✔ 最佳答案
以下解法沒用到"無窮等比級數"的概念
0.8+0.088+0.00888+0.0008888+0.000088888+...
=(8/9)*(0.9+0.099+0.00999+...)
=(8/9)*[(1-0.1)+(0.1-0.001)+(0.01-0.00001)+....]
=(8/9)*[(1+0.1+0.01+0.001+0.0001+...)-(0.1+0.001+0.00001+...)]
=(8/9)*[1.1111...(循環小數)- 0.1010101...(循環小數)]
=(8/9)*[10/9- 10/99] (將10/9, 10/99即得上述之循環小數)
=(8/9)*(100/99)
=800/891
2010-09-13 15:47:27 補充:
(補充)求1.11111....(國中解法)
設x=1.1111...., 則10x=11.1111..., 相減得9x=11-1=10, 故x=10/8
求0.1010101....(循環)
設x=0.1010101..., 則100x=10.101010..., 相減得99x=10, 故x=10/99
2010-09-13 17:12:35 補充:
(將10/9, 10/99除開,即得上述之循環小數)