組合せの等式(6)

\begin{alignat}{2}
&(1)  \displaystyle\sum_{k=0}^{\infty} (-1)^k \left(
\begin{array}{2}
\,n\\
2k\\
\end{array}\right)=
{}_n \mathrm{C}_0-{}_n \mathrm{C}_2+{}_n \mathrm{C}_4-{}_n \mathrm{C}_6+ \cdots =2^{\frac{n}{2}}\cos \frac{nπ}{4}\\
&(2)   \displaystyle\sum_{k=0}^{\infty} (-1)^k \left(
\begin{array}{2}
  n\\
2k+1\\
\end{array}\right)={}_n \mathrm{C}_1-{}_n \mathrm{C}_3+{}_n \mathrm{C}_5-{}_n \mathrm{C}_7+ \cdots =2^{\frac{n}{2}} \sin \frac{nπ}{4}\\
&(3)  \displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
\,n\\
4k\\
\end{array}\right)=
{}_n \mathrm{C}_0+{}_n \mathrm{C}_4+{}_n \mathrm{C}_8+{}_n \mathrm{C}_{12}+ \cdots =\frac{1}{2}\left(2^{n-1}+2^{\frac{n}{2}}\cos \frac{nπ}{4}\right)\\
&(4)  \displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
  n\\
4k+1\\
\end{array}\right)=
{}_n \mathrm{C}_1+{}_n \mathrm{C}_5+{}_n \mathrm{C}_9+{}_n \mathrm{C}_{13}+ \cdots =\frac{1}{2}\left(2^{n-1}+2^{\frac{n}{2}}\sin \frac{nπ}{4}\right)\\
&(5)  \displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
  n\\
4k+2\\
\end{array}\right)=
{}_n \mathrm{C}_2+{}_n \mathrm{C}_6+{}_n \mathrm{C}_{10}+{}_n \mathrm{C}_{14}+ \cdots =\frac{1}{2}\left(2^{n-1}-2^{\frac{n}{2}}\cos \frac{nπ}{4}\right)\\
&(6)  \displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
  n\\
4k+3\\
\end{array}\right)=
{}_n \mathrm{C}_3+{}_n \mathrm{C}_7+{}_n \mathrm{C}_{11}+{}_n \mathrm{C}_{15}+ \cdots =\frac{1}{2}\left(2^{n-1}-2^{\frac{n}{2}}\sin \frac{nπ}{4}\right)\\
\end{alignat}










<証明>

\((1)(2)\) 次の式を二項展開します。
$$(1+i)^n={}_n \mathrm{C}_0+{}_n \mathrm{C}_1 \cdot i+ {}_n \mathrm{C}_2 \cdot i^2+{}_n \mathrm{C}_3 \cdot i^3+{}_n \mathrm{C}_4 \cdot i^4+{}_n \mathrm{C}_5 \cdot i^5+ \cdots $$$$={}_n \mathrm{C}_0+{}_n \mathrm{C}_1 \cdot i-{}_n \mathrm{C}_2 -{}_n \mathrm{C}_3 \cdot i+{}_n \mathrm{C}_4-{}_n \mathrm{C}_5 \cdot i+ \cdots $$$$=({}_n \mathrm{C}_0- {}_n \mathrm{C}_2+{}_n \mathrm{C}_4-{}_n \mathrm{C}_6+{}_n \mathrm{C}_8- \cdots)$$$$+i({}_n \mathrm{C}_1-{}_n \mathrm{C}_3+{}_n \mathrm{C}_5- {}_n \mathrm{C}_7+{}_n \mathrm{C}_9- \cdots) \cdots (A)$$一方 \(\displaystyle 1+i=\sqrt{2}\left( \cos \frac{π}{4}+i \sin \frac{π}{4}\right)\) であるから
\begin{alignat}{2}
(1+i)^n&=2^{\frac{n}{2}}\left( \cos \frac{π}{4}+i \sin \frac{π}{4}\right)^n\\
&= 2^{\frac{n}{2}}\left( \cos \frac{nπ}{4}+i \sin \frac{nπ}{4}\right)\\
&=2^{\frac{n}{2}} \cos \frac{nπ}{4}+i \cdot 2^{\frac{n}{2}} \sin \frac{nπ}{4} \cdots (B)\\
\end{alignat}
\((A)\) と \((B)\) の実部と虚部を比較すれば
\begin{alignat}{2}
&(1)  \displaystyle\sum_{k=0}^{\infty} (-1)^k \left(
\begin{array}{2}
\,n\\
2k\\
\end{array}\right)=
{}_n \mathrm{C}_0-{}_n \mathrm{C}_2+{}_n \mathrm{C}_4-{}_n \mathrm{C}_6+ \cdots =2^{\frac{n}{2}}\cos \frac{nπ}{4}\\
&(2)   \displaystyle\sum_{k=0}^{\infty} (-1)^k \left(
\begin{array}{2}
  n\\
2k+1\\
\end{array}\right)={}_n \mathrm{C}_1-{}_n \mathrm{C}_3+{}_n \mathrm{C}_5-{}_n \mathrm{C}_7+ \cdots =2^{\frac{n}{2}} \sin \frac{nπ}{4}\\
\end{alignat}







\begin{alignat}{2}
(3)  \displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
\,n\\
4k\\
\end{array}\right)&={}_n \mathrm{C}_0+{}_n \mathrm{C}_4+{}_n \mathrm{C}_8+{}_n \mathrm{C}_{12}+ \cdots \\
&=\frac{1}{2}(2{}_n \mathrm{C}_0+2{}_n \mathrm{C}_4+2{}_n \mathrm{C}_8+2{}_n \mathrm{C}_{12}+ \cdots)\\
&=\frac{1}{2}\left\{({}_n \mathrm{C}_0+{}_n \mathrm{C}_2+{}_n \mathrm{C}_4+{}_n \mathrm{C}_6+ \cdots)+({}_n \mathrm{C}_0-{}_n \mathrm{C}_2+{}_n \mathrm{C}_4-{}_n \mathrm{C}_{6}+ \cdots)\right\}\\
&=\frac{1}{2}\left\{\displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
\,n\\
2k\\
\end{array}\right)+\displaystyle\sum_{k=0}^{\infty} (-1)^k\left(
\begin{array}{2}
\,n\\
2k\\
\end{array}\right)\right\}\\
&=\frac{1}{2}\left(2^{n-1}+2^{\frac{n}{2}}\cos \frac{nπ}{4}\right)
\end{alignat}以上より$$\displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
\,n\\
4k\\
\end{array}\right)=
{}_n \mathrm{C}_0+{}_n \mathrm{C}_4+{}_n \mathrm{C}_8+{}_n \mathrm{C}_{12}+ \cdots =\frac{1}{2}\left(2^{n-1}+2^{\frac{n}{2}}\cos \frac{nπ}{4}\right)$$







\begin{alignat}{2}
(4)  \displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
  n\\
4k+1\\
\end{array}\right)&={}_n \mathrm{C}_1+{}_n \mathrm{C}_5+{}_n \mathrm{C}_9+{}_n \mathrm{C}_{13}+ \cdots \\
&=\frac{1}{2}(2{}_n \mathrm{C}_1+2{}_n \mathrm{C}_5+2{}_n \mathrm{C}_9+2{}_n \mathrm{C}_{13}+ \cdots)\\
&=\frac{1}{2}\left\{({}_n \mathrm{C}_1+{}_n \mathrm{C}_3+{}_n \mathrm{C}_5+{}_n \mathrm{C}_7+ \cdots)+({}_n \mathrm{C}_1-{}_n \mathrm{C}_3+{}_n \mathrm{C}_5-{}_n \mathrm{C}_{7}+ \cdots)\right\}\\
&=\frac{1}{2}\left\{\displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
  n\\
2k+1\\
\end{array}\right)+\displaystyle\sum_{k=0}^{\infty} (-1)^k\left(
\begin{array}{2}
  n\\
2k+1\\
\end{array}\right)\right\}\\
&=\frac{1}{2}\left(2^{n-1}+2^{\frac{n}{2}}\sin \frac{nπ}{4}\right)
\end{alignat}以上より$$\displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
  n\\
4k+1\\
\end{array}\right)=
{}_n \mathrm{C}_1+{}_n \mathrm{C}_5+{}_n \mathrm{C}_9+{}_n \mathrm{C}_{13}+ \cdots =\frac{1}{2}\left(2^{n-1}+2^{\frac{n}{2}}\sin \frac{nπ}{4}\right)$$









\begin{alignat}{2}
(5)  \displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
  n\\
4k+2\\
\end{array}\right)&={}_n \mathrm{C}_2+{}_n \mathrm{C}_6+{}_n \mathrm{C}_{10}+{}_n \mathrm{C}_{14}+ \cdots \\
&=\frac{1}{2}(2{}_n \mathrm{C}_2+2{}_n \mathrm{C}_6+2{}_n \mathrm{C}_{10}+2{}_n \mathrm{C}_{14}+ \cdots)\\
&=\frac{1}{2}\left\{({}_n \mathrm{C}_2+{}_n \mathrm{C}_4+{}_n \mathrm{C}_{6}+{}_n \mathrm{C}_{8}+ \cdots)-({}_n \mathrm{C}_0-{}_n \mathrm{C}_2+{}_n \mathrm{C}_4-{}_n \mathrm{C}_{6}+ \cdots)\right\}\\
&=\frac{1}{2}\left\{\displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
\,n\\
2k\\
\end{array}\right)-\displaystyle\sum_{k=0}^{\infty} (-1)^k\left(
\begin{array}{2}
\,n\\
2k\\
\end{array}\right)\right\}\\
&=\frac{1}{2}\left(2^{n-1}-2^{\frac{n}{2}}\cos \frac{nπ}{4}\right)
\end{alignat}以上より$$\displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
\,n\\
4k\\
\end{array}\right)=
{}_n \mathrm{C}_2+{}_n \mathrm{C}_6+{}_n \mathrm{C}_{10}+{}_n \mathrm{C}_{14}+ \cdots =\frac{1}{2}\left(2^{n-1}-2^{\frac{n}{2}}\cos \frac{nπ}{4}\right)$$







\begin{alignat}{2}
(6)  \displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
  n\\
4k+3\\
\end{array}\right)&={}_n \mathrm{C}_3+{}_n \mathrm{C}_7+{}_n \mathrm{C}_{11}+{}_n \mathrm{C}_{15}+ \cdots \\
&=\frac{1}{2}(2{}_n \mathrm{C}_3+2{}_n \mathrm{C}_7+2{}_n \mathrm{C}_{11}+2{}_n \mathrm{C}_{15}+ \cdots)\\
&=\frac{1}{2}\left\{({}_n \mathrm{C}_1+{}_n \mathrm{C}_3+{}_n \mathrm{C}_5+{}_n \mathrm{C}_7+ \cdots)-({}_n \mathrm{C}_1-{}_n \mathrm{C}_3+{}_n \mathrm{C}_5-{}_n \mathrm{C}_{7}+ \cdots)\right\}\\
&=\frac{1}{2}\left\{\displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
  n\\
2k+1\\
\end{array}\right)-\displaystyle\sum_{k=0}^{\infty} (-1)^k\left(
\begin{array}{2}
  n\\
2k+1\\
\end{array}\right)\right\}\\
&=\frac{1}{2}\left(2^{n-1}-2^{\frac{n}{2}}\sin \frac{nπ}{4}\right)
\end{alignat}以上より$$\displaystyle\sum_{k=0}^{\infty} \left(
\begin{array}{2}
  n\\
4k+1\\
\end{array}\right)=
{}_n \mathrm{C}_3+{}_n \mathrm{C}_7+{}_n \mathrm{C}_{11}+{}_n \mathrm{C}_{15}+ \cdots =\frac{1}{2}\left(2^{n-1}-2^{\frac{n}{2}}\sin \frac{nπ}{4}\right)$$

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