每日一题[1238]对影成三人

已知 $n\in\mathbb N^{\ast}$，则 ${\rm C}_ {4n+1}^1+{\rm C}_ {4n+1}^5+{\rm C}_ {4n+1}^9+\cdots+{\rm C}_ {4n+1}^{4n+1}=$ _______．

答案    $2^{4n-1}+(-1)^n\cdot 2^{2n-1}$．

解析    记\[\begin{split} A_0&={\rm C}_ {4n+1}^0+{\rm C}_ {4n+1}^4+{\rm C}_ {4n+1}^8+\cdots+{\rm C}_ {4n+1}^{4n},\\ A_1&={\rm C}_ {4n+1}^1+{\rm C}_ {4n+1}^5+{\rm C}_ {4n+1}^9+\cdots+{\rm C}_ {4n+1}^{4n+1},\\ A_2&={\rm C}_ {4n+1}^2+{\rm C}_ {4n+1}^6+{\rm C}_ {4n+1}^{10}+\cdots+{\rm C}_ {4n+1}^{4n-2},\\ A_3&={\rm C}_ {4n+1}^3+{\rm C}_ {4n+1}^7+{\rm C}_ {4n+1}^{11}+\cdots+{\rm C}_ {4n+1}^{4n-1},\end{split}\]且\[f(x)=(1+x)^{4n+1},\]即\[f(x)={\rm C}_ {4n+1}^0+{\rm C}_ {4n+1}^1x+{\rm C}_ {4n+1}^2x^2+\cdots+{\rm C}_ {4n+1}^{4n+1}x^{4n+1},\]则\[\begin{cases} f(1)=A_0+A_1+A_2+A_3,\\ f({\rm i})=A_0+A_1{\rm i}-A_2-A_3{\rm i},\\ f(-1)=A_0-A_1+A_2-A_3,\\ f(-{\rm i})=A_0-A_1{\rm i}-A_2+A_3{\rm i},\end{cases}\]进而可得\[\begin{cases} A_0+A_1+A_2+A_3=2^{4n+1},\\ -A_0{\rm i}+A_1+A_2{\rm i}-A_3=-{\rm i}(1+{\rm i})^{4n+1},\\ -A_0+A_1-A_2+A_3=0,\\ A_0{\rm i}+A_1-A_2{\rm i}-A_3={\rm i}(1-{\rm i})^{4n+1},\end{cases}\]因此\[A_1=2^{4n-1}+(-1)^n\cdot 2^{2n-1}.\]