(1)当n=1时:S 1 =a 1 =2a 1 ﹣2 1+1 ,解得a=4当n≥2时,由Sn=2a n ﹣2n+1 …①且S n ﹣1=2an﹣1﹣2n …②①﹣②得:a n =2a n ﹣2a n ﹣1﹣2n,有:a n =2a n ﹣1+2n得 ,∴b n ﹣b n﹣1 =1, ,故数列...
设数列{a n }的前n项和为S n ,已知S n =2a n ﹣2n+1(n∈N*).
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