数列{a n }中,n≥2,且a n =a n-1 -2,其前n项和是S n ,则有( ) A.na n <S n <
1个回答
∵a
n
=a
n-1
-2,
∴a
n
-a
n-1
-2,
∴数列是一个递减的等差数列,
∴na
n
<s
n
<na
1
,
故选A.
相关问题
设数列{a n }的首项a 1 =1,前n项和为S n ,且na n -S n =2n(n-1),n∈N*,
已知数列{a n }的前n项和为S n ,且a 1 =4,S n =na n +2- (n≥2,n∈N*)
已知数列{a n }的前n项和为s n ,且a 1 =1,na n+1 =(n+2)s n (n∈N * )
设数列{a n }的前n项和为S n ,a 1 =1,S n =na n -2n(n-1),
数列a_n前n项和S_n已知a_1=1,且3(S_n)^2=a_n(3S_n)-a_n,n>=2
已知数列{a n }中,a 1 =3,a 2 =5,S n 为其前n项和,且满足S n +S n-2 =2S n-1 +
已知数列{a n }的前n项和为S n ,且a 1 =1,a 2 =3,2S n ﹣(n+1)a n =A n +B(其
已知数列{a n }中,a 1 =2,a 2 =3,其前n项和S n 满足S n+1 +S n-1 =2S n +1((
已知数列{a n }满足 S n = n 2 a n (n∈N * ),其中S n 是{a n }的前n项和,且a 1