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T

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)