Question

已知数列an中,a1=1,a(n+1)=an+2^(n-1),则an?

已知数列an中,a1=1,a(n+1)=an+2^(n-1),则an?

Answer 1

a2=a1+1
a3=a2+2^1
a4=a3+2^2
a5=a4+2^3
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an=a(n-1)+2^(n-2)

上式相加得 an=a1+1+2+2^2+2^3+…+2^(n-2)
=1+[1-2^(n-1)]/(1-2)
=1+2^(n-1)-1
=2^(n-1)

Answer 2

2*a(n+1)/2^n=an/2^(n-1)+1
令an/2^(n-1)=bn b1=1

2b(n+1)=bn+1 2b(n+1)-2=bn-1
令bn-1=cn
c1=0 cn是首项是0的等比数列 也解释说cn是常数列 为0
也就是说bn-1=0 an/2^(n-1)-1=0 an=2^(n-1)