title: 请教高一数列题已知数列{an}满足:a1=1,an+1=2an+ 
desc: 
  已知数列 {an} 满足: a1=1, an+1 =2an+3×2n-1, 求 {an} 的通项公式.
  
  
answer: 解：a2=5
  a(n+1)=2an+3×2^(n-1)
  两边除以2^(n+1)
  得a(n+1)/2^(n+1)-an/2^n=3/4
  an/2^n-a(n-1)/2^(n-1)=3/4
  ……
  a2/2^2-a1/2^1=3/4
  两边分别累加：a(n+1)/2^(n+1)-a1/2^1=3n/4
  a(n+1)=2^(n+1)·(3n/4+1/2)=(3n+2)×2^(n-1)
  an=(3n-1)×2^(n-2)
  满足a1=1
  故an=(3n-1)×2^(n-2)
  