首页
已知等差数列{an}的前n项和为Sn,且(2n-1)Sn+1-(2n+1)Sn=4n2-1...
试题详情
已知等差数列{a
n}的前n项和为S
n,且(2n-1)S
n+1-(2n+1)S
n=4n
2-1(n∈N
*).
(Ⅰ)求数列{a
n}的通项公式;
(Ⅱ)求证:
.
相关试题
-
已知数列{an}的前项的和Sn满足Sn=2n-1(n∈N*),则数列{an2}的前项的和为( )
A.4n-1
B.(4n-1)
C.(4n-1)
D.(2n-1)2
-
已知等差数列{an}的前n项和为Sn,且(2n-1)Sn+1-(2n+1)Sn=4n2-1(n∈N*).
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)求证:.
-
已知等差数列{an}的前n项和为Sn,且(2n-1)Sn+1-(2n+1)Sn=4n2-1(n∈N*).
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)求证:.
-
已知等差数列{an}的前n项和为Sn,且(2n-1)Sn+1-(2n+1)Sn=4n2-1(n∈N*).
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)求证:.
-
已知数列{an}满足a1=-1,an+1-2an-3=0数列{bn}满足bn=log2(an+3).
(1)求{bn}的通项公式;
(2)若数列{2n+1bn}的前n项的和为sn,试比较sn与8n2-4n的大小.
-
(理)已知等比数列{an}的前n项和Sn=2n-1,则a12+a22+…+an2等于( )
A.(2n-1)2
B.
C.4n-1
D.
-
(理)已知等比数列{an}的前n项和Sn=2n-1,则a12+a22+…+an2等于( )
A.(2n-1)2
B.
C.4n-1
D.
-
(理)已知等比数列{an}的前n项和Sn=2n-1,则a12+a22+…+an2等于( )
A.(2n-1)2
B.
C.4n-1
D.
-
(理)已知等比数列{an}的前n项和Sn=2n-1,则a12+a22+…+an2等于( )
A.(2n-1)2
B.
C.4n-1
D.
-
(理)已知等比数列{an}的前n项和Sn=2n-1,则a12+a22+…+an2等于( )
A.(2n-1)2
B.
C.4n-1
D.