二項定理
\[ (a+b)^n \] を展開したときの各項\(a^k b^{n-k},k=0,1,\cdots,n\)の係数を考える. この係数は,\(n\)個の内,\(k\)個の\(a\)を取り,残りを\(b\)に取るような,組み合わせになるので,\({}_n C _k\)で表すことができる. であるので, \[ (a+b)^n = \sum ^{n} _{k=0} \left( \begin{array}{l} n \\ k \end{array} \right) a^k b^{n-k} \] と表すことができる.
- 証明 \[ \begin{align} \left( \begin{array}{c} n \\ k \end{array} \right) + \left( \begin{array}{c} n \\ k-1 \end{array} \right) &= \frac{n!}{(n-k)!k!} + \frac{n!}{(n-(k-1))!(k-1)!} \\ &= \frac{n!}{(n-k)!k!} + \frac{n!}{(n-k+1)!(k-1)!} \\ &= (n-k+1)\frac{n!}{(n-k+1)!k!} + k \frac{n!}{(n-k+1)!k!} \\ &= (n+1)\frac{n!}{(n+1-k)!k!} \\ &= \frac{(n+1)!}{(n+1-k)!k!} = \left( \begin{array}{c} n+1 \\ k \end{array} \right) \end{align} \] であるので,帰納法により証明する. まず,\(n=0\)のときは, \[ (a+b)^0 = \sum ^{0} _{k=0} \left( \begin{array}{l} 0 \\ 0 \end{array} \right) a^0 b^0 = 1 \] \(n=1\)のときは, \[ (a+b)^1 = \sum ^{1} _{k=0} \left( \begin{array}{l} 1 \\ k \end{array} \right) a^k b^{1-k} = \left( \begin{array}{l} 1 \\ 0 \end{array} \right)a^0 b^1 + \left( \begin{array}{l} 1 \\ 1 \end{array} \right)a^1 b^0 = a + b \] \(n=2\)のときは, \[ (a+b)^2 = \sum ^{2} _{k=0} \left( \begin{array}{l} 2 \\ k \end{array} \right) a^k b^{1-k} = \left( \begin{array}{l} 2 \\ 0 \end{array} \right)a^0 b^2 + \left( \begin{array}{l} 2 \\ 1 \end{array} \right)a^1 b^1 + \left( \begin{array}{l} 2 \\ 2 \end{array} \right) a^2 b^0 = a^2 + 2ab + b^2 \] \(n=l\)のときに以下が成立するとする, \[ (a+b)^l = \sum ^{l} _{k=0} \left( \begin{array}{c} l \\ k \end{array} \right) a^k b^{l-k} \] \(n=l+1\)のとき, \[ \begin{align} (a+b)^l(a+b) &= (a+b) \left\{ \sum ^{l} _{k=0} \left( \begin{array}{c} l \\ k \end{array} \right) a^k b^{l-k} \right\} \\ &= a \sum ^{l} _{k=0} \left( \begin{array}{c} l \\ k \end{array} \right) a^k b^{l-k} + b \sum ^{l} _{k=0} \left( \begin{array}{c} l \\ k \end{array} \right) a^k b^{l-k} \\ &= \sum ^{l} _{k=0} \left( \begin{array}{c} l \\ k \end{array} \right) a^{k+1} b^{l-k} + \sum ^{l} _{k=0} \left( \begin{array}{c} l \\ k \end{array} \right) a^k b^{l+1-k} \\ &= \left( \begin{array}{c} l \\ l \end{array} \right)a^{l+1} b^{l-l} + \sum ^{l-1} _{k=0} \left( \begin{array}{c} l \\ k \end{array} \right) a^{k+1} b^{l-k} + \sum ^{l} _{k=1} \left( \begin{array}{c} l \\ k \end{array} \right) a^k b^{l+1-k} + \left( \begin{array}{c} l \\ 0 \end{array} \right) a^0 b^{l+1-0} \\ &= \left( \begin{array}{c} l+1 \\ l+1 \end{array} \right)a^{l+1} + \sum ^{l} _{k=1} \left( \begin{array}{c} l \\ k-1 \end{array} \right) a^{k} b^{l-(k-1)} + \sum ^{l} _{k=1} \left( \begin{array}{c} l \\ k \end{array} \right) a^{k} b^{l+(k-1)} + \left( \begin{array}{c} l+1 \\ 0 \end{array} \right) b^{l+1} \\ &= \left( \begin{array}{c} l+1 \\ l+1 \end{array} \right)a^{l+1} + \sum ^{l} _{k=0} \left \{ \left( \begin{array}{c} l \\ k \end{array} \right) + \left( \begin{array}{c} l \\ k-1 \end{array} \right) \right \}a^{k} b^{l+1-k} + \left( \begin{array}{c} l+1 \\ 0 \end{array} \right) b^{l+1} \\ &= \left( \begin{array}{c} l+1 \\ l+1 \end{array} \right)a^{l+1} + \sum ^{l} _{k=1} \left( \begin{array}{c} l + 1 \\ k \end{array} \right) a^{k} b^{l+1-k} + \left( \begin{array}{c} l+1 \\ 0 \end{array} \right) b^{l+1} \\ &= \sum ^{l+1} _{k=0} \left( \begin{array}{c} l + 1 \\ k \end{array} \right) a^{k} b^{l+1-k} = (a+b)^{l+1} \end{align} \] から, \(n=l+1\)のときも成立し,二項定理が成立することが証明された.