关于a12Sn2ank的百科

• 已知数列{an}的前n项和为Sn,且a1=2,Sn=2an+k,等差数列{bn}的前n项和为Tn,且Tn=n2...

• 问题详情：已知数列{an}的前n项和为Sn,且a1=2,Sn=2an+k,等差数列{bn}的前n项和为Tn,且Tn=n2.(1)求k和Sn;(2)若cn=an·bn,求数列{cn}的前n项和Mn.【回答】解(1)∵Sn=2an+k,∴当n=1时,S1=2a1+k.∴a1=-k=2,即k=-2.∴Sn=2an-2.∴当n≥2时,Sn-1=2an-1-2.∴an=Sn-Sn-1=2an-2an-1....
• 2020-09-1911868