Re: [中學] 組合數連加

[JohnMash (Paul)]

※ 引述《Intercome (今天的我小帥)》之銘言：
:    C(n,1) + C(n,2)/2 + C(n,3)/3 + ... + C(n,n)/n = A
:    這是不是沒有辦法求出n的一般式呢? 再請各位高手解答，謝謝~~

   Let

   f(x) = Σ[k;1,n] (1/k) C(n,k) x^k

   f(0) = 0

   then A = f(1)

   f'(x) =Σ[k,1,n] C(n,k) x^{k-1}

   x f'(x) = (1+x)^n - 1

     f'(x) = (1+x)^n / x - 1/x

     let 1+x = u

     f'(u-1) = -u^n / (1-u) + 1/(1-u)

           = (1-u^n)/(1-u)

           = 1+u+u^2+...+u^{n-1}

     f(u-1) = u + u^2/2 + u^3/3 + ... + u^n / n + C

     f(0) = 0 then C = -(1 + 1/2 + ... + 1/n)

     denote g_n (x) =  x + x^2/2 + x^3/3 + ... + x^n/n

     then A = f(1) = g_n(2) - g_n(1)



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推 Intercome   :原來如此 感謝J大喔