(1)已知等差数列{an},满足a1+a2+…+a101=0,则有A.a1+a101>0 B.a1+a1010(2)设Sn是数列{an}的前n项和,且Sn=n^2,则{an}A.等比数列,但不是等差数列B.等差数列,但不是等比数列C.等差数列,而且也是等比

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/29 17:01:06
(1)已知等差数列{an},满足a1+a2+…+a101=0,则有A.a1+a101>0 B.a1+a1010(2)设Sn是数列{an}的前n项和,且Sn=n^2,则{an}A.等比数列,但不是等差数列B.等差数列,但不是等比数列C.等差数列,而且也是等比

(1)已知等差数列{an},满足a1+a2+…+a101=0,则有A.a1+a101>0 B.a1+a1010(2)设Sn是数列{an}的前n项和,且Sn=n^2,则{an}A.等比数列,但不是等差数列B.等差数列,但不是等比数列C.等差数列,而且也是等比
(1)已知等差数列{an},满足a1+a2+…+a101=0,则有
A.a1+a101>0 B.a1+a1010
(2)设Sn是数列{an}的前n项和,且Sn=n^2,则{an}
A.等比数列,但不是等差数列
B.等差数列,但不是等比数列
C.等差数列,而且也是等比数列
D.既非等比数列又非等差数列
(3)设Sn是等差数列{an}的前n项和,若S3/S6=1/3,则S6/S12=
A.3/10 B.1/3 C.1/8 D.1/9

(1)已知等差数列{an},满足a1+a2+…+a101=0,则有A.a1+a101>0 B.a1+a1010(2)设Sn是数列{an}的前n项和,且Sn=n^2,则{an}A.等比数列,但不是等差数列B.等差数列,但不是等比数列C.等差数列,而且也是等比
1:C (a1+a101)*101/2=0 ,so.
2:D if等差S(n+1)-Sn=(n+1)^2-n^2 if等比S(n+1)/Sn=(n+1)^2/n^2
二者解的的结果都非常数~so.
3:B 我忘了 好像用S(6-3) 总之用三项和再做每一项.

1c2b3a

c

已知等差数列{an}满足a(n+1)=an+3n+2,且a1=2,求an.雪地跪拜! 已知数列an满足 a1=1/2,an+1=3an/an+3求证1/an为等差数列已知数列an满足 a1=1/2,an+1=3an/an+3求证1/an为等差数列 已知{an}满足a1=2,a(n+1)=(2an)/(an+2),求证{1/an}为等差数列. 已知数列{an}满足a1=2,a(n+1)=2an/(an+2),证明:数列{1/an}为等差数列 已知等差数列〔an〕满足a1=1,an+1 已知等差数列{an}满足a1+a2+a3+……+a99=0则( ) A.a1+a99>0 B.a已知等差数列{an}满足a1+a2+a3+……+a99=0则( )A.a1+a99>0 B.a1+a99 已知数列{an}满足2an/an+2=an+1(n属于正整数),a1=1/1006.求证:数列{1/an}是等差数列,并求通项an已知数列{an}满足(2an)/(an+2)=a(n+1)(n属于正整数),a1=1/1006.求证:数列{1/an}是等差数列,并求通项an, 已知数列{an}满足:a1=2a,an=2a-a*a/an-1(n(-N*,n>=2).bn=1/an-a1.求证BN是等差数列.2.求数列AN的通项公式. 已知函数f(X)=X/(3x+1),数列{an}满足a1=1,a(n+1)=f(an),证明数列{1/an}是等差数列 已知数列an满足an+a(n+1)=2n+1,求证数列an为等差数列的充要条件为a1=1过程要详细 【数学证明】已知数列an满足an+a(n+1)=2n+1,求证数列an为等差数列的充要条件为a1=1 已知数列{An}满足A1=1,An+1=2An+2^n.求证数列An/2是等差数列 已知递增的等差数列{An}满足A1=1且A1,A2,A5成等比数列.(1)求等差数列{An}的通项An 已知等差数列{an}满足a1+a2+a3+...+a11=0,则有A、a1+a11>0B、a2+a10 【紧急--高一数学】已知数列{an}是首项a1=a,公差为2的等差数列;数列{bn}满足2bn=(n+1)an已知数列{an}是首项a1=a,公差为2的等差数列;数列{bn}满足2bn=(n+1)an(1)若a1,a3,a4成等比数列,求数列{an}的通项 (高二数学)已知数列{an}是首项a1=a,公差为2的等差数列;数列{bn}满足2bn=(n+1)an已知数列{an}是首项a1=a,公差为2的等差数列;数列{bn}满足2bn=(n+1)an(1)若a1,a3,a4成等比数列,求数列{an}的通项公式(2 已知数列an满足(an+1-an)(an+1+an)=9),且a1=2,a>0.求证:{an²}为等差数列求{an}的通项公式. 已知数列{an}满足a1=4且a(n+1),an,3成等差数列,其中.已知数列{an}满足a1=4且a(n+1),an,3成等差数列,其中n属于正自然数.1.求证:数列{an-3}是等比数列2.令bn=2n×an-6n,求数列{bn}的通项公式以及前n项