Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{u_{n + 1}} = {u_n} + {2^n}\\
\Leftrightarrow {u_{n + 1}} - {u_n} = {2^n}\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_2} - {u_1} = {2^1}\\
{u_3} - {u_2} = {2^2}\\
{u_4} - {u_3} = {2^3}\\
....\\
{u_n} - {u_{n - 1}} = {2^{n - 1}}
\end{array} \right.\\
\Rightarrow \left( {{u_2} - {u_1}} \right) + \left( {{u_3} - {u_2}} \right) + \left( {{u_4} - {u_3}} \right) + .... + \left( {{u_n} - {u_{n - 1}}} \right) = {2^1} + {2^2} + {2^3} + .... + {2^{n - 1}}\\
\Leftrightarrow {u_n} - {u_1} = {2^1} + {2^2} + {2^3} + .... + {2^{n - 1}}\\
\Leftrightarrow {u_n} = 1 + 2 + {2^2} + {2^3} + ... + {2^{n - 1}}\\
\Leftrightarrow 2{u_n} = 2 + {2^2} + {2^3} + ... + {2^{n - 1}} + {2^n}\\
\Rightarrow 2{u_n} - {u_n} = \left( {2 + {2^2} + {2^3} + ... + {2^{n - 1}} + {2^n}} \right) - \left( {1 + 2 + {2^2} + {2^3} + ... + {2^{n - 1}}} \right)\\
\Leftrightarrow {u_n} = {2^n} - 1 = {2.2^{n - 1}} - 1 = {2^{n - 1}} + {2^{n - 1}} - 1
\end{array}\)
