等差sn等于
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/11 01:58:54
an+1=2√Sn令n=1,解得a1=1平方得,an²+2an+1=4Sn当n≥2时,a(n-1)²+2a(n-1)+1=4Sn-1两式相减得,an²-a(n-1)&su
∵an与2的等差中项等于Sn与2的等比中项,∴12(an+2)=2Sn,即Sn=18(an+2)2. …(2分)当n=1时,S1=18(a1+2)2⇒a1=2; …(3
(1)∵an与2的等差中项等于Sn与2的等比中项由﹛an﹜是正项数列∴(an+2)/2=√2Sn∴8Sn=(an+2)²∴n=1时,8a1=(a1+2)²,a1=2n=2时,8(2
an+2Sn*S(n-1)=0而an=Sn-S(n-1)∴Sn-S(n-1)+2Sn*S(n-1)=0同除以Sn*S(n-1)整理:1/Sn-1/S(n-1)=2∴{1/Sn}为等差数列,公差2,首项
(2)2,6,10(2)由题意,2sn=[(an+2)/2]的平方,sn=an平方/8+an/2+1/2,则s(n-1)=a(n-1)平方+a(n-1)/2+1/2,两式相减得:sn-s(n-1)=a
看图片:前三项2,6,10(2)由题意,2sn=[(an+2)/2]的平方,sn=an平方/8+an/2+1/2,则s(n-1)=a(n-1)平方+a(n-1)/2+1/2,两式相减得:sn-s(n-
①依题意,得(an+2)/2=根号下(2Sn),∴a1+2=2根号下(2S1)=2根号下(2a1),∴(a1-2)的平方=0,∴a1=2,由a2+2=2根号下(2S2)=2根号下[2(2+a2)],得
{an},{√sn}都是等差数列,∴2√S2=√S1+√S3,即2√(2a1+d)=√a1+√(3a1+3d),平方得4(2a1+d)=a1+3a1+3d+2√[a1(3a1+3d)],4a1+d=2
(an+2)/2=√(2Sn)8Sn=(an+2)²n=1时,8S1=8a1=(a1+2)²(a1-2)²=0a1=2n≥2时,8Sn=(an+2)²8S(n-
前三项之和为9.A1=1,A2=3,A3=5猜想An=2n-1经验证符合题意.
Sn与2的等比中项为√(2Sn),an与2的等差中项为(an+2)/2由题目可知,8Sn=(an+2)^2,所以8S_(n-1)=[a_(n-1)+2]^2.两者相减,得8an=an^2+4an-[a
由已知条件可得(an+1)/2=√Sn下面就是逐步化解an^2+2an+1=4Sna(n-1)^2+2a(n-1)+1=4S(n-1)所以4an=an^2+2an+1-[a(n-1)^2+2a(n-1
打字好麻烦!还是写给你吧,第一问我不写了啊,自己带依题有:(an+2)/2=根号(2Sn),两边平方得,(an+2)²=an²+4an+4=8Sn,所以8Sn-8Sn-1=8an=
给你做成了一张图,做成详细的word比较麻烦
an与1的等差中项为:(an+1)/2因为{an}是正数组成的数列,所以Sn与1的等比中项为根号Sn那么根号Sn=(an+1)/2所以Sn=(an+1)^2/4当n1=,a1=(a1+1)^2/4即a
2)(an+2)/2=sqrt(2Sn)an^2+4an+4=8Sna(n+1)^2+4a(n+1)+4=8S(n+1)a(n+1)^2-an^2=4a(n+1)+4ana(n+1)-an=4数列{a
由已知an与1的等差中项等于Sn与1的等比中项得(an+1)/2=√SnSn=(an+1)²/4n=1时,S1=a1=(a1+1)²/4,整理,得(a1-1)²=0a1=
(1)由题意得2Sn=[(an+2)/2]^2,且an>0.取n=1,得2a1=[(a1+2)/2]^2,可解得a1=2;取n=2,得2(1+a2)=[(a2+2)/2]^2,且a2=6或a2=-2(
由已知条件可得(an+1)/2=√Sn下面就是逐步化解an^2+2an+1=4Sna(n-1)^2+2a(n-1)+1=4S(n-1)所以4an=an^2+2an+1-[a(n-1)^2+2a(n-1
解题思路:利用等差数列的通项公式及前n项和公式进行计算。解题过程: