2.
因應武陵高中校慶,校方準備了\(m\)個禮物要在\(n\)天發完,發法如下;第一天先發一個,再從剩餘的\(m-1\)個禮物選\(\displaystyle \frac{1}{7}\)送出去;第二天先送出兩個,再從剩餘的禮物中挑\(\displaystyle \frac{1}{7}\)發出去。按照此發法,在第\(n\)天的時候發出\(n\)個剛好全部發完,請問數對\((m,n)=\)?
[解答]
設第\(k\)天頒完後,剩\({{a}_{k}}\)個禮物
\(\begin{align}
& {{a}_{0}}=m,{{a}_{n}}=0 \\
& {{a}_{k}}=\frac{6}{7}\left( {{a}_{k-1}}-k \right) \\
& {{a}_{k-1}}=\frac{7}{6}{{a}_{k}}+k \\
& m={{a}_{0}}=1+2\times \frac{7}{6}+3\times {{\left( \frac{7}{6} \right)}^{2}}+\cdots \cdots +n{{\left( \frac{7}{6} \right)}^{n-1}}=\left( n-6 \right)\times \frac{{{7}^{n}}}{{{6}^{n-1}}}+36 \\
& \left( {{7}^{n}},{{6}^{n-1}} \right)=1,n-6<{{6}^{n-1}},m\in N \\
& n=6,m=36 \\
\end{align}\)