早教吧作业答案频道 -->数学-->
已知等差数列(an)的前n项和为sn=n^2+pn+q(p,q∈r),且a2,a3 ,a5成等比数列(1)求p,q的值(2)若数列bn满足an+log2n=log2bn,求数列bn的前n项和Tn
题目详情
已知等差数列(an)的前n项和为sn=n^2+pn+q(p,q∈r),且a2,a3 ,a5成等比数列(1)求p,q的值(2)若数列bn满足an+log2n=log2bn,求数列bn的前n项和Tn
▼优质解答
答案和解析
(1)
Sn = n^2+pn+q
n=1 , a1= p+q+1
an = Sn - S(n-1)
= 2n-1 + p (1)
a1= 1+p = p+q+1
=>q=0
from (1)
a2 = 3+p
a3 =5+p
a5 = 9+p
a2,a3 ,a5成等比数列
a2.a5 = (a3)^2
(3+p)(9+p)=(5+p)^2
12p+27 = 10p+25
p=-1
(2)
an = 2n-2
an+logn=logbn
2n-2 = log(bn/n)
bn/n = 2^(2n-2)
bn = n.2^(2n-2)
let
S = 1.2^0 + 2.2^2 + .+n.2^(2n-2) (2)
4S = 1.2^2 + 2.2^4 + .+n.2^(2n) (3)
(3)-(2)
3S = n.2^(2n) - [ 1 + 2^2+2^4+...+2^(2n-2) ]
= n.2^(2n) - (1/3)(2^(2n) -1)
S = 3n.2^(2n) - (2^(2n) -1)
= 1 + (3n-1).2^(2n)
Tn = b1+b2+...+bn=S =1 + (3n-1).2^(2n)
Sn = n^2+pn+q
n=1 , a1= p+q+1
an = Sn - S(n-1)
= 2n-1 + p (1)
a1= 1+p = p+q+1
=>q=0
from (1)
a2 = 3+p
a3 =5+p
a5 = 9+p
a2,a3 ,a5成等比数列
a2.a5 = (a3)^2
(3+p)(9+p)=(5+p)^2
12p+27 = 10p+25
p=-1
(2)
an = 2n-2
an+logn=logbn
2n-2 = log(bn/n)
bn/n = 2^(2n-2)
bn = n.2^(2n-2)
let
S = 1.2^0 + 2.2^2 + .+n.2^(2n-2) (2)
4S = 1.2^2 + 2.2^4 + .+n.2^(2n) (3)
(3)-(2)
3S = n.2^(2n) - [ 1 + 2^2+2^4+...+2^(2n-2) ]
= n.2^(2n) - (1/3)(2^(2n) -1)
S = 3n.2^(2n) - (2^(2n) -1)
= 1 + (3n-1).2^(2n)
Tn = b1+b2+...+bn=S =1 + (3n-1).2^(2n)
看了 已知等差数列(an)的前n项...的网友还看了以下:
已知数列an满足a1=1╱4an=an-1(-1)n╱an-1-2设bn=1╱an2,求数列bn的 2020-05-17 …
数列问题(速求!)此处的a和S旁的1,n+1,n都是下标1.已知数列满足a1=2,an+1=3的n 2020-06-06 …
已知数列{an}满足a1=1/4,a2=3/4,a(n+1)=2an-a(n-1)(n>等于2,n 2020-06-27 …
已知数列{an}满足a1=1/4,a2=3/4,a(n+1)=2an-a(n-1)(n>等于2,n 2020-06-27 …
已知数列{an}满足a1=1/4,a2=3/4,a(n+1)=2an-a(n-1)(n>等于2,n 2020-06-27 …
已知数列{an}满足a1=1/4,a2=3/4,a(n+1)=2an-a(n-1)(n>等于2,n 2020-06-27 …
设n维列向量组a1,a2,...ar线性无关A是m*n列满秩矩阵,证明Aa1,...Aar也线性无 2020-06-30 …
已知数列{an}的前n项和为Sn=n*(a1+a2)/2,数列{bn}满足b1+3b2+3^2b3 2020-07-03 …
数列{an}与{bn}满足关系:a1=2,a(n+1)=(an^2+1)/2an,bn=(an+1 2020-07-22 …
1、已知数列{An}满足:A1=1,A2=1/2,且[3+(-1)^n]A-2An+2[(-1)^ 2020-08-01 …