设an为正级数,sn为其部分和.an sn收敛
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![设an为正级数,sn为其部分和.an sn收敛](/uploads/image/f/7246150-70-0.jpg?t=%E8%AE%BEan%E4%B8%BA%E6%AD%A3%E7%BA%A7%E6%95%B0%2Csn%E4%B8%BA%E5%85%B6%E9%83%A8%E5%88%86%E5%92%8C.an+sn%E6%94%B6%E6%95%9B)
设∑an收敛到SS,n->∞∴1/Sn->1/S≠0,∴∑(1/Sn)发散
思路如下,缩放题要自己慢慢领悟归纳,慢慢来,思想就那几个的,关键注意通项的放缩化简
Un=S(n+1)-Sn=1/(2n+2)+1/(2n+1)-1/(n+1)=1/(2n+1)-1/(2n+2)Un的部分和=1/3-1/(2n+2)收敛于1/3再问:un不是应该等于sn-s(n-1
数列各项均为正,Sn>0.2√Sn是a(n+2)与an的等比中项,则(2√Sn)²=(an+2)an4Sn=an²+2ann=1时,4a1=4S1=a1²+2a1a1
(1)2Sn=an^2+an2Sn-1=a(n-1)^2+a(n-1)2an=2Sn-2Sn-1=an^2-a(n-1)^2+an-a(n-1)an^2-a(n-1)^2=an+a(n-1)[an+a
1因为Sn=3/2an-1/2所以S(n-1)=3/2a(n-1)-1/2两式相减得:an=3/2(an-a(n-1))化简得:an=3a(n-1)当n=1时,S1=3/2a1-1/2,S1=a1,解
1)由题意得,a1=1,当n>1时,sn=an^2/2+an/2sn-1=a(n-1)^2/2+a(n-1)/2,∴sn-sn-1=an^2/2-a(n-1)^2/2+an/2-a(n-1)/2即(a
(1)(an+2)/2=根号下2Sn所以8Sn=(an+2)^2n=1,S1=a1.8a1=(a1+2)^2,得a1=2n=2,8S2=(a2+2)^2,8(a1+a2)=(a2+2)^2,得a2=6
由题意可知;an=log2n+1n+2(n∈N*),设{an}的前n项和为Sn=log223+log234+…+log2nn+1+log2n+1n+2,=[log22-log23]+[log23-lo
(一)(1)由a1=1,S(n+1)=4an+2.可得:a1=1,a2=5,a3=16.a4=44.∴由bn=(an)/2^n得:b1=2/4,b2=5/4,b3=8/4,b4=11/4.显然,b1,
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
Sn=a1n+n(n-1)d/2S4=4a1+6d=-62S6=6a1+15d=-75a1=-20,d=3an=a1+(n-1)d=3n-23当n<8时,an<0当n≥8时,an>0|a1|+|a2|
因为an,Sn,an^2成等差数列所以2Sn=an^2+an2an=2Sn-2S(n-1)=an^2+an-a(n-1)^2-a(n-1)得:(an-a(n-1))(an+a(n-1))-(an+a(
a2a4=64=(a3)^2,所以a3=8(都是正数)S3=a3/q^2+a3/q+a3=14,解得q=2,所以an=2^n,bn=(n+1)log2C(n+1)=C1+log2[2/2+3/2^2+
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
an,Sn,an^2成等差数列2Sn=an^2+an2Sn-1=an-1^2+an-1相减2an=an^2+an-(an-1^2+an-1)(an+an-1)=(an-an-1)(an+an-1)若{
24=S4=a1+a2+a3+a4=2(a2+a3)=>a2+a3=12a2*a3=35=>a2=5,a3=7=>a1=3=>an=3+(n-1)*2=2n+1bn=1/an*a(n+1)=1/((2
an=log2(n+1)-log2(n+2)Sn=log2(2)-log2(3)+log2(3)-log2(4)+.+log2(n)-log2(n+1)+log2(n+1)-log2(n+2)=log