有关数列第一道题~谢谢

题目

有关数列第一道题~谢谢
数列{an}的前n项和Sn,已知a1=1/2,Sn=n^2an-n(n-1),n∈N
(1)写出Sn与S(n-1)(n≥2)的递推关系,并求Sn的表达式
(2)设bn=[(n+1)Snp^n]/n(p∈R),求数列{bn}的前n项和Tn.
这道题有些繁琐,希望大家帮忙详细解答一下,谢谢了~

答案

(1)
S(n) = n^2 * a(n) - n(n-1)
a(n) = S(n) - S(n-1)带入上式
S(n) = n^2 * ( S(n) - S(n-1) ) - n(n-1)
( n^2 - 1 ) * S(n) = n^2 * S(n-1) + n(n-1)
S(n) = n^2 / ( n^2 - 1 ) * S(n-1) + n / ( n + 1 )
( n + 1 ) * S(n) / n = n * S(n-1) / ( n - 1 ) + 1
数列{( n + 1 ) * S(n) / n}为等差数列
( n + 1 ) * S(n) / n = ( 1 + 1 ) * S(1) / 1 + n - 1
S(1) = a(1) = 0.5,带入上式
S(n) = n^2 / ( n + 1 )
(2)
b(n) = ( ( n + 1 ) * S(n) * p^n ) / n
= n * p^n
T(n) = ∑(k=1->n) b(k) = ∑(k=1->n) k * p^k
∫ T(n) / p dp = ∑(k=1->n) k * ∫ p^(k-1)dp
= ∑(k=1->n) p^k
= ( p - p^(n+1) ) / ( 1 - p )
T(n) = d( ∫ T(n) / p dp ) / dp * p
= d( ( p - p^(n+1) ) / ( 1 - p ) ) / dp
= ( ( 1 - (n+1) * p^n ) * ( 1 - p ) + p - p^(n+1) ) / ( 1 - p )^2
= ( p - (n+1) * p^(n+1) + n * p^(n+2) ) / ( 1 - p )^2