已知数列{an}满足a1=1,an=2an-1+1,则an= ___ .
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已知数列{an}满足a1=1,an=2an-1+1,则an= ___ .
![已知数列{an}满足a1=1,an=2an-1+1,则an= ___ .](/uploads/image/z/17394018-42-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%EF%BC%8Can%3D2an-1%2B1%EF%BC%8C%E5%88%99an%3D+___+%EF%BC%8E)
由an=2an-1+1可得an+1=2an-1+2=2(an-1+1),
故可得
an+1
an-1+1=2,故数列{an+1}为公比为2的等比数列,
由题意可得该数列的首项为:a1+1=2,
故可得an+1=2×2n-1,故an=2n-1,
故答案为:2n-1
故可得
an+1
an-1+1=2,故数列{an+1}为公比为2的等比数列,
由题意可得该数列的首项为:a1+1=2,
故可得an+1=2×2n-1,故an=2n-1,
故答案为:2n-1
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