Question

已知an是为正数的等比数列,a1=1,a5=256,Sn为等差数列bn的前n项和,b1=2,5S5=2S8

1.求an和bn的通项公式
2.设Tn=a1b1++a2b2+......+anbn

Answer 1

(1)
an=a1q^(n-1)
a1=1,a5=256
256=q^(4)
q=4
ie
an=4^(n-1)

bn= b1+(n-1)d
Sn = (2b1+(n-1)d)n/2
b1=2, 5S5=2S8
25[4+4d]/2 = 8(4+7d)
100+100d=64+112d
d= 3
ie
bn=2+(n-1)3
= 3n-1
(2)
Tn=a1b1+a2b+... +anbn
= summation ( (3n-1).4^(n-1))
= summation [ 3(n.4^(n-1)) - 4^(n-1) ]
consider
(x^(n+1)-1)/(x-1) = 1+x +x^2+..+x^n
[(x^(n+1)-1)/(x-1)]' = 1+2x+3x^2+... +n(x^(n-1))
1+2x+3x^2+... +n(x^(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=4
1(4^0)+2(4^1)+3(4^2)+..+n(4^(n-1))
= [n4^(n+1)-(n+1)4^n +1]/9
Therefore
Tn
=summation [ 3(n.4^(n-1)) - 4^(n-1) ]
= [n4^(n+1)-(n+1)4^n +1]/3 - 4^n +1
= [ (3n-4).4^n + 4 ]/3