已知数列{an}的前n项和Sn满足Sn=2an+(-1)n(n∈N*)
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已知数列{an}的前n项和Sn满足Sn=2an+(-1)n(n∈N*)
(1)求数列{an}的前三项a1,a2,a3;
(2)求证:数列{a
(1)求数列{an}的前三项a1,a2,a3;
(2)求证:数列{a
![已知数列{an}的前n项和Sn满足Sn=2an+(-1)n(n∈N*)](/uploads/image/z/1449306-18-6.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E6%BB%A1%E8%B6%B3Sn%3D2an%2B%EF%BC%88-1%EF%BC%89n%EF%BC%88n%E2%88%88N%2A%EF%BC%89)
(1)对于Sn=2an+(−1)n(n∈N*),令n=1,可得a1=S1=2a1-1,解得a1=1.
令n=2,则a1+a2=S2=2a2+1,把a1=1代入解得a2=0.
令n=3,则a1+a2+a3=S3=2a3-1,把a1=1,a2=0代入解得a3=2.
(2)当n≥2时,an=Sn-Sn-1=2an+(−1)n-[2an−1+(−1)n−1],化为an−2an−1+2(−1)n=0.
∴an+
2
3(−1)n=2[an−1+
2
3(−1)n−1],
∴数列{an+
2
3(−1)n}是首项为a1+
2
3×(−1)1=
1
3,2为公比的等比数列.
∴an+
2
3(−1)n=
1
3×2n−1.
∴an=
1
3×2n−1+
2
3×(−1)n−1.
令n=2,则a1+a2=S2=2a2+1,把a1=1代入解得a2=0.
令n=3,则a1+a2+a3=S3=2a3-1,把a1=1,a2=0代入解得a3=2.
(2)当n≥2时,an=Sn-Sn-1=2an+(−1)n-[2an−1+(−1)n−1],化为an−2an−1+2(−1)n=0.
∴an+
2
3(−1)n=2[an−1+
2
3(−1)n−1],
∴数列{an+
2
3(−1)n}是首项为a1+
2
3×(−1)1=
1
3,2为公比的等比数列.
∴an+
2
3(−1)n=
1
3×2n−1.
∴an=
1
3×2n−1+
2
3×(−1)n−1.
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