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1/2+1/4+1/8+1/16+1/32+...+1/2的100次方的值
1/2+1/4+1/8+1/16+···+1/2^100 =(1/2)[1-(1/2)^100]/(1-1/2) =1-(1/2)^100 ≈1。
设A=(1/2)+(1/4)+(1/8)+(1/16)+(1/32)+……+1/(2^100)………(1) 则: 2A=1+(1/2)+(1/4)+(1/8)+(1/16)+……+1/(2^99)………………(2) (2)-(1)得到: 2A-A=1-[1/(2^100)] 所以,A=1-[1/(2^100)] 即,原式=1-[1/(2^100)].
原式=(1-1/2)+(1/2-1/4)+(1/4-1/8)+…+[1/(2)^98-1/(2)^99]-[1/(2)^99-1/(2)^100] =1-(1/2)^100
利用(1/2)^n+(1/2)^n=(1/2)^(n-1),添一项1/2的100次方,再减一项1/2的100次方! 1/2+1/4+1/8+1/16+1/32+...+1/2的100次方 =[1/2+1/4+1/8+1/16+...+1/2的100次方+(1/2)^100]-(1/2)^100 =1-(1/2)^100 =(2^100-1)/2^100
答:1-1/2-1/4-1/8-1/16-1/32 =1/2-1/4-1/8-1/16-1/32 =1/4-1/8-1/16-1/32 =1/8-1/16-1/32...详情>>
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