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战巡
发表于 2024-8-29 03:41
本帖最后由 战巡 于 2024-8-30 10:20 编辑 分子其实一看就是切比雪夫多项式相关的东西，不过还是用更普通一点的做法好了
所以只有笨办法，直接上母函数
先看分子
\[f(n)=\sum_{k=0}^{[\frac{n}{2}]}C_{n+1}^{n-2k}(\frac{1}{2})^k=\sum_{k=0}^{[\frac{n}{2}]}C_{n+1}^{2k+1}(\frac{1}{2})^k\]
令
\[F(t)=\sum_{n=0}^\infty f(n)t^n=\sum_{n=0}^\infty\sum_{k=0}^{[\frac{n}{2}]}C_{n+1}^{2k+1}(\frac{1}{2})^k\cdot t^n\]
这个先不管它那么多，就假设它收敛，于是交换求和次序
\[=\sum_{k=0}^\infty\sum_{n=2k}^\infty C_{n+1}^{2k+1}(\frac{1}{2})^k\cdot t^n\]
\[=\sum_{k=0}^\infty(1-t)^{-2(k+1)}t^{2k}(\frac{1}{2})^k=\frac{2}{t^2-4t+2}=\frac{1}{\sqrt{2}}\left(\frac{1}{t-(2+\sqrt{2})}-\frac{1}{t-(2-\sqrt{2})}\right)\]
注意到
\[\sum_{n=0}^\infty(at)^n=\frac{1}{1-at}\]
即有
\[-\frac{1}{a}\sum_{n=0}^\infty(\frac{t}{a})^n=\frac{1}{t-a}\]
那也就有
\[上面=\frac{1}{\sqrt{2}}\left(-\frac{1}{A}\sum_{n=0}^\infty(\frac{t}{A})^n+\frac{1}{B}\sum_{n=0}^\infty(\frac{t}{B})^n\right)=\frac{1}{\sqrt{2}}\sum_{n=0}^\infty\left(\frac{1}{B^{n+1}}-\frac{1}{A^{n+1}}\right)t^n\]
其中$A=2+\sqrt{2},B=2-\sqrt{2}$
这一比对就知道有
\[f(n)=\frac{1}{\sqrt{2}}\left(\frac{1}{B^{n+1}}-\frac{1}{A^{n+1}}\right)\]
分母采用类似的办法
\[g(n)=\sum_{k=0}^{[\frac{n}{2}]}C_{n-k}^{n-2k}(-\frac{1}{8})^k=\sum_{k=0}^{[\frac{n}{2}]}C_{n-k}^k(-\frac{1}{8})^k\]
令
\[G(t)=\sum_{n=0}^\infty\sum_{k=0}^{[\frac{n}{2}]}C_{n-k}^k(-\frac{1}{8})^kt^n\]
\[=\sum_{k=0}^\infty\sum_{n=2k}^\infty C_{n-k}^k(-\frac{1}{8})^kt^n\]
\[=\sum_{k=0}^\infty(1-t)^{-k-1}t^{2k}(-\frac{1}{8})^k\]
\[=\frac{8}{t^2-8t+8}=\sqrt{2}\left(\frac{1}{t-2(2+\sqrt{2})}-\frac{1}{t-2(2-\sqrt{2})}\right)=\sqrt{2}\left(\frac{1}{t-2A}-\frac{1}{t-2B}\right)\]
于是跟上面同理，会得到
\[g(n)=\sqrt{2}\left(\frac{1}{(2B)^{n+1}}-\frac{1}{(2A)^{n+1}}\right)=\frac{\sqrt{2}}{2^{n+1}}\left(\frac{1}{B^{n+1}}-\frac{1}{A^{n+1}}\right)\]
故此
\[\frac{f(n)}{g(n)}=2^n\] |
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