a12Sn2ank的精選

問題詳情：已知數列{an}的前n項和爲Sn,且a1=2,Sn=2an+k,等差數列{bn}的前n項和爲Tn,且Tn=n2.(1)求k和Sn;(2)若cn=an·bn,求數列{cn}的前n項和Mn.【回答】解(1)∵Sn=2an+k,∴當n=1時,S1=2a1+k.∴a1=-k=2,即k=-2.∴Sn=2an...