若数列{an}满足a1=1,且a(n+1)=4an+2的n次方,求an的通项公式急·······

Question

若数列{an}满足a1=1,且a(n+1)=4an+2的n次方,求an的通项公式
急·······

Answer 1

a(n+1)=4an+2ⁿ
a(n+1)+2ⁿ=4an+2×2ⁿ=4an+4×2^(n-1)
[a(n+1)+2ⁿ]/[an+2^(n-1)]=4,为定值.
a1+2^0=1+1=2
数列{an+2^(n-1)}是以2为首项,4为公比的等比数列.
an+2^(n-1)=2×4^(n-1)=2^(2n-1)
an=2^(2n-1)-2^(n-1)
n=1时,a1=2-1=1,同样满足.
数列{an}的通项公式为an=2^(2n-1)-2^(n-1)