已知数列An满足An>0,其前n项和为Sn为满足2Sn=An的平方+An(1)求An(2)设数列Bn满足An/2的n次方
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已知数列An满足An>0,其前n项和为Sn为满足2Sn=An的平方+An(1)求An(2)设数列Bn满足An/2的n次方,Tn=b1+b2+
(1)
2Sn=an^2+an
2Sn-1=a(n-1)^2+a(n-1)
2an=2Sn-2Sn-1=an^2-a(n-1)^2+an-a(n-1)
an^2-a(n-1)^2=an+a(n-1)
[an+a(n-1)][an-a(n-1)]=an+a(n-1)
an>0 an+a(n-1)>0,等式两边同除以an+a(n-1)
an-a(n-1)=1
an=a(n-1)+1
2a1=2S1=a1^2+a1
a1^2-a1=0
a1(a1-1)=0
a1=1
数列为首项是1,公差是1的等差数列.an=1+(n-1)*1=n
an=n
(2)
bn=n/2^n
Tn=b1+b2+...+bn=1/2+2/2^2+3/2^3+...n/2^n
Tn/2=(b1+b2+...+bn)=1/2^2+2/2^3+...+(n-1)/2^n
Tn-Tn/2=1/2+1/2^2+1/2^3+...+1/2^n
=(1/2)[(1/2)^(n+1)-1]/(1/2-1)
=1-(1/2)^(n+1)
Tn=2[1-(1/2)^(n+1)]
=2-(1/2)^n
=2-1/2^n
Tn=2-1/2^n
2Sn=an^2+an
2Sn-1=a(n-1)^2+a(n-1)
2an=2Sn-2Sn-1=an^2-a(n-1)^2+an-a(n-1)
an^2-a(n-1)^2=an+a(n-1)
[an+a(n-1)][an-a(n-1)]=an+a(n-1)
an>0 an+a(n-1)>0,等式两边同除以an+a(n-1)
an-a(n-1)=1
an=a(n-1)+1
2a1=2S1=a1^2+a1
a1^2-a1=0
a1(a1-1)=0
a1=1
数列为首项是1,公差是1的等差数列.an=1+(n-1)*1=n
an=n
(2)
bn=n/2^n
Tn=b1+b2+...+bn=1/2+2/2^2+3/2^3+...n/2^n
Tn/2=(b1+b2+...+bn)=1/2^2+2/2^3+...+(n-1)/2^n
Tn-Tn/2=1/2+1/2^2+1/2^3+...+1/2^n
=(1/2)[(1/2)^(n+1)-1]/(1/2-1)
=1-(1/2)^(n+1)
Tn=2[1-(1/2)^(n+1)]
=2-(1/2)^n
=2-1/2^n
Tn=2-1/2^n
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