已知数列{an}的前n项和Sn=2an-2n+1.
(1)证明数列{
(2)若不等式2n2-n-3<(5-λ)an对n∈N*恒成立,求λ的取值范围.
秦颖回答:
(1)证明:当n=1时,S1=2a1−22,解得a1=4,Sn=2an−2n+1,当n≥2时,an=Sn−Sn−1=2an−2an−1−2n,∴an=2an−1+2n,∴an2n−an−12n−1=2an−1+2n2n−an−12n−1=an−12n−1+1−an−12n−1=1,又a121=2,...
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已知数列{an}的前n项和Sn=2an-2n+1.
(1)证明数列{
(2)若不等式2n2-n-3<(5-λ)an对n∈N*恒成立,求λ的取值范围.
秦颖回答:
(1)证明:当n=1时,S1=2a1−22,解得a1=4,Sn=2an−2n+1,当n≥2时,an=Sn−Sn−1=2an−2an−1−2n,∴an=2an−1+2n,∴an2n−an−12n−1=2an−1+2n2n−an−12n−1=an−12n−1+1−an−12n−1=1,又a121=2,...