Question

已知数列{an}中,a1=1,a2=1/4,且a(n+1)=(n-1)an/(n-an)

Answer 1

(1)

a_(n+1)=(n-1)a_n/(n-a_n)

(n-a_n)/a_n=(n-1)/a_(n+1)

n/(a_n)-1=(n-1)/a_(n+1)

1/a_(n+1)=(n/n-1)(1/a_n)-1/(n-1)

1/a_(n+1)-1=(n/n-1)(1/(a_n)-1)

带入a_2=1/4

a_n=1/(3n-2)

(2)

S_n=1+1/4^2+1/7^2+...+1/(3n-2)^2