HMMT 二月 2015 · 团队赛 · 第 10 题

英文解析

10. [ 40 ] The sequences of real numbers { a } and { b } satisfy a = ( a − 1)( b + 1) andi i n +1 n − 1 ni =1 i =1
    b = a b − 1 for n ≥ 2, with a = a = 2015 and b = b = 2013. Evaluate, with proof, then +1 n n − 1 1 2 1 2
    infinite sum
    ( )
    ∞
    ∑
    1 1
    b − .
    a ann +1 n +3
    n =1
    1 4058211
    Answer: 1 + OR First note that a and b are weakly increasing and tend ton n
    2014 · 2015 4058210
    infinity. In particular, a , b ∈ { / 0 , − 1 , 1 } for all n .
    n n
    For n ≥ 1, we have a = ( a − 1)( b + 1) = ( a − 1)( a b ), son +3 n +1 n +2 n +1 n +1 nb 1 1 1 = = − .na a ( a − 1) a − 1 an +3 n +1 n +1 n +1 n +1
    Therefore,
    ( )
    ∞ ∞
    ∑ ∑
    b b b 1 1
    n n n − = − −
    a a a a − 1 an +1 n +3 n +1 n +1 n +1
    n =1 n =1
    ∞
    ∑
    b + 1 1 = − .na a − 1
    n +1 n +1
    n =1
    Teaman +1
    Furthermore, b + 1 = for n ≥ 2. So the sum over n ≥ 2 isna − 1
    n − 1
    ( ) ( )
    ∞ N
    ∑ ∑
    1 1 1 1 − = lim −
    a − 1 a − 1 N →∞ a − 1 a − 1
    n − 1 n +1 n − 1 n +1
    n =2 n =2
    ( )
    1 1 1 1 = + − lim +
    N →∞
    a − 1 a − 1 a − 1 a − 1
    1 2 N N +1
    1 1 = + .
    a − 1 a − 1
    1 2
    Hence the final answer is
    ( ) ( )
    b + 1 1 1 1 − + + .1
    a a − 1 a − 1 a − 1
    2 2 1 2
    Cancelling the common terms and putting in our starting values, this equals
    2014 1 1 1 1 + = 1 − + = 1 + .
    2015 2014 2015 2014 2014 · 2015
    Team