数列{an}满足a1=2,a(n+1)=2an+n+2,求an

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数列{an}满足a1=2,a(n+1)=2an+n+2,求an数列{an}满足a1=2,a(n+1)=2an+n+2,求an数列{an}满足a1=2,a(n+1)=2an+n+2,求an由an+1=2

数列{an}满足a1=2,a(n+1)=2an+n+2,求an
数列{an}满足a1=2,a(n+1)=2an+n+2,求an

数列{an}满足a1=2,a(n+1)=2an+n+2,求an
由a n+1=2an+n+2
得an+1 + (n+1)+3 = 2(an+n+3)
所以{an+n+3}构成公比为2的等比数列,其首项为a1+1+3=6
所以an+n+3=6×2^(n-1)=3×2^n
所以an=3×2^n-n-3