a) \(u_1=1+\left(1-1\right).2^1=1\); \(u_2=1+\left(2-1\right).2^2=1+2^2=5\); \(u_3=1+\left(3-1\right).2^3=1+2.2^3=17\); \(u_4=1+\left(4-1\right).2^4=1+3.2^4=49\); \(u_5=1+\left(5-1\right).2^5=1+4.2^5=129\). b) \(u_n=1+\left(n-1\right).2^n\). \(u_{n+1}=1+\left(n+1-1\right).2^{n+1}=1+n.2^{n+1}\) \(=1+\left(n-1\right).2^{n+1}+2^{n+1}\)\(=2\left[1+\left(n-1\right).2^n\right]+2^{n+1}-1\) \(=2.u_n+2^{n+1}-1\). Vậy công thức truy hồi của dãy số là: \(\left\{{}\begin{matrix}u_1=1\\u_n=2u_{n-1}+2^n-1\end{matrix}\right.\). c) Có \(u_n=1+\left(n-1\right).2^n\ge1+\left(1-1\right).2^n=1\). Vậy \(u_n\ge1,\forall n\in N^{\circledast}\). Nên dãy \(\left(u_n\right)\) bị chặn dưới bởi 1. Xét . \(u_n-u_{n-1}=2u_{n-1}+2^n-1-u_{n-1}=u_{n-1}+2^n-1\)\(\ge1+2^n-1=2^n>0,\forall n\in N^{\circledast}\). Vậy \(u_n-u_{n-1}>0,\forall n\in N^{\circledast}\) nên dãy \(\left(u_n\right)\) là dãy số tăng.