已知数列{an}中,a1=1,a2=1/4,且a(n+1)=(n-1)an/(n-an)求证：对一切n属于正整数,有6(a1)^2+6(a2)^2+6(a3)^2+...+6(an)^2＜7

Question

已知数列{an}中,a1=1,a2=1/4,且a(n+1)=(n-1)an/(n-an)
求证：对一切n属于正整数,有6(a1)^2+6(a2)^2+6(a3)^2+...+6(an)^2＜7

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