特征根方法两道

hbghlyj

(1)证明：对任何非负整数n，$\sum_{k=0}^{n}C_{2n+1}^{2k+1}2^{3k}$不能被35整除
(2)n元集的幂等变换数为i(n)=$\sum_{k=1}^nC_n^kk^{n-k}$，且$1+\sum_{n=1}^\infty i(n)\frac{x^n}{n!}=e^{xe^x}$

hbghlyj

回复 1# hbghlyj
(1)$\sum_{k=0}^{n}C_{2n+1}^{2k+1}2^{3k}={2^{ - \frac{5}{2}}}\left( {{{\left( {{2^{\frac{3}{2}}} + 1} \right)}^{2n + 1}} + {{\left( {{2^{\frac{3}{2}}} - 1} \right)}^{2n + 1}}} \right)=\frac{\left(44+25 \sqrt{2}\right) \left(9+4 \sqrt{2}\right)^n+7 \left(4+\sqrt{2}\right) \left(9-4 \sqrt{2}\right)^n}{8 \left(9+4 \sqrt{2}\right)}$
得递推公式$a_{n+2}=18a_{n+1}+49a_{n},a_1=11,a_2=149$，在模35意义下是周期为6的循环数列11, 9, 1, 4, 16, 29...

(2)不会啊