设正项数列{an}的前n项和为Sn,向量a=(根号Sn,1),b=(an+1,2),(n∈N*)满足a∥b(1)求数列{an}的通项公式(2)设数列{bn}的通项公式为bn=an/(an+t)(t∈N*),若b1,b2,bm(m≥3,m∈N*)成等差数列,
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![设正项数列{an}的前n项和为Sn,向量a=(根号Sn,1),b=(an+1,2),(n∈N*)满足a∥b(1)求数列{an}的通项公式(2)设数列{bn}的通项公式为bn=an/(an+t)(t∈N*),若b1,b2,bm(m≥3,m∈N*)成等差数列,](/uploads/image/z/9300003-51-3.jpg?t=%E8%AE%BE%E6%AD%A3%E9%A1%B9%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%90%91%E9%87%8Fa%3D%EF%BC%88%E6%A0%B9%E5%8F%B7Sn%2C1%EF%BC%89%2Cb%3D%EF%BC%88an%2B1%2C2%EF%BC%89%2C%EF%BC%88n%E2%88%88N%2A%EF%BC%89%E6%BB%A1%E8%B6%B3a%E2%88%A5b%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%882%EF%BC%89%E8%AE%BE%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E4%B8%BAbn%3Dan%2F%EF%BC%88an%2Bt%EF%BC%89%EF%BC%88t%E2%88%88N%2A%EF%BC%89%2C%E8%8B%A5b1%2Cb2%2Cbm%EF%BC%88m%E2%89%A53%2Cm%E2%88%88N%2A%EF%BC%89%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C)
设正项数列{an}的前n项和为Sn,向量a=(根号Sn,1),b=(an+1,2),(n∈N*)满足a∥b(1)求数列{an}的通项公式(2)设数列{bn}的通项公式为bn=an/(an+t)(t∈N*),若b1,b2,bm(m≥3,m∈N*)成等差数列,
设正项数列{an}的前n项和为Sn,向量a=(根号Sn,1),b=(an+1,2),(n∈N*)满
足a∥b
(1)求数列{an}的通项公式
(2)设数列{bn}的通项公式为bn=an/(an+t)(t∈N*),若b1,b2,bm(m≥3,m∈N*)成等差数列,求t和m的值
(3)如果等比数列{cn}满足c1=a1,公比q满足0<q<1/2,且对任意正整数k,ck-(c(k+1)+c(k+2))仍是该数列中的某一项,求公比q的取值范围
设正项数列{an}的前n项和为Sn,向量a=(根号Sn,1),b=(an+1,2),(n∈N*)满足a∥b(1)求数列{an}的通项公式(2)设数列{bn}的通项公式为bn=an/(an+t)(t∈N*),若b1,b2,bm(m≥3,m∈N*)成等差数列,
(1)
a=(√Sn,1),b=(an +1,2)
a//b
√Sn/(an +1)=1/2
2√Sn = an + 1
4Sn = (an + 1)^2
n=1,
(a1)^2-2a1=1=0
a1=1
an = Sn-S(n-1)
4an = (an + 1)^2 - (a(n-1) + 1)^2
(an)^2- [a(n-1)]^2 - 2[an + a(n-1)]=0
[an + a(n-1)].[an - a(n-1)-2]=0
an - a(n-1)-2=0
an-a(n-1) =2
an-a1=2(n-1)
an = 2n-1
(2)
bn= an/(an +t)
b1,b2,bm成等差数列
b1+bm = 2b2
a1/(a1 +t) + am/(am +t) = 2[a2/(a2 +t)]
1/(1 +t) + (2m-1)/(2m-1 +t) = 2[3/(3 +t)]
2- t/(1+t) - t/(2m-1 +t) = 2 - 2t/(3 +t)
1/(1+t) + 1/(2m-1 +t) = 2/(3+t)
2(1+t)(2m-1+t) = 2(m +t)(3+t)
t^2+2mt+(2m-1) = t^2+(m+3)t+3m
(m-3)t = m+1
t = (m+1)/(m-3)
m=4
t= 5
(3)
cn=c1*q^(n-1)=q^(n-1)
q^(k-1)-[q^k+q^(k+1)]
=q^(k-1)*[1-q-q²]
由于cn都是q的几次方的形式
所以1-q-q²应该也是q的几次方的形式
而0