设正整数等比数列为一个等比数列,切a2=4,a4=16
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第一题很简单,用带特殊值的方法做,a2=2a1+2a3=2(a1+a2)+2因为an是等比数列,可以分q是1和q不是1来设,然后代入,只剩q和a1两个未知数,易解第二题不会......
设{an}公差为d,{bn}公比为q,q>0a3+b3=a1+2d+b1q^2=1+2d+3q^2=17d=(16-3q^2)/2T3-S3=b1(1+q+q^2)-(a1+a1+d+a1+2d)=3
解题思路:数列解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.php?
⑴若q=1,显然所求极限为na1/(n-5)a1=n/(n-5)的极限,易知极限是1q≠1时,所求的实际是Sn/(Sn-S5)的极限Sn/(Sn-S5)=a1(q^n-1)/[a1(q^n-1)/(q
S2=a1+a2S4-S2=a3+a4=(a1+a2)*q^2S6-S4=a5+a6=(a3+a4)*q^2所以公比为Q=q^2=1/9
s1=a1=2a1+1^2-3*1-2a1=2a1-4a1=4sn=2an+n^2-3n-2s(n-1)=2a(n-1)+(n-1)^2-3(n-1)-2sn-s(n-1)=2an+n^2-3n-2-
a(n+1)=2S(n-1)(1)a(n)=2S(n-2)(2)a(n+1)-an=2a(n-1)a(n+2)-a(n+1)-2an=0Theauxilaryequationx^2-x-2=0(x-2
解题思路:先向a1,q转换,再解方程组,然后分类讨论。解题过程:同学好最终答案:略
(1)∵Sn=2an-3n,对于任意的正整数都成立∴S(n-1)=2a(n-1)-3n-3两式相减,得a(n+1)=2a(n+1)-2an-3,即a(n+1)=2an+3∴a(n+1)+3=2(an+
1.a1a2=a1+da4=a1+3da2^2=a1*a4(a1+d)^2=a1*(a1+3d)2a1*d+d^2=3a1*dd=a1S10=110=10a1+10*9*d/2a1=d=2{an}=2
解题思路:等比数列解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.ph
(I)由a1,a2,a4成等比数列可得:(a1+2)2=a1(6+a1)∴4=2a1即a1=2∴an=2+2(n-1)=2n(II)∵bn=n•2an,=n•22n=n•4n∴Sn=1•4+2•42+
f(a1)=lga1+lgq,f(a2)=lga1+2lgq,…,f(a的第2m+1项)=lga1+(2m+1)lgq,加起来合并得:(2m+1)lga1+m(2m+1)lgq=(2m+1)(lga1
因为f(A1)+f(A2)+……+f(A2m+1)=1,所以lgA1+lgA2+...+lgA2m+1=1,即A1*A2*...*A(2m+1)=10,Am+1为该等比数列的中间项,所以Am+1=10
1.证:a(n+1)=an²+4an+2a(n+1)+2=an²+4an+4=(an+2)²log3[a(n+1)+2]=log3[(an+2)²]=2log3
q≠0,一、当q≠1时,Sn=a1((q^n)-1)/(q-1)>0等价于a1((q^n)-1)(q-1)>0,这是一式∵设等比数列{an}的公比为q,对任意正整数n,前n项的和Sn>0∴S1=a1>
1、an+1=2sn+2an=2s(n-1)+2相减a(n+1)-an=2ana(n+1)=3ann=1代入a2=2a1+2,a1=2an=2*3^(n-1)2、dn=(a(n+1)-an)/(n+1
解题思路:根据题目条件,由等比数列的知识可求解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/inc
k=b1+(k-1)d(d为公差,常数)设An=a1*q^(n-1)(q为公比,常数)则Abk=a1*q^[b1+(k-1)d]Ab(k-1)=a1*q^[b1+(k-2)d]所以Abk:Ab(k-1
①由题,有Sn==m+1-m*anS(n-1)=m+1-m*a(n-1)上式对应作差,可得:an=-m*an+m*a(n-1),即:an/a(n-1)=m/(m+1)故,数列{an}是以m/(m+1)