中四math指數

不知道你要多简化，我的答案：

如题
=[(n+3) log2] - 1/6[ (n+1) log 2 - n log 2]
=[(n+3) log2] - 1/6[ 1log2]
= 2^(n+3) - (2^1)/6
= 2^(n+3) - 1/3

希望能帮到你.

=((2^n)*(2^3)-(2^n)*(2^1))/((2^n)*6)
抽(2^n)
=(2^n)((2^3-2^1))/((2^n)*6)
消(2^n)
=((2^3)-(2^1))/6
=6/6
=1

另一個解
2^(n+3)-((2^n)*(2^1))/((2^n)*6)
消(2^n)
=2^(n+3)-(2/6)
=2^(n+3)-(1/3)

本人認為你抄錯題目，所以另一個解是根據先乘除後加減計

2^n+3-2^n+1/6X2^n
=(2^n)3-1/6x2^n
=2^2n/6X2^n
=2^2n-n/6
=2^n/6