Question

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=？
详解

Answer 1

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)