解:A(n+1)=An^2+6An+6A(n+1)+3=(An+3)^2所以A(n+1)+3=(A1+3)*2^n=5*2^n所以A(n+1)=5*2^n-3An=5*2^(n-1)-3Bn=1/(An -6)-1/(An^2+6An)=1/[5*2^(n-1)-9]-1/[25*2^(2n-2)-9]Tn={[1/(-4)-1/16]+[1/1-1/91]+...+1/[5*2^(n-1)-9] 展开
解:A(n+1)=An^2+6An+6A(n+1)+3=(An+3)^2所以A(n+1)+3=(A1+3)*2^n=5*2^n所以A(n+1)=5*2^n-3An=5*2^(n-1)-3Bn=1/(An -6)-1/(An^2+6An)=1/[5*2^(n-1)-9]-1/[25*2^(2n-2)-9]Tn={[1/(-4)-1/16]+[1/1-1/91]+...+1/[5*2^(n-1)-9]-1/[25*2^(2n-2)-9]}因为当n≥2时,1/[5*2^(n-1)-9]-1/[25*2^(2n-2)-9]>0所以Tn>1/(-4)-1/16=-5/16 收起
