设函数 已知项数为2m 1 且各项均为正数的等比数列 ,
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![设函数 已知项数为2m 1 且各项均为正数的等比数列 ,](/uploads/image/f/7255560-48-0.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0+%E5%B7%B2%E7%9F%A5%E9%A1%B9%E6%95%B0%E4%B8%BA2m+1+%E4%B8%94%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B0%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97+%2C)
Sk=ak×a(k+1)/2Sk-1=ak×a(k-1)/2ak=Sk-Sk-1=ak[a(k+1)-a(k-1)]/2a1=1≠0,又ak=ak[a(k+1)-a(k-1)]/2对于任意正整数k均成
(1)∵2Sn=an2+n-4(n∈N*).∴2Sn+1=an+12+n+1-4.两式相减得2Sn+1-2Sn=an+12+n+1-4-(an2+n-4),即2an+1=an+12-an2+1,则an
2S2=b2(a1+a2)=b1*q*(2a1+d)=32,b3S3=b3(a1+a2+a3)=b1*q²*(3a1+3d)=120,得d=2(都是正数),q=2.∴an=a1+d(n-1)
a3^2=9a2*a6=a3*9a59a5=a3比值为1/3所以2a1+3a2=2a1+a1=3a1=1a1=1/3通项公式为an=1/3^n
a1=2,a2=1,等比1/2,an=2×(1/2)^(n-1).a1=2a2=1,a1=1,a2=1/2,等比1/2,an=1×(1/2)^(n-1).
设正项等比数列{an}公比为q(q>0)由a3^2=9a2a6=>(a3)^2=9(a3/q)*(a3*q^3)=>(a3)^2=9(a3)^2*q^2=>q=1/3又2a1+3a2=1=>2a1+3
(1)若λ=1,则(S(n+1)+λ)an=(Sn+1)a(n+1)两边除以ana(n+1)得S(n+1)/a(n+1)+1/a(n+1)=Sn/an+1/an∴Sn/an+1/an,是常数列.Sn/
sn=an(an+1)/2---------------1s(n-1)=a(n-1)(a(n-1)+1)/2-----------------21减2得an=an(an+1)/2-a(n-1)(a(n
sn=2n^2-n,bn=sn/(n+p)=(2n^2-n)/(n+p)b1=1/(1+p),b2=6/(2+p),b3=15/(3+p).bn是等差数列,则b1+b3=2b2,即1/(1+p)+15
设bn=根号an所以A(n-1)-An=(2倍根号An)+1等于根号[b(n-1)]^2-bn^2=2bn+1即[b(n-1)]^2=(bn+1)^2因为{a}中各项为正数,且a1=2所以b(n-1)
1.n=1时,2a1=2S1=a1²+1-4a1²-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=
n=1时,2a1=2S1=a1^2+1-4a1^2-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=an^2+n-4-a
6Sn=an^2+3an+26S(n-1)=a(n-1)^2+3a(n-1)+26Sn-6S(n-1)=6an=an^2+3an+2-a(n-1)^2-3a(n-1)-26an=an^2+3an-a(
设公比为q,数列各项均为正,q>0若q=1则S(2n)/Sn=(2na1)/(na1)=2≠6560/80,与已知不符,因此q≠1S(2n)/Sn=6560/80[a1(q^(2n)-1)/(q-1)
由题意得1S3=a1+a2+a3=7……1;6a2=a1+1+a3+6……22式+1式得a2=2……3将3式代入12得q=2或1/2a1=4或1an=4*(1/2)^(n-1)或an=2^(n-1)2
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
a1+a2+...+an=(1/2)(an²+an)a1+a2+...+a(n-1)=(1/2)(a(n-1)²+a(n-1))两式相减得an=(1/2)(an²+an)
B因为等差Sn=(a1+an)n/2=17*17,17已经是质数,所以要么a1+an/2=17,要么n/2=17(舍),所以d=0时,a1=an=17.d=16时,a1=9,an=25,d=32时,a
由题意2an=Sn+1/2Sn=2an-1/2n=1时,S1=a1a1=2a1-1/2a1=1/2S(n+1)-Sn=a(n+1)2a(n+1)-1/2-[2an-1/2]=a(n+1)a(n+1)=