已知数列an满足a1=3,An+1=2An+2^n (1)求证数列[An/2^n]是等差数列 (2)

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(1)证：a(n+1)=2an+2ⁿ
等式两边同除以2^(n+1)
a(n+1)/2^(n+1)=an/2ⁿ +1/2
a(n+1)/2^(n+1)-an/2ⁿ=1/2,为定值.
a1/2=3/2,数列{an/2ⁿ}是以3/2为首项,1/2为公差的等差数列.
(2)an/2ⁿ=(3/2)+(n-1)/2=n/2 +1
an=2ⁿ(n/2 +1)=n×2^(n-1) +2ⁿ
n=1时,a1=1×2^0 +2=1+2=3,同样满足.
数列{an}的通项公式为an=n×2^(n-1) +2ⁿ.