Arithmetic Sequences..........

4, 8, 16, 32
係呢條問題，你要用呢條式 Geometric Sequence︰
aRn-1
Where 'a' is the [first term]
R needs to be calculated in order to find the answer
Use the second term to divide the first term.
4(8/4)n-1
= 4(2)n-1
First term : 4(2)0 = 4 x 1 = 4
Second term : 4(2)2-1 = 4(2)1 = 4 x 2 = 8
Third term : 4(2)3-1 = 4(2)2 = 4 x 4 = 16
Fourth term : 4(2)4-1 = 4(2)3 = 4 x 8 = 32

The first question :
log5, log10, log20, log40....
Substitute the formula :
T(n) = log [5(10/5)n-1]
T(n)=log [5(2)n-1]
因為這條是等比關係的問題，所以不能用等差數列。

T(n)=a+(n-1)d 只是用來找 等差數列 的第n項的項數

例如:
3, 15, 27, 39, ......
每項數之間都是相差12 ( a=3, d=12 )

而你的例子全都不是等差數列, 當然用不到 T(n)=a+(n-1)d 啦!
只能靠找出每個項數之間的共通點 而想出 T(n)
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log5, log10, log20, log40....
共通點 : 5, 10, 20, 40 都是5的倍數

5 = 5x1 = 5 x 2的零次方 = 5 x 2的 (1-1) 次方
10 = 5x2 = 5 x 2的 1 次方 = 5 x 2的 (2-1) 次方
20 = 5x4 = 5 x 2的 2 次方 = 5 x 2的 (3-1) 次方
40 = 5x8 = 5 x 2的 3 次方 = 5 x 2的 (4-1) 次方

所以 : T(n)=log [5 x 2^(n-1)]
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2/3, 3/4, 4/5, 5/6, ....
共通點 :
分子 : 2, 3 ,4 ,5 , ....., (n+1)
分母 : 3, 4 ,5 ,6 , ....., (n+2)

所以 : T(n) = (n+1)/(n+2)
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4, 8, 16, 32 , .....
共通點 : 4的倍數

4 = 4x1 = 4x 2^0
8 = 4x2 = 4x 2^1
16 = 4x4 = 4x 2^2
32 = 4x8 = 4x 2^3

所以 : T(n) = 4x2^(n-1)