HMMT 二月 2006 · 冲刺赛 · 第 35 题

英文解析

35. A sequence is defined by A = 0 , A = 1 , A = 2, and, for integers n ≥ 3,
    0 1 2
    A + A + A 1
    n − 1 n − 2 n − 3
    A = +
    4 2 n
    3 n − n
    Compute lim A .
    N →∞ N
    13 π2
    Answer: − .
    6 12
    Solution: If we sum the given equation for n = 3 , 4 , 5 , . . . , N , we obtain
    N N
    ∑ ∑
    A + A + A 1
    n − 1 n − 2 n − 3
    A = +
    4 2 n
    3 n − nn =3 n =3
    This reduces dramatically to
    ∑N
    2 A A 2 A A 1
    N − 1 N − 2 1 0
    A + + = A + + + ()
    N 2
    4 2
    3 3 3 3 n − nn =3
    Let lim A = L . Under this limit, the left hand side of () is simply 2 L . We
    N →∞ Ncompute the sum on the right with the help of partial fractions
    N ∞
    ∑ ∑
    1 1 1
    lim = −
    4 2 2 2
    N →∞
    n − n n − 1 nn =3 n =3
    ( )
    ( )
    ∞ ∞
    ∑ ∑
    1 1 1 1 1 1 = − + + −
    2 2 2
    2 n − 1 n + 1 1 2 nn =3 n =1
    ( )
    1 1 1 5 π2 = + + −
    2 2 3 4 6
    5 π2 = −
    3 6
    ( )
    2 2
    1 2 1 5 π 13 π
    With this we easily find L = · 2 + · 1 + · 0 + − = − , and we are done.
    2 3 3 3 6 6 12