题目内容
数列{an}的前n项和为Sn,已知Sn=
.
(1)求数列{an}的通项公式;
(2)若数列{cn}满足cn=
,求数列{cn}的前n项和为Tn.
n2+3n |
2 |
(1)求数列{an}的通项公式;
(2)若数列{cn}满足cn=
|
(1)当n=1时,a1=s1=2
n≥2时,an=sn-sn-1=
-
=n+1
当n=1时,a1=2适合上式
故an=n+1
(2)当n为偶数时,Tn=(a1+a3+…+an-1)+(a2+a4+…+an)
=(2+4+…+n)+(22+24+…+2n)
=
+
=
+
当n为奇数时,n-1为偶数
Tn=(a1+a3+…+an)+(a2+a4+…+an-1)
=(2+4+…+n+1)+(22+24+…+2n-1)
=
+
=
+
∴Tn=
n≥2时,an=sn-sn-1=
n2+3n |
2 |
(n-1)2+3(n-1) |
2 |
当n=1时,a1=2适合上式
故an=n+1
(2)当n为偶数时,Tn=(a1+a3+…+an-1)+(a2+a4+…+an)
=(2+4+…+n)+(22+24+…+2n)
=
(2+n)•
| ||
2 |
4(1-4
| ||
1-4 |
=
n(n+2) |
4 |
4(2n-1) |
3 |
当n为奇数时,n-1为偶数
Tn=(a1+a3+…+an)+(a2+a4+…+an-1)
=(2+4+…+n+1)+(22+24+…+2n-1)
=
(3+n)•
| ||
2 |
4(1-4
| ||
1-4 |
=
n2+4n+3 |
4 |
4(2n-1-1) |
3 |
∴Tn=
|
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