Denote $a_n = b_n + n(n-1)2^{n-2} - n - 1$; then we have $b_1=3$ and $b_{n+1} = 2b_n$, hence $b_n = 2^{n-1}b_1 = 3\cdot 2^{n-1}$, and so $a_n = 3\cdot 2^{n-1} + n(n-1)2^{n-2} - n - 1$.

Equivalent to $a_{n+4}-7a_{n+3}+18a_{n+2}-20a_{n+1}+8a_{n}=0$ $x^4-7x^3+18x^2-20x+8=0 \Rightarrow (x-2)^3(x-1)=0$ So $a_{n}=\lambda_{1}+\lambda_{2}.2^n+\lambda_{3}.n.2^n+\lambda_{4}.n^2.2^n$ We know $a_{1}=1, a_{2}=5, a_{3}=20, a_{4}=67$ Therefore, we can find $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4}$