二項定理の証明

二項定理を使うと面白い式を導くことができます。

   

\begin{eqnarray*}
& \int_0^1\left(\ _n C_0+\ _n C_1 x+\ _n C_2 x^2+\cdots \cdots+\ _n C_n x^n\right) d x \\
& =\left[\ _n C_0 x+\frac{\ _n C_1}{2} x^2+\frac{\ _n C_2}{3} x^3+\cdots \cdots+\frac{\ _n C_n}{n+1} x^{n+1}\right]_0^1 \\
& =\ _n C_0+\frac{\ _n C_1}{2}+\frac{\ _n C_2}{3}+\cdots \cdots+\frac{\ _n C_n}{n+1} \\
& \therefore \quad \frac{\mathbf{2}^{\boldsymbol{n}+1}-\mathbf{1}}{\boldsymbol{n}+\mathbf{1}}=\ _n C_{\mathbf{0}}+\frac{\ _{\boldsymbol{n}} C_{\mathbf{1}}}{f\mathbf{2}}+\frac{\ _{\boldsymbol{n}} C_{\mathbf{2}}}{\mathbf{3}}+\cdots+\frac{\ _n C_{\boldsymbol{n}}}{\boldsymbol{n}+\mathbf{1}}
\end{eqnarray*}

\begin{eqnarray*}m_X(t)&=&E(\mathrm{e}^{tX})\\ &=&\displaystyle \int_{ – \infty }^{ \infty }\mathrm{e}^{tx}f(x)dx\\ &=&\displaystyle \int_{ – \infty }^{ \infty }\frac{1}{\sqrt{2πσ^2}}\exp{[-\frac{(x-\mu)^2}{2σ^2}+tx]}dx\\ &=&\displaystyle \int_{ – \infty }^{ \infty }\frac{1}{\sqrt{2πσ^2}}\exp{[-\frac{1}{2σ^2}(x^2-2\mu x+{\mu}^{2}-2{\sigma}^{2}tx)]}dx\\ &=&\displaystyle \int_{ – \infty }^{ \infty }\frac{1}{\sqrt{2πσ^2}}\exp{[-\frac{1}{2σ^2}({(x-(\mu+{\sigma}^{2}t))}^{2}+2\mu{\sigma}^{2}t+{\sigma}^{4}t^{2})]}dx\\ &=&\displaystyle \int_{ – \infty }^{ \infty }\frac{1}{\sqrt{2πσ^2}}\exp{[-\frac{(x-(\mu+{\sigma}^{2}t))^2}{2σ^2}+\mu t+\frac{{\sigma}^{2}t^2}{2}]}dx\\ &=&{\mathrm{e}}^{\mu t+\frac{{\sigma}^{2}t^2}{2}}\displaystyle \int_{ – \infty }^{ \infty }\frac{1}{\sqrt{2πσ^2}}\exp{[-\frac{(x-(\mu+{\sigma}^{2}t))^2}{2σ^2}]}dx\\ &=&{\mathrm{e}}^{\mu t+\frac{{\sigma}^{2}t^2}{2}}\end{eqnarray*}

\begin{eqnarray*}

M_X(t)&=&E(\mathrm{e}^{tk})\\ &=&\sum_{k=0}^{n}\mathrm{e}^{tk}P(X=k)\\

 &=&\sum_{k=0}^{n}\mathrm{e}^{tk}\begin{pmatrix}n \\

  k\end{pmatrix} p^{k}{(1-p)}^{n-k}\\

   &=&\sum_{k=0}^{n}\mathrm{e}^{tk}\ _n C_k p^{k}{(1-p)}^{n-k}\\

    &=&\sum_{k=0}^{n}\ _n C_k {(\mathrm{e}^tp)}^{k}{(1-p)}^{n-k}\\

     &=&{(\mathrm{e}^tp+1-p)}^n

      \end{eqnarray*}

\begin{eqnarray*} E(X^2)&=&\sum_{k=0}^{n}k^2P(X=k)\\ &=&\sum_{k=0}^{n}k^2\begin{pmatrix}n \\ k\end{pmatrix} p^{k}{(1-p)}^{n-k}\\ &=&\sum_{k=0}^{n}k^2\ _n C_k p^{k}{(1-p)}^{n-k}\\ &=&\sum_{k=0}^{n}(k(k-1)+k)\frac{n!}{{(n-k)}!k!}p^{k}{(1-p)}^{n-k}\\ &=&\sum_{k=0}^{n}k(k-1)\frac{n!}{{(n-k)}!k!}p^{k}{(1-p)}^{n-k}+\sum_{k=0}^{n}k\frac{n!}{{(n-k)}!k!}p^{k}{(1-p)}^{n-k}\tag{1} \end{eqnarray*}

\begin{eqnarray*}M_X(t)&=&E(\mathrm{e}^{tk})\\ &=&\sum_{k=0}^{n}\mathrm{e}^{tk}P(X=k)\\ &=&\sum_{k=0}^{n}\mathrm{e}^{tk}\begin{pmatrix}n \\ k\end{pmatrix} p^{k}{(1-p)}^{n-k}\\ &=&\sum_{k=0}^{n}\mathrm{e}^{tk} p^{k}{(1-p)}^{n-k}\\ &=&\sum_{k=0}^{n}\ _n C_k {(\mathrm{e}^tp)}^{k}{(1-p)}^{n-k}\\ &=&{(\mathrm{e}^tp+1-p)}^n \end{eqnarray*}

kekkyoku

\begin{eqnarray*}\ _{(n+1)+r} C_{k-1}\end{eqnarray*}

$\ _{(n+1)+r} C_{k-1}$

\begin{eqnarray*}

M_X(t)&=&E(\mathrm{e}^{tk})\\ &=&\sum_{k=0}^{n}\mathrm{e}^{tk}P(X=k)\\

 &=&\sum_{k=0}^{n}\mathrm{e}^{tk}\begin{pmatrix}n \\

  k\end{pmatrix} p^{k}{(1-p)}^{n-k}\\

   &=&\sum_{k=0}^{n}\mathrm{e}^{tk}\ _n C_k p^{k}{(1-p)}^{n-k}\\

    &=&\sum_{k=0}^{n}\ _n C_k {(\mathrm{e}^tp)}^{k}{(1-p)}^{n-k}\\

     &=&{(\mathrm{e}^tp+1-p)}^n

      \end{eqnarray*}

\begin{equation} \int \int_{B} \varphi \frac{\partial \psi}{\partial n}
d S = \int \int \int_{D} (\varphi \Delta \psi +
\mbox{grad} \varphi \cdot \mbox{grad} \varphi) d V \end{equation}

$\displaystyle (1+x)^n=\sum_{k=0}^{n} \ _nC_k x^k=\ _nC_0x^0+\ _nC_1x^1+\ _n C_2x^2+\ _nC_3x^3+\cdots \ _n C _n x^n$

$x=1$を代入すると

$(1+1)^n=\sum\limits_{k=0}^{n}\ _nC_k=\ _nC_0+\ _nC_1+\ _nC_2+\ _nC_3+\cdots\ _nC_n=2^n$

$x=-1$を代入すると

$(1-1)^n=\displaystyle \sum_{k=0}^{n}\ _nC_k=\ _nC_0-\ _nC_1+\ _nC_2-\ _nC_3+ \cdots\ _n C _n=0$

$x=-2$を代入すると

$\displaystyle n\cdot2^{n-1} = 1_n C _1 + \ 2_n C_2+ 3_n C _3 +4_n C _4 + \cdots n _n C _n $

$= 0_n C _0+1_n C _1 + \ 2_n C_2+ 3_n C _3 +4_n C _4 + \cdots n _n C _n$

さらに期待値の計算で使えそうな式です。

$\displaystyle \sum_{k=0}^{n}k _n C_k = 0_n C _0 + 1_n C _1 + \ 2_n C_2+ 3_n C _3 +4_n C _4 + \cdots n _n C _n　＝n 2^{n-1}$

証明は次の通りです。