题目

已知数列{bn}的前n项和为Sn=【(3n^2)-n】/2,(1)求数列{bn}的通项公式 (2)已知数列{an}满足：a(n+1)=2an+3
已知数列{bn}的前n项和为Sn=【(3n^2)-n】/2,(1)求数列{bn}的通项公式
(2)已知数列{an}满足：a(n+1)=2an+3n-5,a1=1,n属于正数,求证：数列{an+3n-2}为等比数列,
(3)在(1)、(2)的条件下,求数列{(an+bn)×bn}的前n项和Tn

提问时间：2021-01-05

答案

(1)
Sn=【(3n^2)-n】/2
bn= Sn -S(n-1)
= 3n-1
Sn=【(3n^2)-n】/2
n=1,=> b1=1
ie
bn=1 ; n=1
=3n-1 ; n>=2
(2)
a(n+1)=2an+3n-5
a(n+1)+3(n+1)-2 = 2(an+3n-2)
[a(n+1)+3(n+1)-2]/(an+3n-2)=2
=>{an+3n-2}为等比数列
[a(n+1)+3(n+1)-2]/(an+3n-2)=2
(an+3n-2)/(a1+3-2)=2^(n-1)
an+3n-2=2^n
an = -3n+2+2^n
(3)
n=1
c1 = (a1+b1)b1 =(1+1)1= 2
for n>=2
cn = (an+bn)bn
= (-3n+2+2^n+3n-1)(3n-1)
= (1+2^n)(3n-1)
= (3n-1) + 6[n.2^(n-1)] - 2^n
c2+c3+...+cn= (3n+4)(n-1)/2 - 4(2^(n-1)-1) + 6[∑(i:2->n) i.2^(i-1)]
= (3n+4)(n-1)/2 - 4(2^(n-1)-1) + 6[∑(i:1->n) i.2^(i-1)] - 6
= (3n+4)(n-1)/2 - 2^(n+1) -2 + 6[∑(i:1->n) i.2^(i-1)]
c1+c2+...+cn =(3n+4)(n-1)/2 - 2^(n+1) -1 + 6[∑(i:1->n) i.2^(i-1)]
consider
1+x+x^2+...+x^n = (x^(n+1)-1)/(x-1)
1+2x+.+nx^(n-1) = [(x^(n+1)-1)/(x-1)]'
= [nx^(n+1)-(n+1)x^n+1]/(x-1)^2
put x=2
∑(i:1->n) i.2^(i-1) = n.2^(n+1)-(n+1).2^n+1
= 1+(n-1).2^n
Tn=c1+c2+.+cn
=(3n+4)(n-1)/2 - 2^(n+1) -1 + 6[∑(i:1->n) i.2^(i-1)]
=(3n+4)(n-1)/2 - 2^(n+1) -1 + 6[1+(n-1).2^n]
=(3n+4)(n-1)/2 +5 +(6n-8).2^n
= (3n^2+n+6)/2 + (6n-8).2^n