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
试确定b1的值,使数列{bn}是等差数列
![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](/uploads/image/z/19754130-66-0.jpg?t=an%3D%E2%88%9A%281%2F%284n-3%29%29%2C%E8%8B%A5%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E4%B8%94%E6%BB%A1%E8%B6%B3T%28n%2B1%29%2Fan%5E2%3DTn%2Fa%28n%2B1%29%5E2%2B%284)
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)
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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