已知数列{an}满足an+1=2an+3*2^n,a1=2,求数列{an}的通项公式

2025-02-25 05:56:45
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回答1:

a(n+1)=2an+3*2^n
两边同时除以2^(n+1)
a(n+1)/2^(n+1)=an/2^n+3/2
数列{an/2^n}是以a1/2=1为首项,3/2为公差的等差数列
an/2^n=1+(n-1)3/2=(3n-1)/2
an=(3n-1)/2*2^n