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,对任意n∈N*,有a1+3*a2+5*a3+.+(2n-1)*a= 2020-05-13 …
设数列{an}的通项公式为an=pn+q(n∈N*,p>0).数列{bn}定义如下:对于正整数m, 2020-05-13 …
下列程序段的输出结果是B.int*p,*q,k=1,j=10;p=&j;q=&k;p=q;(*p) 2020-05-14 …
(p-q)(a+b)²+2(p-q)(a+b)-(q-p)=? 2020-05-14 …
下面语句中完全正确的是A.inta,*p;*p=&a;B.inta,*p,*q=&a;p=q;C. 2020-06-12 …
设命题p和命题q,“p∨q”的否定是真命题,则必有()A.p真q真B.p假q假C.p真q假D.p假 2020-08-01 …
Q=(V×D)/(10×A)错误应该是Q×A=V×D×10配制及稀释漂白水时应考虑所使用漂白水之有效 2020-11-06 …
真理逻辑题v表示或n表示与A=(Pn(-Q))v((-P)nQ)B=-((PnQ)v((-P)n(- 2020-11-21 …
对于命题p和命题q,若p真q假,则命题p∧q和命题p∨q的真假为()A.p∧q和p∨q都为真B.p∧ 2020-12-13 …
设p:(3x2+ln3)′=6x+3;q:(3-x2)ex的单调增区间是(-3,1),则下列复合命题 2020-12-13 …