F.3數學指數 2條

第1條要有括號先計到
2^(n+3)-5(2^n)/5(2^n+1)
↓
[2^(n+3)-5(2^n)]/5(2^n+1)
=[(2^n+2^3)-5(2^n)]/5(2^n+2^1)
=[2^n(2^3-5)]/(2^n)(5乘2)
=(2^3-5)/10<<<<<(2^n)係分線度約左
2^3=8
所以
(8-5)/10
=3/10

不過第二條我計到-1/23-.-""唔知點計

答案maybe 係錯 !!!

Maybe 係 1/23

1) 其實我念條題目應該係[ 2^(n+3)-5(2^n) ] / 5(2^n+1)

[ 2^(n+3)-5(2^n) ] / 5(2^n+1)
= 2^(n+3) / 5(2^n+1) - 5(2^n) / 5(2^n+1) ~拆做兩個分數~
=2^2/5 - 1/2 ~兩個分數的2的次方約簡後得出的結果~
=3/10

2) 方法同上一題差唔多, 不過唔知係因為我計錯數定係你抄錯題目/答案,所以我計唔到個答案......

P.S.
3^(n+3)-12(3^n+1)
= (3^2 - 12)3^n+1 ~揪common factor~
= - 3x3^n+1
= - 3^n+2

應該會計到負數......

第二條你可唔可以再對一下題目？