已知{an}为等比数列,a1=1,a5=256,S n为等差数列{bn}的前n项和,b1=2,5S5=2S8 设Tn=a1b1+a2b2+...+anbn,求Tn

∵{an}为等比数列,a1=1,a5=256,
∴a1*q^4=256
∴q=4
∴an=4^n-1
又∵{bn}为等差数列，b1=2,5S5=2S8
Sn=na1+n(n-1)/2*d
∴d=3
∴bn=3n-1
∴Tn=2*4^0+5*4^1+8*4^2+...+(3n-1)*q^n-1①
将等式两边同乘以4
4Tn=2*4^1+5*4^2+...+(3n-1)*q^n②
由①-②，得到-3Tn=2*4^0+3*4^1+3*4^2+...+(3n-3)*4^n-1-(3n-1)*4^n
∴-3Tn=2*4^0+3(4^1+4^2+...+4^n)-(3n-1)*4^n
∴-3Tn=2*4^0+4^(n-1)-4-(3n-1)*4^n
∴Tn=-2/3+4^(n-1)/3-(3n-1)*4^n/3
O(∩_∩)O，希望对你有帮助

设公比为q,公差为d,
a5=a1q^4=q^4=256,q=土4.
由5S5=2S8 得
5[10+10d]=2[16+28d],
50+50d=32+56d,18=6d,d=3.
bn=3n-1.
Tn=2a1+5a2+……+（3n-1)an,
qTn=.....2a2+……+（3n-4)an+(3n-1)a,
相减得（1-q)Tn=-a1+3(a1+a2+……+an)-(3n-1)a
=3(1-q^n)/(1-q)-1-(3n-1)q^n
=[3-3*q^n-1-(3n-1)q^n+q+(3n-1)q^(n+1)]/(1-q)
=[2+q-(3n+2)q^n+(3n-1)q^(n+1)]/(1-q),
∴Tn=[2+q-(3n+2)q^n+(3n-1)q^(n+1)]/(1-q)^2,
q=4时Tn=[6-(3n+2)*4^n+(3n-1)*4^(n+1)]/9
=[2-(3n-2)*4^n]/3;
q=-4时Tn=[-2-(3n+2)*(-4)^n+(3n-1)*(-4)^(n+1)]/25
=[-2-(15n-2)*(-4)^n]/25.

求出通项公式 是第一步：a5/a1=q^4=256 q=4,an=a1*q^(n-1)=4^(n-1) n∈N*；5S5=2S8 推出：5（b1+b5）*5=2（b1+b8）*825b1+25b5=16b1+16b89b1=16b8-25b5得公差12d=18b1 d =3 bn=3n-1 n∈N*an*bn=（3n-1）*4^(n-1) n∈...

{an}为等比数列,a1=1, a5=256
==>q^4=256,q=4
an=4^(n-1)
等差数列{bn},b1=2,5S5=2S8
==>5(5b1+10d)=2(8b1+28d)
==>d=3, ∴bn=3n-1
Tn=2×1+5×4+8×16+...+(3n-1)×4^(n-1) (1)
4Tn=2×4+5×16+......+(3n-4)×4^(n-1)+(3n-1)×4^n (2)
(1)-(2): -3Tn=2+3[4+16+......+4^(n-1)]-(3n-1)×4^n
=2+12[4^(n-1)-1]/3-(3n-1)×4^n
=-2 -(3n-2)×4^n
Tn=2/3+(n-2/3)×4^n