已知数列{an}的前n项和为Sn

因为2Sn=(n+2)an-1 n为任意正整数
2S(n-1)=(n+1)a(n-1)-1
2an=(n+2)an-(n+1)a(n-1)
nan-(n+1)a(n-1)=0
an/a(n-1)=(n+1)/n
2s1=3a1-1
s1=a1
a1=1
an/a1=an/a(n-1)*a(n-1)/a(n-2)....*a2/a1
=(n+1)/n*n/(n-1)......*3/2
an=(n+1)/2
(2) Tn=1/(a1*a2)+1/(a2*a3)+1/(a3*a4)…+1/(an*an+1)
=4/2*3+4/3*4+....4/(n+1)(n+2)
=4(1/2-1/3+1/3-1/4....+1/(n+1)-1/(n+2))
=4(1/2-1/(n+2))
=2n/(n+2)

(1)因为s1=a1，所以2a1=3a1-1，得到a1=1
2Sn=(n+2)an-1
2Sn-1=(n+1)an-1-1
故2an=(n+2)an-(n+1)an-1
nan=(n+1)an-1
an=[(n+1)/n]an-1
=[(n+1)/n][n/(n-1)]an-2
=...=
=[(n+1)/n][n/(n-1)]...[3/2]a1
=(n+1)/2
(2)
Tn=2/3+1/((3/2)(4/2))+1/((4/2) (5/2))+...+1/((n+1)(n+2)/4)
=2/3+4/3-4/4+4/4-4/5+...+4/(n+1)-4/(n+2)]
=1/a1^2*[2-4/(n+2)
=2n/(n+2)

2a1=3a1-1 a1=1
2Sn=[(n+2)an]-1
2S（n-1）=[(n-1+2)a（n-1）]-1
2Sn-2S（n-1）=nan+2an-na（n-1）-