已知数列{an}中,a1=1,a1+2a2+3a3+...+nan=（n+1)(an)/2+1（n∈正整数）（1）求数列{an}的通项公式an

Question

已知数列{an}中,a1=1,a1+2a2+3a3+...+nan=（n+1)(an)/2 +1（n∈正整数）（1）求数列{an}的通项公式an

Answer 1

a1+2a2+3a3+...+nan=（n+1)(an)/2 (1)
a1+2a2+3a3+...+nan+ (n+1)a(n+1)=（n+2)a(n+1)/2 (2)
(2)-(1)
(n+1)a(n+1) = (1/2)[ (n+2)a(n+1) -(n+1)an ]
na(n+1) = -(n+1)an
a(n+1)/an = - (n+1)n
an/a(n-1) = - n/(n-1)
an/a1 = (-1)^(n-1) .n
an=(-1)^(n-1) .n