各项为正数的数列{an},其前n项和为Sn,且Sn=(√(Sn-1)+√a1)^2(n≥2),数列{bn}的前n项和为Tn,(见下)若bn=an+1/an+an/an+1,则Tn=?
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![各项为正数的数列{an},其前n项和为Sn,且Sn=(√(Sn-1)+√a1)^2(n≥2),数列{bn}的前n项和为Tn,(见下)若bn=an+1/an+an/an+1,则Tn=?](/uploads/image/z/1741111-7-1.jpg?t=%E5%90%84%E9%A1%B9%E4%B8%BA%E6%AD%A3%E6%95%B0%E7%9A%84%E6%95%B0%E5%88%97%7Ban%7D%2C%E5%85%B6%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94Sn%3D%28%E2%88%9A%EF%BC%88Sn-1%EF%BC%89%2B%E2%88%9Aa1%29%5E2%28n%E2%89%A52%29%2C%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%28%E8%A7%81%E4%B8%8B%29%E8%8B%A5bn%3Dan%2B1%2Fan%2Ban%2Fan%2B1%2C%E5%88%99Tn%3D%3F)
各项为正数的数列{an},其前n项和为Sn,且Sn=(√(Sn-1)+√a1)^2(n≥2),数列{bn}的前n项和为Tn,(见下)若bn=an+1/an+an/an+1,则Tn=?
各项为正数的数列{an},其前n项和为Sn,且Sn=(√(Sn-1)+√a1)^2(n≥2),数列{bn}的前n项和为Tn,(见下)
若bn=an+1/an+an/an+1,则Tn=?
各项为正数的数列{an},其前n项和为Sn,且Sn=(√(Sn-1)+√a1)^2(n≥2),数列{bn}的前n项和为Tn,(见下)若bn=an+1/an+an/an+1,则Tn=?
当n≥2是
Sn=(√(Sn-1)+√a1)^2
√Sn=√(Sn-1)+√a1
√Sn-√(Sn-1)=√a1
∴{√Sn}为等差数列,公差为√a1
∴√Sn=n√a1
∴Sn=n^2 *a1
S(n+1)=(n+1)^2*a1
a(n+1)=(2n+1)a1
an=(2n-1)a1
bn=an+1/an+an/an+1
=(2n+1)/(2n-1)+(2n-1)/(2n+1)
=[(2n-1)+3]/(2n-1)+[(2n+1)-3]/(2n+1)
=1+3/(2n-1)+1-3/(2n+1)
=2+3/(2n-1)-3/(2n+1)
Tn=b1+b2+b3+.+bn
=(2+3/1-3/3)+(2+3/3-3/5)+.+[2+3/(2n-1)-3/(2n+1)]
=2n+3[1-1/3+1/3-1/5+1/5-1/7+.+1/(2n-1)-1/(2n+1)]
=2n+3[1-1/(2n+1)]=2n+6n/(2n+1)=4n(n+2)/(2n+1)