等比数列｛an｝的前n项和为Sn,Sn=2的n次方+m,（1）求m（2）bn=4an分之n+1,求bn前n项和等比数列｛an｝的前n项和为Sn,Sn=2的n次方+m,（1）求m（2）bn=4an分之n+1,求bn前n项和

Question

等比数列｛an｝的前n项和为Sn,Sn=2 的n次方+m,（1）求m（2）bn=4an分之n+1,求bn前n项和
等比数列｛an｝的前n项和为Sn,Sn=2
的n次方+m,（1）求m（2）bn=4an分之n+1,求bn前n项和

Answer 1

设公比为q,q=1时,Sn=na1≠2ⁿ+m,因此q≠1
Sn=a1(qⁿ-1)/(q-1)=[a1/(q-1)]qⁿ -[a1/(q-1)]
Sn=2ⁿ+m
q=2
a1/(q-1)=1
m=-a1/(q-1)
解得m=-1 a1=1 q=2
m=-1
an=a1q^(n-1)=1×2^(n-1)=2^(n-1)
bn=(n+1)/4an=(n+1)/2^(n+1)
Tn=b1+b2+...+bn
=2/2²+3/2³+4/2⁴+...+(n+1)/2^(n+1)
Tn /2=2/2³+3/2⁴+...+n/2^(n+1) +(n+1)/2^(n+2)
Tn- Tn /2=Tn /2=2/2²+1/2³+1/2⁴+...+1/2^(n+1) -(n+1)/2^(n+2)
=1/2+1/2²+...+1/2^(n+1) -(n+1)/2^(n+2) -1/4
=(1/2)×[1-(1/2)^(n+1)]/(1-1/2) -(n+1)/2^(n+2) -1/4
=3/4- (n+3)/2^(n+2)
Tn=3/2 -(n+3)/2^(n+1)