an=√(1/(4n-3)),若数列{bn}的前n项和为Tn,且满足T(n+1)/an^2=Tn/a(n+1)^2+(4

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an=√(1/(4n-3)),若数列{bn}的前n项和为Tn,且满足T(n+1)/an^2=Tn/a(n+1)^2+(4n+3)(4n+1)
试确定b1的值,使数列{bn}是等差数列

As far as I am concerned,this problem is false before "(4n+3)(4n+1)" is changed as "(4n-3)(4n+1)".
再问：恩，那该怎么做
再答：代入，得（4n-3）T(n+1)=(4n+1)Tn+(4n-3)(4n+1) 变形，得T(n+1)/(4n+1)-Tn/(4n-3)=1 即{Tn/(4n-3)}是以1为公差的等差数列设Tn/(4n-3)=Xn 那么X1=T1=b1 Xn=b1+n-1 Tn=(4n-3)(b1+n-1) T(n+1)=(4n+1)(b1+n-1) T(n+1)-Tn=b(n+1)=4b1+8n-3=b1+8n（由该式可知公差为8）解得b1=1 (严密的解题过程还需检验)

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