问题标题:
已知数列{an}的前n项和为Sn,若a1=2,nan+1=Sn+n(n+1),{bn}为等比数列,且b1=1,b4=127.(1)求数列{an},{bn}的通项公式;(2)设cn=anbn,求数列{cn}的前n项和Tn.(3)求和:Mn=12a1+13a2+…+1(n+1)an.
问题描述:
已知数列{an}的前n项和为Sn,若a1=2,nan+1=Sn+n(n+1),{bn}为等比数列,且b1=1,b4=
(1)求数列{an},{bn}的通项公式;
(2)设cn=anbn,求数列{cn}的前n项和Tn.
(3)求和:Mn=
甘俊英回答:
(1)∵nan+1=Sn+n(n+1)∴(n-1)an=Sn-1+n(n-1)(n≥2)两式相减可得,nan+1-(n-1)an=Sn-Sn-1+2n即nan+1-(n-1)an=an+2n,(n≥2)整理可得,an+1=an+2(n≥2)(*)由a1=2,可得a2=S1+2=4,a2-a1=2适合(*...
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