4+1/8+1/16+32/1+1/128没有1/64简算
1--1/2-1/4-1/8-1/16-1/32-1/64怎么简便计算
1-1/2-1/4-1/8-1/16-1/32-1/64简便计算
1/2+1/4+1/8+1/16+1/32+1/64+1/128