作业帮高二数-已知数列『an』是各项为正的等比数列,且a1=1,a3a5=64..设bn=a(n+1)·log2 a(n+1),求数列『bn』的和
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高二数-已知数列『an』是各项为正的等比数列,且a1=1,a3a5=64..设bn=a(n+1)·log2 a(n+1),求数列『bn』的和
高二数-已知数列『an』是各项为正的等比数列,且a1=1,a3a5=64..设bn=a(n+1)·log2 a(n+1),求数列『bn』的和
高二数-已知数列『an』是各项为正的等比数列,且a1=1,a3a5=64..设bn=a(n+1)·log2 a(n+1),求数列『bn』的和
a3*a5=a1^q^2*a1*q^4=q^6=64 ∴q=2 ∴an=2^(n-1) ∴bn=a(n+1)*log2[a(n+1)]=n*2^n ∴{bn}的前n项和:Sn=2+2*2^2+3*2^3+…+n*2^n ∴2Sn=2^2+2*2^3+…+(n-1)*2^(n-1)+n*2^(n+1) ∴下式减上式,可得 Sn=-2-(2^2+2^3…+2^n)+n*2^(n+1) =(n-1)*2^(n+1)+2
an =a1q^(n-1)=q^(n-1)
a3.a5=64
q^6=64
q=2
an = 2^(n-1)
bn =a(n+1).log<2>a(n+1)
= n.2^n
let
S = 1.2+2.2^2+.....+n.2^n (1)
2S = ...
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an =a1q^(n-1)=q^(n-1)
a3.a5=64
q^6=64
q=2
an = 2^(n-1)
bn =a(n+1).log<2>a(n+1)
= n.2^n
let
S = 1.2+2.2^2+.....+n.2^n (1)
2S = 1.2^2+2.2^3+.....+n.2^(n+1) (2)
(2)-(1)
S = n.2^(n+1) -( 2+2^2+...+2^n)
=n.2^(n+1) - 2(2^n-1 )
= 2+ (n-1) .2^(n+1)
bn =n.2^n
Sn =b1+b2+...+bn = S =2+ (n-1) .2^(n+1)
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