数列{AN}满足a1=2,An+1=an2+6an+6
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数列{AN}满足a1=2,An+1=an2+6an+6
1.求数列{AN}的通项公式
2.设bn=1/(an-6)-1/(an2+6an),{BN}前N项和为TN,
求证:-5/16
1.求数列{AN}的通项公式
2.设bn=1/(an-6)-1/(an2+6an),{BN}前N项和为TN,
求证:-5/16
(1)a(n+1)=(an)^2+6an+6 两边同加3得到
a(n+1)+3=(an)^2+6an+9=(an+3)^2;
所以an+3=[a(n-1)+3]^2=[a(n-2)+3]^4=……=(a1+3)^(2^(n-1))=5^(2^(n-1));
所以 an=5^(2^(n-1))-3;
(2)bn=1/(an-6)-1/(an2+6an)
=1/(an-6)-1/(a(n+1)-6);
所以
Tn=1/(a1-6)-1/(a2-6)+1/(a2-6)-1/(a3-6)+…+1/(an-6)-1/(a(n+1)-6)
=1/(a1-6)-1/(a(n+1)-6) 代入a1=2 和 a(n+1)=5^(2^n)-3;
= -1/4-1/[5^(2^n)-9];
显然对任意自然数1/[5^(2^n)-9]>0,所以Tn< -1/4;
又因为[5^(2^n)-9]单增,即1/[5^(2^n)-9]单减,故-1/[5^(2^n)-9]也单增,所以Tn在n=1时取最小值,即
Tn>=T1= -1/4-1/16= -5/16
a(n+1)+3=(an)^2+6an+9=(an+3)^2;
所以an+3=[a(n-1)+3]^2=[a(n-2)+3]^4=……=(a1+3)^(2^(n-1))=5^(2^(n-1));
所以 an=5^(2^(n-1))-3;
(2)bn=1/(an-6)-1/(an2+6an)
=1/(an-6)-1/(a(n+1)-6);
所以
Tn=1/(a1-6)-1/(a2-6)+1/(a2-6)-1/(a3-6)+…+1/(an-6)-1/(a(n+1)-6)
=1/(a1-6)-1/(a(n+1)-6) 代入a1=2 和 a(n+1)=5^(2^n)-3;
= -1/4-1/[5^(2^n)-9];
显然对任意自然数1/[5^(2^n)-9]>0,所以Tn< -1/4;
又因为[5^(2^n)-9]单增,即1/[5^(2^n)-9]单减,故-1/[5^(2^n)-9]也单增,所以Tn在n=1时取最小值,即
Tn>=T1= -1/4-1/16= -5/16
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