问题
解答题
已知函数f(x)=-
( I)求数列{an}的通项公式; ( II)数列{bn}的前n项和为Tn且满足bn=an2an+12,求Tn. |
答案
(1)-
=f(an)=-1 an+1
且an>04+ 1 a 2n
∴
=1 an+1
∴4+ 1 a 2n
-1 a 2n+1
=41 a 2n
∴数列{
}是等差数列,首项1 a 2n
=1,公差d=41 a 21
∴
=1+4(n-1)∴1 a 2n
=a 2n 1 4n-3
∵an>0∴an=1 4n-3
(2)bn=
•1 4n-3
=1 4n+1
(1 4
-1 4n-3
)1 4n+1
Tn=
(1-1 4
+1 5
-1 5
+…+1 9
-1 4n-3
)=1 4n+1
(1-1 4
)=1 4n+1 n 4n+1