Własności liczb
: 19 maja 2020, o 20:54
Niech \(\displaystyle{ P(x)=\sum^{n}_{k=0}a_{k}\cdot \left(x \atop k \right)}\) będzie wielomianem. Następujące warunki są równoważne:
\(\displaystyle{
a_{1},a_{2},\ldots, a_{n}\in \mathbb{Z}}\)
\(\displaystyle{ \forall_{x\in \mathbb{Z}}\ P(x)\in \mathbb{Z}}\)
Udowodnij poniższe równości
\(\displaystyle{ \left\lbrace n+1 \atop m+1 \right\rbrace=\sum_{k}\left( n \atop k \right)\left\lbrace k \atop m\right\rbrace\\
\left[n+1 \atop m+1 \right]=\sum_{k}\left[n\atop k\right]\left(k\atop m\right)\\
\left\lbrace n\atop m \right\rbrace=\sum_{k}\left(n\atop k\right)\left\lbrace k+1\atop m+1 \right\rbrace (-1)^{n-k}\\
\left[n\atop m\right]=\sum_{k}\left[n+1 \atop k+1\right]\left(k\atop m \right)(-1)^{m-k}\\
m!\left\lbrace n\atop m\right\rbrace=\sum_{k}\left(m\atop k\right)k^{n}(-1)^{m-k}\\
\left\lbrace n+1\atop m+1 \right\rbrace=\sum^{n}_{k=0}\left\lbrace k\atop m\right\rbrace (m+1)^{n-k}\\
\left[n+1\atop m+1 \right]=\sum^{n}_{k=0}\left[k\atop m\right]n^{\underline{n-k}}=n!\sum^{n}_{k=0}\left[k\atop m\right]/k!\\
\left\lbrace m+n+1 \atop m\right\rbrace=\sum^{m}_{k=0}k\left\lbrace n+k \atop k\right\rbrace\\
\left[m+n+1 \atop m \right]=\sum^{m}_{k=0}(n+k)\left[n+k\atop k\right]\\
\left(n\atop m\right)=\sum_{k}\left\lbrace n+1 \atop k+1\right\rbrace\left[k\atop m\right](-1)^{m-k}}\)
Bardzo proszę o pomoc...
\(\displaystyle{
a_{1},a_{2},\ldots, a_{n}\in \mathbb{Z}}\)
\(\displaystyle{ \forall_{x\in \mathbb{Z}}\ P(x)\in \mathbb{Z}}\)
Udowodnij poniższe równości
\(\displaystyle{ \left\lbrace n+1 \atop m+1 \right\rbrace=\sum_{k}\left( n \atop k \right)\left\lbrace k \atop m\right\rbrace\\
\left[n+1 \atop m+1 \right]=\sum_{k}\left[n\atop k\right]\left(k\atop m\right)\\
\left\lbrace n\atop m \right\rbrace=\sum_{k}\left(n\atop k\right)\left\lbrace k+1\atop m+1 \right\rbrace (-1)^{n-k}\\
\left[n\atop m\right]=\sum_{k}\left[n+1 \atop k+1\right]\left(k\atop m \right)(-1)^{m-k}\\
m!\left\lbrace n\atop m\right\rbrace=\sum_{k}\left(m\atop k\right)k^{n}(-1)^{m-k}\\
\left\lbrace n+1\atop m+1 \right\rbrace=\sum^{n}_{k=0}\left\lbrace k\atop m\right\rbrace (m+1)^{n-k}\\
\left[n+1\atop m+1 \right]=\sum^{n}_{k=0}\left[k\atop m\right]n^{\underline{n-k}}=n!\sum^{n}_{k=0}\left[k\atop m\right]/k!\\
\left\lbrace m+n+1 \atop m\right\rbrace=\sum^{m}_{k=0}k\left\lbrace n+k \atop k\right\rbrace\\
\left[m+n+1 \atop m \right]=\sum^{m}_{k=0}(n+k)\left[n+k\atop k\right]\\
\left(n\atop m\right)=\sum_{k}\left\lbrace n+1 \atop k+1\right\rbrace\left[k\atop m\right](-1)^{m-k}}\)
Bardzo proszę o pomoc...