設無窮數列\( a_1,a_2,a_3,\ldots \)滿足遞迴關係\( (n+1)a_n=2a_{n-1}+n-1 \)(\(n=2,3,4,\ldots\))。若\(a_3=5\)，則下列哪些選項是正確的？
(1)\(a_4=4\)
(2)數列\( \langle\; a_n \rangle\; \)中不可能有一項的值等於1
(3)\( \displaystyle a_{10}=\frac{2^{10}}{10!}-1 \)
(4)數列\( \langle\; a_n \rangle\; \)中恰有3項是整數，其餘各項都不是整數
(5)數列\( \langle\; a_n \rangle\; \)一定收斂且\( \displaystyle \lim_{n \to \infty}a_n=1 \)


想請教選項三四五
判斷關鍵..事實上我不會解這遞迴@@''
先謝了

\(\begin{align}
  & {{a}_{1}}=13 \\
& \left( n+1 \right){{a}_{n}}=2{{a}_{n-1}}+n-1 \\
& {{a}_{n}}=\frac{2}{n+1}{{a}_{n-1}}+\frac{n-1}{n+1}=\frac{2}{n+1}{{a}_{n-1}}-\frac{2}{n+1}+1=\frac{2}{n+1}\left( {{a}_{n-1}}-1 \right)+1 \\
& {{a}_{n}}-1=\frac{2}{n+1}\left( {{a}_{n-1}}-1 \right)=\frac{2}{n+1}\times \frac{2}{n}\left( {{a}_{n-2}}-1 \right)=...=\frac{2}{n+1}\times \frac{2}{n}\times ...\times \frac{2}{3}\left( {{a}_{1}}-1 \right)=\frac{{{2}^{n}}}{\left( n+1 \right)!}\times 12 \\
& {{a}_{n}}=\frac{{{2}^{n}}}{\left( n+1 \right)!}\times 12+1 \\
\end{align}\)