已知数列{an}是首项a1=1的等差数列,其前n项和为Sn,数列{bn}是首项b1=2的等比数列,且把S2=16,b1b3=b4

2025-05-13 02:19:19
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回答(1):

(1)设数列{an}的公差为d,数列{bn}的公比为q,
则an=1+(n-1)d,bn=2qn?1
由b1b3=b4,得q=

b4
b3
=b1=2,
∴an=2n-1,bn2n
(2)T2n+1=c1+a1+(a2+b1)+a3+(a4+2?b2)+…+a2n-1+(a2n+nbn
=1+S2n+(b1+2b2+…+nbn),
令A=b1+2b2+…+nbn
则A=2+2?22+…+n?2n
2A=22+2?23+…+(n-1)?2n+n?2n+1
∴-A=2+22+…+2n-n?2n+1
A=?
2(1?2n)
1?2
+n?2n+1?2n+1+2

S2n
2n(1+a2n)
2
=4n2
T2n+1=1+4n2+n?2n+1?2n+1+2
=3+4n2+(n-1)?2n+1