HMMT 二月 2008 · 代数 · 第 10 题

英文解析

10. [ 8 ] Evaluate the infinite sum
    ( )
    ∞
    ∑
    2 n 1
    .
    n 5 nn =0
    √
    Answer: 5 First Solution: Note that
    ( )
    2 n (2 n )! (2 n )(2 n − 2)(2 n − 4) · · · (2) (2 n − 1)(2 n − 3)(2 n − 5) · · · (1) = = ·
    n n ! · n ! n ! n !
    ( ) ( ) ( ) ( )
    ( − 2) 1 1 1 1 nn = 2 · − − − 1 − − 2 · · · − − n + 1
    n ! 2 2 2 2
    ( ) −1
    2 n = ( − 4) .
    1 n
    Then, by the binomial theorem, for any real x with | x | < , we have
    ( ) ( )4
    ∞ ∞
    ∑ ∑1 − 2 n − 1 / 2 n n
    (1 − 4 x ) = ( − 4 x ) = x .2
    n nn =0 n =0
    Therefore,
    ( ) ( )
    ∞ n
    ∑
    √
    2 n 1 1 = √ = 5 .
    n 5
    1 −4
    n =0
    Second Solution: Consider the generating function 5
    ( )
    ∞
    ∑
    2 nnf ( x ) = x .
    n =0 n
    It has formal integral given by
    ( )
    ∞ ∞ ∞
    ∑ ∑ ∑
    1 2 nn +1 n +1 ng ( x ) = I ( f ( x )) = x = C x = x C x ,
    n nn + 1 nn =0 n =0 n =0
    ( )
    ∑
    ∞
    2 nn 1
    where C = is the n th Catalan number. Let h ( x ) = C x ; it suffices to compute thisn nn =0
    n +1 ngenerating function. Note that
    ( )
    ∑ ∑ ∑ ∑k
    2 i + j k k +1
    1 + xh ( x ) = 1 + x C C x = 1 + x C C x = 1 + C x = h ( x ) ,
    i j i k − i k +1
    i,j ≥ 0 i =0
    k ≥ 0 k ≥ 0
    where we’ve used the recurrence relation for the Catalan numbers. We now solve for h ( x ) with thequadratic equation to obtain
    √
    √
    1 /x ± 1 /x − 4 /x 2
    1 ± 1 − 4 xh ( x ) = = .
    2 2 x
    Note that we must choose the − sign in the ± , since the + would lead to a leading term of for h (by 1
    √xexpanding 1 − 4 x into a power series). Therefore, we see that
    √
    ( )
    1 − 1 − 4 x 1
    f ( x ) = D ( g ( x )) = D ( xh ( x )) = D = √
    1 − 4 x 2
    √
    and our answer is hence f (1 / 5) = 5. 3