Question

b＞0，数列{an}满足：a1=b，an=nban-1/(an-1+n-1)(n≥2). ⑴求数列{an}的通项公式

⑵证明：对于一切正整数n，2an≤bn+1+1

Answer 1

an=nba(n-1)/(a(n-1)+n-1)
an.a(n-1) +(n-1)an = nba(n-1)
1+(n-1)[ 1/a(n-1)] = nb (1/an)
(n-1)( 1/a(n-1) +[1/(1-b)]/(n-1)) = nb( 1/an + [1/(1-b)]/n )
( 1/an + [1/(1-b)]/n ) /( 1/a(n-1) +[1/(1-b)]/(n-1)) = (1/b) (n-1)/n
( 1/an + [1/(1-b)]/n )/(1/a1- 1/(1-b)) = (1/b) 1/n
( 1/an + [1/(1-b)]/n ) = (1-2b)/[b^2(1-b)] (1/n)
1/an = (1/n) [1/(1-b)] [ (1-2b)/b^2 - 1]
an = n(1-b)/ [ (1-2b)/b^2 - 1]
= n(1-b) b^2/ (1-2b-b^2)