\( (n+1)a_n=2a_{n-1}+(n-1) \)，\( n=2,3,4,... \)，\( a_3=5 \)，求\( a_n \)？

先給答案
\( \displaystyle a_n=\frac{96 \times 2^{n-3}}{(n+1)!}+1 \)

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好厲害,如何算的?

\(\left(n+1\right)a_n = 2a_{n-1}+\left(n-1\right)\)

\(\Rightarrow \left(n+1\right)a_n = 2\left(a_{n-1}-1\right)+\left(n+1\right)\)

\(\Rightarrow \left(n+1\right)\left(a_n -1\right) = 2\left(a_{n-1}-1\right)\)

後面用「累乘法」～