# 二項定理

${\left(a+b\right)}^{n}$$={}_{n}{C}_{0}{a}^{n}+{}_{n}{C}_{1}{a}^{n-1}b$$+{}_{n}{C}_{2}{a}^{n-2}{b}^{2}$$+\cdots \cdots +{}_{n}{C}_{r}{a}^{n-r}{b}^{r}$$+\cdots \cdots +{}_{n}{C}_{n}{b}^{n}$

$=\sum _{r=0}^{n}{}_{n}{C}_{r}{a}^{n-r}{b}^{r}$

${}_{n}{C}_{r}={}_{n-1}{C}_{r-1}+{}_{n-1}{C}_{r}\text{ }\left(n\ge 2\right),$$\text{ }{}_{n}{C}_{r}={}_{n}{C}_{n-r}$

## ■二項定理の導出

${\left(a+b\right)}^{n}$ の二項展開を$n=1$ から順に計算してみる．

${\left(a+b\right)}^{1}={1}·a+{1}·b$

${\left(a+b\right)}^{2}=\left(a+b\right)\left(a+b\right)$

$=\left(a+b\right)a+\left(a+b\right)b$

$=aa+ba+ab+bb$（単項式の数は$2×2=4$

$={\mathbf{1}}·{a}^{2}+{\mathbf{1}}·ab+{\mathbf{1}}·ab+{\mathbf{1}}·{b}^{2}$

$={\mathbf{1}}·{a}^{2}+\left({\mathbf{1}}+{\mathbf{1}}\right)ab+1·{b}^{2}$

$={\mathbf{1}}·{a}^{2}+{\mathbf{2}}·ab+{\mathbf{1}}·{b}^{2}$

${\left(a+b\right)}^{3}={\left(a+b\right)}^{2}\left(a+b\right)$

$=\left\{\left(a+b\right)a+\left(a+b\right)b\right\}\left(a+b\right)$

$=\left\{\left(a+b\right)a+\left(a+b\right)b\right\}a$$+\left\{\left(a+b\right)a+\left(a+b\right)b\right\}b$

（単項式の数は$2×2×2=8$

$=aaa+baa+aba+bba+aab$$+bab+abb+bbb$

$=\left\{{\mathbf{1}}·{a}^{2}+{\mathbf{2}}·ab+{\mathbf{1}}·{b}^{2}\right\}a$$+\left\{{\mathbf{1}}·{a}^{2}+{\mathbf{2}}·ab+{\mathbf{1}}·{b}^{2}\right\}b$

$={\mathbf{1}}·{a}^{3}+{\mathbf{2}}·{a}^{2}b+{\mathbf{1}}·a{b}^{2}$$+{\mathbf{1}}·{a}^{2}b+{\mathbf{2}}·a{b}^{2}+{\mathbf{1}}·{b}^{3}$

$={\mathbf{1}}·{a}^{3}+\left({\mathbf{2}}+{\mathbf{1}}\right){a}^{2}b+\left({\mathbf{1}}+{\mathbf{2}}\right)a{b}^{2}+{\mathbf{1}}·{b}^{3}$

$={\mathbf{1}}·{a}^{3}+{\mathbf{3}}·{a}^{2}b+{\mathbf{3}}·a{b}^{2}+{\mathbf{1}}·{b}^{3}$

${\left(a+b\right)}^{4}={\left(a+b\right)}^{3}\left(a+b\right)$

$=\left[\left\{\left(a+b\right)a+\left(a+b\right)b\right\}a$$+\left\{\left(a+b\right)a+\left(a+b\right)b\right\}b\right]\left(a+b\right)$

$={\mathbf{1}}·{a}^{4}+{\mathbf{4}}·{a}^{3}b+{\mathbf{6}}·{a}^{2}{b}^{2}$$+{\mathbf{4}}·a{b}^{3}+{\mathbf{1}}·{b}^{4}$

$=aaaa+baaa+abaa+bbaa+aaba$$+baba+abba+bbba+aaab+baab$$+abab+bbab+aabb+babb+abbb+bbbb$

（単項式の数は$2×2×2×2=16$

$=\left({\mathbf{1}}·{a}^{3}+{\mathbf{3}}·{a}^{2}b+{\mathbf{3}}·a{b}^{2}+{\mathbf{1}}·{b}^{3}\right)a$$+\left({\mathbf{1}}·{a}^{3}+{\mathbf{3}}·{a}^{2}b+{\mathbf{3}}·a{b}^{2}+{\mathbf{1}}·{b}^{3}\right)b$

$={\mathbf{1}}·{a}^{4}+{\mathbf{3}}·{a}^{3}b+{\mathbf{3}}·{a}^{2}{b}^{2}+{\mathbf{1}}·a{b}^{3}$$+{\mathbf{1}}·{a}^{3}b+{\mathbf{3}}·{a}^{2}{b}^{2}+{\mathbf{3}}·a{b}^{3}+{\mathbf{1}}·{b}^{4}$

$={\mathbf{1}}·{a}^{4}+\left({\mathbf{3}}+{\mathbf{1}}\right){a}^{3}b+\left({\mathbf{3}}+{\mathbf{3}}\right){a}^{2}{b}^{2}$$+\left({\mathbf{1}}+{\mathbf{3}}\right)·a{b}^{3}+{\mathbf{1}}·{b}^{4}$

$={\mathbf{1}}·{a}^{4}+{\mathbf{4}}·{a}^{3}b+{\mathbf{6}}·{a}^{2}{b}^{2}+{\mathbf{4}}·a{b}^{3}$$+{\mathbf{1}}·{b}^{4}$

${\left(a+b\right)}^{n}=\sum _{r=0}^{n}{}_{n}{C}_{r}{a}^{n-r}{b}^{r}$　･･････(1)

となることがわかる．これを数学的帰納法を用いて証明する．

$n=1$ のとき，

${\left(a+b\right)}^{1}{=}_{1}{C}_{0}{a}^{1}{b}^{0}{+}_{1}{C}_{1}{a}^{0}{b}^{1}=a+b$

となり，(1)は成り立つ．

$n=k$のとき，(1)が成りたつと仮定すると，すなわち

${\left(a+b\right)}^{k}=\sum _{r=0}^{k}{}_{k}{C}_{r}{a}^{k-r}{b}^{r}$

が成り立つと仮定する．

${\left(a+b\right)}^{k+1}=\left(a+b\right){\left(a+b\right)}^{k}$

$=\left(a+b\right)\sum _{r=0}^{k}{}_{k}{C}_{r}{a}^{k-r}{b}^{r}$

$=\sum _{r=0}^{k}{}_{k}{C}_{r}a·{a}^{k-r}{b}^{r}+\sum _{r=0}^{k}{}_{k}{C}_{r}{a}^{k-r}b·{b}^{r}$

$=\sum _{r=0}^{k}{}_{k}{C}_{r}{a}^{k-r+1}{b}^{r}+\sum _{r=0}^{k}{}_{k}{C}_{r}{a}^{k-r}{b}^{r+1}$

$=\sum _{r=0}^{k}{}_{k}{C}_{r}{a}^{k-r+1}{b}^{r}$$+\sum _{s=1}^{k+1}{}_{k}{C}_{s-1}{a}^{k-\left(s-1\right)}{b}^{s}$

$=\sum _{r=0}^{k}{}_{k}{C}_{r}{a}^{\left(k+1\right)-r}{b}^{r}$$+\sum _{r=1}^{k+1}{}_{k}{C}_{r-1}{a}^{\left(k+1\right)-r}{b}^{r}$

$={}_{k}{C}_{0}{a}^{k+1}+\sum _{r=1}^{k}{}_{k}{C}_{r}{a}^{\left(k+1\right)-r}{b}^{r}$$+\sum _{r=1}^{k}{}_{k}{C}_{r-1}{a}^{\left(k+1\right)-r}{b}^{r}+{}_{k+1}{C}_{k+1}{b}^{k+1}$

ここで

${}_{k}{C}_{r}{+}_{k}{C}_{r-1}$$=\frac{k\left(k-1\right)\left(k-2\right)\cdots \left(k-r+1\right)}{r!}$$+\frac{k\left(k-1\right)\left(k-2\right)\cdots \left(k-r+2\right)}{\left(r-1\right)!}$

$=\frac{k\left(k-1\right)\left(k-2\right)\cdots \left(k-r+2\right)}{\left(r-1\right)!}\left(\frac{k-r+1}{r}+1\right)$

$=\frac{\left(k+1\right)k\left(k-1\right)\left(k-2\right)\cdots \left(k-r+2\right)}{r!}$

${=}_{k+1}{C}_{r}$

より （この関係からパスカルの三角形が得られる）

$={}_{k}{C}_{0}{a}^{k+1}+\sum _{r=1}^{k}{}_{k+1}{C}_{r}{a}^{\left(k+1\right)-r}{b}^{r}$$+{}_{k+1}{C}_{k+1}{b}^{k+1}$

$=\sum _{r=0}^{k+1}{}_{k+1}{C}_{r}{a}^{\left(k+1\right)-r}{b}^{r}$

よって，$n=k+1$ のときも(1)が成り立ち，数学的帰納法により，(1)はすべての自然数$a$  に対して成り立つ．

${\left(a+b\right)}^{n}$ の係数は，上述した具体的な計算事例と${}_{k}{C}_{r}+{}_{k}{C}_{r-1}={}_{k+1}{C}_{r}$ の関係から，パスカルの三角形が得られる．

パスカルの三角形を下に示す．

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