题目内容
在公差为d(d≠0)的等差数列{an}和公比为q的等比数列{bn}中,已知a1=b1=1,a2=b2,a8=b3.
(1)求数列{an}与{bn}的通项公式;
(2)令cn=an•bn,求数列{cn}的前n项和Tn.
(1)求数列{an}与{bn}的通项公式;
(2)令cn=an•bn,求数列{cn}的前n项和Tn.
(1)由条件得:
?
?an=5n-4,bn=6n-1
(2)Tn=c1+c2+c3+…+cnTn=a1b1+a2b2+a3b3+…+an-1bn-1+anbn①qTn=a1b2+a2b3+a3b4+…+an-1bn+anbn+1②
①-②:(1-q)Tn=a1b1+db2+db3+…+dbn-1+dbn-anbn+1=a1b1+d
-anbn+1
即 -5Tn=1+5
-(5n-4)6n
∴Tn=(n-1)6n+1
|
|
(2)Tn=c1+c2+c3+…+cnTn=a1b1+a2b2+a3b3+…+an-1bn-1+anbn①qTn=a1b2+a2b3+a3b4+…+an-1bn+anbn+1②
①-②:(1-q)Tn=a1b1+db2+db3+…+dbn-1+dbn-anbn+1=a1b1+d
| b2(1-qn-1) |
| 1-q |
即 -5Tn=1+5
| 6(1-6n-1) |
| -5 |
∴Tn=(n-1)6n+1
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