设数列{An}的前n项和Sn=An-1(n=1,2,3…),数列{bn}满足条件b1=3,bk+1=Ak+bk(k=1,
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设数列{An}的前n项和Sn=An-1(n=1,2,3…),数列{bn}满足条件b1=3,bk+1=Ak+bk(k=1,2,3)
求数列bn的前n项和sn‘
(2)设cn=3/(n+2)(Sn’-Sn),且数列cn的前n项和为Tn,求证Tn
错了,求证Tn
求数列bn的前n项和sn‘
(2)设cn=3/(n+2)(Sn’-Sn),且数列cn的前n项和为Tn,求证Tn
错了,求证Tn
S1=a1=2(a1)-1 a1=1
S(n-1)=2a(n-1)-1
an=Sn-S(n-1)=2an-2a(n-1)
an=2a(n-1)
an=a1*2^(n-1)=2^(n-1)
b(k+1)=2^(k-1)+bk
bk-b(k-1)=2^(k-2)
b(k-1)-b(k-2)=2^(k-3)
…………………………
b2-b1=2^0=1
以上各式左右相加:
bk-b1=1+2+2^2+……+2^(k-2)=2^(k-1)-1
bk=2^(k-1)+2
前n项和:
Tn=b1+b2+……+bn
=2n+[1+2+2^2+……+2^(n-1)]
=2n+2^n-1
=2^n+2n-1
S(n-1)=2a(n-1)-1
an=Sn-S(n-1)=2an-2a(n-1)
an=2a(n-1)
an=a1*2^(n-1)=2^(n-1)
b(k+1)=2^(k-1)+bk
bk-b(k-1)=2^(k-2)
b(k-1)-b(k-2)=2^(k-3)
…………………………
b2-b1=2^0=1
以上各式左右相加:
bk-b1=1+2+2^2+……+2^(k-2)=2^(k-1)-1
bk=2^(k-1)+2
前n项和:
Tn=b1+b2+……+bn
=2n+[1+2+2^2+……+2^(n-1)]
=2n+2^n-1
=2^n+2n-1
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