Question

高中数学题求详解谢谢

Answer 1

(1)
an=2+(n-1)x
bn=3*y^(n-1)
则有2+2x+3*y^4=56
2+4x+3*y^2=26
得（2*y^2+7）*（y^2-4）=0
得y=2(由于bn为正数数列，故y=2)
得x=3
得an=2+3*(n-1)=3n-1
bn=3*2^(n-1)
(2)
取m=n+1
则有an=a(m-1)=3m-4=am-3
a(n+2)=a(m+1)=3m+2=am+3
得an*a(n+2)=(am-3)*(am+3)
1/an*a(n+2)=1/(am-3)*(am+3)=[1/(am-3)-1/(am+3)]*(1/6)
Tn=(1/6)*[T(1/(am-3))-T(1/(am+3))]
T(1/(am-3))=T(1/(a(n+1)-3))=1/(a(2)-3)+1/(a(3)-3)+…………+1/(a(n+1)-3)
T(1/(am+3))=T(1/(a(n+1)+3))=1/(a(2)+3)+1/(a(3)+3)+…………+1/(a(n+1)+3)