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=2S81.求an和bn的通项公式2.设Tn=a1b1++a2b2+.+anbn
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![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=2S81.求an和bn的通项公式2.设Tn=a1b1++a2b2+.+anbn](/uploads/image/z/5218958-38-8.jpg?t=%E5%B7%B2%E7%9F%A5an%E6%98%AF%E4%B8%BA%E6%AD%A3%E6%95%B0%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2Ca1%3D1%2Ca5%3D256%2CSn%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2Cb1%3D2%2C5S5%3D2S81.%E6%B1%82an%E5%92%8Cbn%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F2.%E8%AE%BETn%3Da1b1%2B%2Ba2b2%2B.%2Banbn)
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=2S81.求an和bn的通项公式2.设Tn=a1b1++a2b2+.+anbn
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
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=2S81.求an和bn的通项公式2.设Tn=a1b1++a2b2+.+anbn
(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