Đáp án:
a) ${v_1} + {v_2} + {v_3} + .... + {v_n} = 1 - {\left( {\frac{1}{2}} \right)^n}$
b) ${u_n} = 2 - {\left( {\dfrac{1}{2}} \right)^{n - 1}}$
c) $\lim {u_n} = 2$
Giải thích các bước giải:
a,
Ta có:
\(\begin{array}{l}
{v_n} = {u_{n + 1}} - {u_n} = \frac{1}{{{2^n}}}\\
\Rightarrow {v_1} + {v_2} + {v_3} + .... + {v_n} = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^4}}} + .... + \frac{1}{{{2^n}}}\\
= \frac{{{{\left( {\frac{1}{2}} \right)}^{n + 1}} - \left( {\frac{1}{2}} \right)}}{{\frac{1}{2} - 1}} = 1 - {\left( {\frac{1}{2}} \right)^n}
\end{array}\)
b,
\(\begin{array}{l}
\left\{ \begin{array}{l}
{u_2} - {u_1} = \frac{1}{2}\\
{u_3} - {u_2} = \frac{1}{{{2^2}}}\\
{u_4} - {u_3} = \frac{1}{{{2^3}}}\\
.....\\
{u_n} - {u_{n - 1}} = \frac{1}{{{2^{n - 1}}}}
\end{array} \right.\\
\Rightarrow \left( {{u_2} - {u_1}} \right) + \left( {{u_3} - {u_2}} \right) + .... + \left( {{u_n} - {u_{n - 1}}} \right) = \frac{1}{2} + \frac{1}{{{2^2}}} + .... + \frac{1}{{{2^{n - 1}}}}\\
\Leftrightarrow {u_n} - {u_1} = \frac{{{{\left( {\frac{1}{2}} \right)}^n} - \frac{1}{2}}}{{\frac{1}{2} - 1}}\\
\Leftrightarrow {u_n} - 1 = 1 - {\left( {\frac{1}{2}} \right)^{n - 1}}\\
\Rightarrow {u_n} = 2 - {\left( {\frac{1}{2}} \right)^{n - 1}}\\
c,\\
\lim {\left( {\frac{1}{2}} \right)^{n - 1}} = 0 \Rightarrow \lim {u_n} = \lim \left( {2 - {{\left( {\frac{1}{2}} \right)}^{n - 1}}} \right) = 2
\end{array}\)
