高一數學問題...好難喔

1.
設 S = 1 + 4/5 + 7/5^2 + 10/5^3 + ...... +(3n-2)/5^(n-1) + ......
則 (1/5)S = 1/5 + 4/5^2 + 7/5^3 + ...... +(3n-2)/5^n + ......

把以上兩式相減：
S - (1/5)S = 1 + (3/5 + 3/5^2 + 3/5^3 + ......)
(4/5)S = 1 + (3/5)/[1 - (1/5)]
(4/5)S = 7/4
S = 35/16

所以，1 + 4/5 + 7/5^2 + 10/5^3 + ...... + (3n-2)/5^(n-1)+ ...... = 35/16


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2.
設 S = 1/2 + (1/4+1/8) + (1/8+1/16+1/32) + (1/16+1/32+1/64+1/128)+ ....
則 (1/2)S = 1/4 + (1/8+1/16) + (1/16+1/32+1/64) + ....

把以上兩式相減：
S - (1/2)S = 1/2 + 1/8 + 1/32 + 1/128
(1/2)S = (1/2) + (1/2)(1/4) + (1/2)(1/4)^2 + (1/2)(1/4)^3 + ......
(1/2)S = (1/2)/[1 - (1/4)]
(1/2)S = 2/3
S = 4/3

所以，1/2 + (1/4+1/8) + (1/8+1/16+1/32) + (1/16+1/32+1/64+1/128)+ .... = 4/3


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3.
設需至小數點以下第 n 位。

1/3 - [0.3 + 0.3*(0.1) + 0.3*(0.1)^2 + ... + 0.3*(0.1)^(n - 1)] < 3/10^4
[0.3 + 0.3*(0.1) + 0.3*(0.1)^2+ ...] - [0.3 + 0.3*(0.1) + ... + 0.3*(0.1)^(n -1)] < 0.0003
0.3*(0.1)^n + 0.3*(0.1)^(n+1) + 0.3*(0.1)^(n+2) + ...... < 0.0003
0.3*(0.1)^n / (1 - 0.1) < 0.0003
0.3*(0.1)^n / 0.9 < 0.0003
0.3*(0.1)^n < 0.00027
(0.1)^n < 0.0009
log(0.1)^n < log(0.0009)
n*log(0.1) < log(0.0009)
n*(-1) < log(0.0009)
n > -log(0.0009)
n > 3.0457
由於 n 為整數，故 n ≥ 4

故需至小數點以下第 4 個位。