Quod Erat Demonstrandum

2016/02/20

nHk

Filed under: NSS — johnmayhk @ 8:33 上午
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之前談過 H^n_k,早前見以下關係:

\displaystyle \sum^n_{k=0}H^n_k=H^{n+1}_n

本來想透過數算來證,但不好想,於是乖乖地用返 H^n_k=C^{n+k-1}_k 來處理。

\displaystyle \sum^n_{k=0}H^n_k

\displaystyle =H^n_0+\sum^n_{k=1}H^n_k

\displaystyle =1+\sum^n_{k=1}C^{n+k-1}_k

\displaystyle =1+\sum^n_{k=1}(C^{n+k}_k-C^{n+k-1}_{k-1})

\displaystyle =1+C^{2n}_n-C^n_0

\displaystyle =C^{2n}_n

\displaystyle =H^{n+1}_n

習題:

1. 知 C^6_3=C^2_0+C^3_1+C^4_2+C^5_3,你能把 C^n_r 表達為類似以上形式的二項式係數和嗎?
2. 試證 \displaystyle \sum^n_{k=0}H^{n+1}_k=H^{n+1}_{n-1}
3. 試製作一些關於 H^n_k 的關係式。

johnmayhk-nhk

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