組合せの等式(8)

\begin{alignat}{2}
&(18)  {}_n \mathrm{C}_0 \cdot {}_m \mathrm{C}_k+{}_n \mathrm{C}_1 \cdot {}_m \mathrm{C}_{k-1}+{}_n \mathrm{C}_2 \cdot  {}_m \mathrm{C}_{k-2}+ \cdots \\
&       \cdots +{}_n \mathrm{C}_k \cdot {}_m \mathrm{C}_0={}_{n+m} \mathrm{C}_k\\
&(19)  {}_{2n+1} \mathrm{C}_0+{}_{2n+1} \mathrm{C}_1+{}_{2n+1} \mathrm{C}_2+ \cdots\\
&        \cdots +{}_{2n+1} \mathrm{C}_{n-1}+{}_{2n+1} \mathrm{C}_n=2^{2n}
\end{alignat}





<証明>
\begin{alignat}{2}
&(18) (1+x)^{n+m}=(1+x)^n(1+x)^m\\
&   =({}_n \mathrm{C}_0+{}_n \mathrm{C}_1 x+{}_n \mathrm{C}_2 x^2+{}_n \mathrm{C}_3 x^3+\cdots+{}_n \mathrm{C}_k x^k+ \cdots +{}_n \mathrm{C}_n x^n) \\
&    ×({}_m \mathrm{C}_0+{}_m \mathrm{C}_1 x+{}_m \mathrm{C}_2 x^2+{}_m \mathrm{C}_3 x^3+ \cdots \\
&          \cdots+{}_m \mathrm{C}_{k-1} x^{k-1}+{}_m \mathrm{C}_k x^k+ \cdots +{}_m \mathrm{C}_m x^m)
\end{alignat}一方
\begin{alignat}{2}
&(1+x)^{n+m}={}_{n+m} \mathrm{C}_0+{}_{n+m} \mathrm{C}_1 x+{}_{n+m} \mathrm{C}_2 x^2+ \cdots \\
&          \cdots+{}_{n+m} \mathrm{C}_k x^k+ \cdots +{}_{n+m} \mathrm{C}_{n+m} x^{n+m}
\end{alignat}
ここで \(x^k\) の係数に着目して \((1+x)^n(1+x)^m\) を展開すれば
\begin{alignat}{2}
&\cdots+({}_n \mathrm{C}_0 \cdot {}_m \mathrm{C}_k+{}_n \mathrm{C}_1 \cdot {}_m \mathrm{C}_{k-1}+{}_n \mathrm{C}_2 \cdot  {}_m \mathrm{C}_{k-2}+ \cdots \\
&\cdots +{}_n \mathrm{C}_k \cdot {}_m \mathrm{C}_0)x^k+\cdots
\end{alignat}以上より \(x^k\) の係数を等号で結べば
\begin{alignat}{2}
&{}_n \mathrm{C}_0 \cdot {}_m \mathrm{C}_k+{}_n \mathrm{C}_1 \cdot {}_m \mathrm{C}_{k-1}+{}_n \mathrm{C}_2 \cdot  {}_m \mathrm{C}_{k-2}+ \cdots \\
&\cdots +{}_n \mathrm{C}_k \cdot {}_m \mathrm{C}_0={}_{n+m} \mathrm{C}_k
\end{alignat}

\begin{alignat}{2}
&(19) (1+x)^{2n+1}={}_{2n+1} \mathrm{C}_0+{}_{2n+1} \mathrm{C}_1 x+{}_{2n+1} \mathrm{C}_2 x^2+{}_{2n+1} \mathrm{C}_3 x^3+ \cdots +{}_{2n+1} \mathrm{C}_{2n+1} x^{2n+1}\\
\end{alignat}\(x=1\) を代入すると$${}_{2n+1} \mathrm{C}_0+{}_{2n+1} \mathrm{C}_1 +{}_{2n+1} \mathrm{C}_2+{}_{2n+1} \mathrm{C}_3+ \cdots +{}_{2n+1} \mathrm{C}_{2n+1}=2^{2n+1}$$この左辺について、次のように括弧の部分は同じ値になるから
\begin{alignat}{2}
&({}_{2n+1} \mathrm{C}_0+{}_{2n+1} \mathrm{C}_1 +{}_{2n+1} \mathrm{C}_2+{}_{2n+1} \mathrm{C}_3+ \cdots +{}_{2n+1} \mathrm{C}_n)+({}_{2n+1} \mathrm{C}_{n+1}+ \cdots +{}_{2n+1} \mathrm{C}_{2n}+{}_{2n+1} \mathrm{C}_{2n+1})=2^{2n+1}\\
&2({}_{2n+1} \mathrm{C}_0+{}_{2n+1} \mathrm{C}_1 +{}_{2n+1} \mathrm{C}_2+{}_{2n+1} \mathrm{C}_3+ \cdots +{}_{2n+1} \mathrm{C}_n)=2^{2n+1}
\end{alignat}以上より$${}_{2n+1} \mathrm{C}_0+{}_{2n+1} \mathrm{C}_1+{}_{2n+1} \mathrm{C}_2+ \cdots +{}_{2n+1} \mathrm{C}_{n-1}+{}_{2n+1} \mathrm{C}_n=2^{2n}$$

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