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基本公式:
\[{n \choose k} = {n \choose n - k} \\ Pascal三角形:{n \choose k} = {n - 1 \choose k - 1} + {n - 1 \choose k}\\ 恆等式:\sum {n \choose i} = 2 ^ n\\ 二項式定理: (x + y)^n = \sum_{i = 0}^{n} {n \choose i} x^i y ^ {n - i} \] -
經驗式:
\[(r + 1) ^ k = \sum_{i = 0}^{k} {k \choose i} r ^ i\\ k {n \choose k} = n {n - 1 \choose k - 1}, 證明: (n - 1) - (k - 1) = n - k\\ 在二項式定理里,令x = 1, y = -1:\\ \sum_{i = 0}^{n} {n \choose i} (-1) ^ n = 0\\ 移項:\sum_{i = 0, i ~is~ odd}^{n} {n \choose i} = \sum_{i = 0, i ~is~ even}^{n} {n \choose i} = 2 ^{n - 1}\\ \]
\[\sum_{i = 0}^{n} i{n \choose i} = n * 2 ^ {n - 1}\\ 證明:取二項式定理:(1 + n) ^ n = \sum_{i = 0}^{n} {n \choose i}\\ 兩邊求導: n(1 + n)^{n - 1} = \sum_{i = 1}^{n} i{n \choose i}n ^ {i - 1}, 取x = 1\\ 對於\sum_{i = 0} ^{ n} i ^k {n \choose i} 求k次導即可\\ \]
\[\sum_{i = 0}^{n} {i \choose k} = {n + 1 \choose k + 1} \\ \sum_{i = 0}^{k} {n + i \choose i} = {n + k + 1 \choose k}\\ 分別對Pascal公式的第一項和最后意向反復迭代展開即可. \]
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范德蒙特卷積:
\[\sum_{k = 0}^{n} {m1 \choose k}{m2 \choose n - k} = {m1 + m2 \choose n} \]考慮組合意義證明, \({m1 + m2 \choose n}\)是\(m1 + m2\)向上向右的步子中向上的步數是\(n\)的方案數.
那么我們把式子展開, 在前\(m1\)步中, 我們走\(k\)步,那么在后面的\(m2\)步中,我們一定要走\(n - k\)步
所以有$$\sum_{k = 0}^{n} {n \choose k} ^ 2 = {2n \choose n} $$
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亂七八糟的式子:
\[\sum_{i = 1} {n \choose i}{i \choose m}(-1)^{i + m} = [n = m] \]