1/2+1/4+1/8+1/16+1/32+1/64+1/128

来源：百度知道编辑：UC知道时间：2024/09/24 02:26:02

解：
设原式=S
则2S=1+1/2+1/4+1/8+1/16+1/32+1/64
2S-S=S=1-128/1
所以原式=127/128

1/2+1/4+1/8+1/16+1/32+1/64+1/128为等比数列公比为1/2
1/2+1/4+1/8+1/16+1/32+1/64+1/128=1/2*[1-(1/2)^7]/(1-1/2)
=1-(1/2)^7
=127/128

令S=1/2+1/4+1/8+1/16+1/32+1/64+1/128
2S=1+1/2+1/4+1/8+1/16+1/32+1/64
S=2S-S=1-1/128=127/128

1/2+1/4+1/8+1/16+1/32+1/64+1/128
=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/128-1/128
=1/2+1/4+1/8+1/16+1/32+1/64+1/64-1/128
=1/2+1/4+1/8+1/16+1/32+1/32-1/128
=1/2+1/4+1/8+1/16+1/16-1/128
=1/2+1/4+1/8+1/8-1/128
=1/2+1/4+1/4-1/128
=1/2+1/2-1/128
=1-1/128=127/128

等比数列前N项和公式

(1+1/2+1/3+1/4)× 1+1/2+1+1/3+1+1/4+......+1/100=? 1/1+1/2+1/3+1/4+。。。。+1/N 是多少 1/1+1/2+1/3+1/4+......1/2002＝？ 1-1/2+1/3-1/4+........1/99-1/100 (1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000) (1/2) (1/4+1/6)= (1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8) (1-1/2^2)*(1-1/3^2)*(1-1/4^2).......(1-1/100^2) s=1/1+1/1!+1/2！+1/3！+1/4！+....+1/25!