已知数列{an}是公比大于1的等比数列,满足a3·a4=128,a2+a5=36;数列{bn}满足b(n+1)已知数列{an}是公比大于1的等比数列,满足a3·a4=128,a2+a5=36；数列{bn}满足b（n+1）=2bn-b（n-1）n属于N+,n大于等于2）且b1=1,b2

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已知数列{an}是公比大于1的等比数列,满足a3·a4=128,a2+a5=36;数列{bn}满足b(n+1)已知数列{an}是公比大于1的等比数列,满足a3·a4=128,a2+a5=36；数列{bn}满足b（n+1）=2bn-b（n-1）n属于N+,n大于等于2）且b1=1,b2
已知数列{an}是公比大于1的等比数列,满足a3·a4=128,a2+a5=36;数列{bn}满足b(n+1)
已知数列{an}是公比大于1的等比数列,满足a3·a4=128,a2+a5=36；数列{bn}满足b（n+1）=2bn-b（n-1）
n属于N+,n大于等于2）且b1=1,b2=2
（1）求{an}及{bn}的通向公式（2）求数列{anbn}的前n项和Sn

已知数列{an}是公比大于1的等比数列,满足a3·a4=128,a2+a5=36;数列{bn}满足b(n+1)已知数列{an}是公比大于1的等比数列,满足a3·a4=128,a2+a5=36；数列{bn}满足b（n+1）=2bn-b（n-1）n属于N+,n大于等于2）且b1=1,b2
1...
a3*a4=a2*a5=128
a2+a5=36
又公比>1 ∴a2=4 a5=32 ∴公比=2 a1=2 ∴an=2^n
2.
b（n+1）=2bn-b（n-1）∴ b(n+1)-bn=bn-b(n-1) ∴{bn}为等差数列又b1=1,b2=2 ∴bn=n
3...
S=1*2+2*4+3*8+...+n(2^n)
2S=1*2^2+2*2^3+..+(n-1)(2^(n))+n(2^(n+1))
相减,有:
-S=1*2+2^2+2^3+...+2^n-n(2^(n+1))
=2(1-2^n)/(1-2)-n(2^(n+1))
=2^(n+1)-2-n(2^(n+1))
S=n(2^(n+1))-2^(n+1)+2

(1)
an=2^n
bn=n
(2)
Sn=(n-1)2^(n+1)+2

(1)
an=a1q^(n-1)
a3.a4=128
(a1)^2. q^5 =128 (1)
a2+a5=36
a1q(1+q^3) =36 (2)
(1)/(2)^2
q^3/(1+q^3)^2 = 8/81
81q^...

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(1)
an=a1q^(n-1)
a3.a4=128
(a1)^2. q^5 =128 (1)
a2+a5=36
a1q(1+q^3) =36 (2)
(1)/(2)^2
q^3/(1+q^3)^2 = 8/81
81q^3 = 8q^6+16q^3+8
8q^6-65q^3+8=0
(8q^3-1)(q^3-8)=0
q=2
a1= 4
an =a1q^(n-1) = 2^(n+1)
b(n+1)=2bn-b(n-1)
b(n+1) -bn = bn-b(n-1)
=b2-b1
=1
bn - b1 = n-1
bn = n
(2)
an.bn = n.2^(n+1)
let
S = 1.2^2+ 2.2^3+....+n.2^(n+1) (1)
2S = 1.2^3+2.2^4+....+n.2^(n+2) (2)
(2)-(1)
S = n.2^(n+1) -( 2^2+2^3+...+2^(n+1) )
=n.2^(n+1) - 4(2^n -1 )
= 4+ (n-2).2^(n+1)
Sn =a1b1+a2b2+..+anbn
=S
=4+ (n-2).2^(n+1)

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