Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
{u_{n + 1}} = {u_n} + \left( {n + 1} \right){.2^n}\\
 \Leftrightarrow {u_{n + 1}} - {u_n} = \left( {n + 1} \right){.2^n}\\
\left( {n + 1} \right){.2^n} > 0,\,\,\,\forall n \ge 1 \Rightarrow {u_{n + 1}} - {u_n} > 0,\,\,\,\forall n \ge 1 \Leftrightarrow {u_{n + 1}} > {u_n},\,\,\forall n \ge 1
\end{array}\)

Do đó, \(\left( {{u_n}} \right)\) là dãy tăng.

\(\begin{array}{l}
b,\\
{u_{n + 1}} = {u_n} + \left( {n + 1} \right){.2^n}\\
 \Rightarrow \left\{ \begin{array}{l}
{u_2} = {u_1} + {2.2^1}\\
{u_3} = {u_2} + {3.2^2}\\
{u_4} = {u_3} + {4.2^3}\\
.....\\
{u_n} = {u_{n - 1}} + n{.2^{n - 1}}
\end{array} \right.\\
 \Rightarrow {u_2} + {u_3} + {u_4} + {u_5} + .... + {u_n} = \left( {{u_1} + {u_2} + {u_3} + .... + {u_{n - 1}}} \right) + \left( {{{2.2}^1} + {{3.2}^2} + {{4.2}^3} + .... + n{{.2}^{n - 1}}} \right)\\
 \Leftrightarrow {u_n} = {u_1} + \left( {{{2.2}^1} + {{3.2}^2} + {{4.2}^3} + .... + n{{.2}^{n - 1}}} \right)\\
 \Leftrightarrow {u_n} = 1 + \left( {{{2.2}^1} + {{3.2}^2} + {{4.2}^3} + .... + n{{.2}^{n - 1}}} \right)\\
 \Leftrightarrow {u_n} = \left( {{2^1} + {{2.2}^2} + {{3.2}^3} + .... + \left( {n - 1} \right){{.2}^{n - 1}}} \right) + \left( {1 + {2^1} + {2^2} + .... + {2^{n - 1}}} \right)\\
S = {2^1} + {2.2^2} + {3.2^3} + .... + \left( {n - 1} \right){.2^{n - 1}}\\
 \Leftrightarrow 2S = {2^2} + {2.2^3} + {3.2^4} + .... + \left( {n - 1} \right){.2^n}\\
 \Rightarrow 2S - S = \left( {{2^2} + {{2.2}^3} + {{3.2}^4} + .... + \left( {n - 1} \right){{.2}^n}} \right) - \left( {{2^1} + {{2.2}^2} + {{3.2}^3} + .... + \left( {n - 1} \right){{.2}^{n - 1}}} \right)\\
 \Leftrightarrow S =  - {2^1} - {2^2} - {2^3} - .... - {2^{n - 1}} + \left( {n - 1} \right){.2^n}\\
 \Rightarrow {u_n} = S + \left( {1 + {2^1} + {2^2} + .... + {2^{n - 1}}} \right)\\
 \Leftrightarrow {u_n} = \left( {n - 1} \right){.2^n} + 1
\end{array}\)