问题
解答题
已知数列{an}满足
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答案
∵当n≥2时,an=2an-1+1,∴an+1=2(an-1+1)
=2,∴数列{an+1}为等比数列,且公比为2,an+1 an-1+1
又∵a1=1,∴a1+1=2
∴an+1=2n,an=2n-1
Sn=21-1+22-1+…+2n-1=
-n=2n+1-2-n2(1-2n) 1-2
已知数列{an}满足
|
∵当n≥2时,an=2an-1+1,∴an+1=2(an-1+1)
=2,∴数列{an+1}为等比数列,且公比为2,an+1 an-1+1
又∵a1=1,∴a1+1=2
∴an+1=2n,an=2n-1
Sn=21-1+22-1+…+2n-1=
-n=2n+1-2-n2(1-2n) 1-2