Question

已知数列｛an｝的前n项和为sn，且sn=2n^2+n，n是正整数，又an=4log（2）bn+3

（1）求an，bn
（2）数列｛an*bn｝的前n项和Tn
我第二小题不会做，求思路

Answer 1

an*bn=n*2^(n+1)-2^(n-1)
后面2^(n-1)直接套等比数列求和公式
令cn=n*2^(n+1)
故1/2cn=n*2^n
Tn=c1+c2+c3……+cn=1*4+2*8+3*16+4*32……+(n-1)*2^n+n*2^(n+1)
1/2Tn=…… =1*2+2*4+3*8+…… +n*2^n（看对齐的2^k的系数都相差1）
Tn-1/2Tn=-(2+4+8……+2^n)+n*2^(n+1)
就可以求了
主要思想就是错位相减得到一个等比数列求和就行了

Answer 2

(1)
Sn= 2n^2+n (1)
S(n-1) = 2(n-1)^2 + (n-1) (2)
(1) -(2)
an= 2n^2+n - 2(n-1)^2 - (n-1)
= 4n-1
an=4log(2)bn+3
4n-1 = 4log(2)bn+3
log(2)bn = n-1
bn = 2^(n-1)
(2)
anbn = (4n-1)(2^(n-1))
= 4[n2^(n-1)] - 2^(n-1)
consider
[x^(n+1)-1]/(x-1) = 1+x+x^2+...+x^n
[(x^(n+1)-1)/(x-1)]' = 1+2x+..+nx^(n-1)
1+2x+..+nx^(n-1) = [ (x-1)(n+1)x^n) -(x^(n+1)-1) ]/(x-1)^2
= [nx^(n+1)-(n+1)x^n+1]/(x-1)^2
put x=2
1+2(2)+3(2)^2+..+n(2)^(n-1)
=n2^(n+1)- (n+1)2^n +1
summation anbn
=summation {4[n2^(n-1)] - 2^(n-1)}
= 4(n2^(n+1)- (n+1)2^n +1) - (2^n-1)
=(4n).2^(n+1) - (4n+5).2^n +5

Answer 3

(1-q)sn