设an是正数组成的数列,an与2的等差中项等于sn与2
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∵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
(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
因为点(an,an+1)(n∈N*)在函数y=x2+1的图象上,所以an+1=(an)2+1=an+1,即an+1-an=1,所以数列{an}是以1为首项,以1为公差的等差数列,则an=a1+(n-1
(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-
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=
/>由已知条件列式:(an+2)/2=√(2Sn)整理,得(an+2)²=8Sn令n=1(a1+2)²=8a1整理,得(a1-2)²=0a1=2令n=2(a2+2)
因为an与2的等差中项等于Sn与2的等比中项所以(an+2)/2=√(2Sn)即Sn=(an+2)^2/8.(1)当n=1时a1=S1=(a1+2)^2/8解得a1=2当n≥2时S(n-1)=(a(n
[(A{n}+2)/2]^2=2S{n}[(A{n+1}+2)/2]^2=2S{n+1}上下相减(A{n+1}-2)=(A{n+}+2)即A{n+1}=An+4再求A1=2An=4×n-2;(2)Sn
1当n=1时易得a1=t(an+t)^2/4=Sn*t展开4t*Sn=(an+t)^24t*S(n-1)=(a(n-1)+t)^2相减,配方an-t=an-1+tan=an-1+2tan=(2n-1)
[a(n)+2]^2=8s(n),[a(1)+2]^2=8s(1)=8a(1),[a(1)-2]^2=0,a(1)=2.[a(2)+2]^2=8s(2)=8[a(1)+a(2)],[a(2)-2]^2
1.8A1=8S1=(A1+2)^2(A1)^2-4A1+4=0A1=28(A1+A2)=8S2=(A2+2)^2(A2)^2-4A1-12=0A2=6A2=-2(舍去)8(A1+A2+A3)=(A3
1.4a1=4S1=(a1+1)²整理,得(a1-1)²=0a1=14S2=4a1+4a2=4+4a2=(a2+1)²整理,得(a2-1)²=4a2=-1(舍去
楼上的你的已经错了好不好啊bn=4/(an*an+1)=1/(4n-2)-1/(4n+2)错了!应该是bn=4/(an*an+1)=4/an-4/an+1Tn=4/a1-4/an+1不要误人子弟好不好
由题意得(an+1)/2=√(Sn×1)Sn=[(an+1)/2]²n=1时,S1=a1=[(a1+1)/2]²,整理,得(a1-1)²=0a1=1n≥2时,Sn=[(a
a1=2,a2=6,a3=10(an+2)/2=√2sn(an+2)^2=8sn(a(n-1)+2)^2=8s(n-1)相减:(an+2)^2-(a(n-1)+2)^2=8sn-8s(n-1)an^2