设级数Σan发散,设Sn为其部分和,证明Σan Sn发散
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1.已知{an}是等差数列,其首项是a1,公差为d,前n项和为Sn,由a11=11,s11=153得a1+10d=11且11a1+[11*(11-1)/2]d=153解得a1=185/11,d=-32
因an=log2[(n+2)/(n+1)]=log2(n+2)-log2(n+1),n应该从1开始.所以Sn=log2(3)-log2(2)+log2(4)-log2(3)+...+log2(n+2)
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
若∑(an平方)收敛,证明∑(an/n)必收敛证明,∑(an)^2收敛,∑(bn)^2=∑(1/n)^2收敛(p级数p>1时收敛)所以∑|anbn|≤∑(1/2)((an)^2+(bn)^2)收敛(因
(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)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
2sn=(an)^2+an,2(sn+1)=(an+1)^2+(an+1)作差((sn+1)-(sn)=an+1)则((an+1)-an-1)((an+1)+an)=0因为数列{an}的各项都是正数所
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
你的数据是这样的吗?感觉数据不好算哦~……等比数列有一性质就是:如果Sn是等比数列(不为常数),则Sm、S2m-Sm,S3m-S2m……也构成等比数列.例如本题中,S3=a1+a1*q+a1*q^2S
设数列{an}的前n项和为Sn,Sn=a1(3n−1)2(对于所有n≥1),则a4=S4-S3=a1(81−1)2−a1(27−1)2=27a1,且a4=54,则a1=2故答案为2
设公差为d,则:a1=2-3da2=2-2da3=2-da4=2a5=2+d∵S5=20∴10-5d=20d=-2∴an=10-2n以上希望对你有所帮助
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|
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=
1.4a1=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
解题思路:考查数列的通项,考查等差数列的证明,考查数列的求和,考查存在性问题的探究,考查分离参数法的运用解题过程: