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问题：数列{AN}满足a1=2,An+1=an2+6an+61.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
答案：↓↓↓ 侯看看的回答： 网友采纳 (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=T1=-1/4-1/16=-5/16 | 
