P5591 小猪佩奇学数学

\[f(n)=\sum_{i=0}^n \binom{n}{i} p^i\lfloor \frac{i}{k} \rfloor \]

\[\begin{aligned} f(n+1)&=p^{n+1}\lfloor \frac{n+1}{k} \rfloor+ \sum_{i=0}^n (\binom{n}{i}+\binom{n}{i-1}) p^i \lfloor \frac{i}{k} \rfloor \\ &=p^{n+1}\lfloor \frac{n+1}{k} \rfloor + f(n)+ \sum_{i=0}^{n-1}\binom{n}{i}p^{i+1}\lfloor \frac{i+1}{k} \rfloor \\ &= f(n)+ \sum_{i=0}^{n}\binom{n}{i}p^{i+1}\lfloor \frac{i+1}{k} \rfloor \\ &= f(n)+p\sum_{i=0}^{n}\binom{n}{i}p^i(\lfloor \frac{i}{k} \rfloor+[k|i+1])\\ &= (1+p)f(n)+p\sum_{i=0}^n \binom{n}{i}p^i [k|i+1] \\ &= (1+p)f(n)+p\sum_{i=0}^n \binom{n}{i}p^i \frac{1}{k}\sum_{j=0}^{k-1}\omega_{k}^{(i+1)j}\\ &= (1+p)f(n)+ \frac{p}{k} \sum_{j=0}^{k-1} \omega_{k}^j\sum_{i=0}^n\binom{n}{i}(p\omega_{k}^j)^i \\ &= (1+p)f(n)+ \frac{p}{k} \sum_{j=0}^{k-1} \omega_{k}^j(1+p\omega_{k}^j)^n \\ \Rightarrow f(n)&=\sum_{i=0}^{n-1}(1+p)^{n-1-i} \frac{p}{k}\sum_{j=0}^{k-1}\omega_{k}^j(1+p\omega_{k}^j)^i\\ &=\frac{p}{k}\sum_{j=0}^{k-1}\omega_{k}^j\sum_{i=0}^{n-1}(1+p)^{n-1-i} (1+p\omega_{k}^j)^i\\ &=\frac{p}{k}\sum_{j=0}^{k-1}\omega_{k}^j(1+p)^{n-1}\sum_{i=0}^{n-1} (\frac{1+p\omega_{k}^j}{1+p})^i\\ \end{aligned} \]

等比数列求和即可