Chủ đề này có 5 trả lời

[chuyensuli]

\lim (\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+...+\frac{2n-1}
{2^{n}})
n \to \infty

[yellow]

\lim (\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+...+\frac{2n-1}
{2^{n}})
n \to \infty

Bạn nhập lại đi bạn ơi

[chuyensuli]

Bạn nhập lại đi bạn ơi

Đây \lim
_{n \to \infty }(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+...+\frac{2n-1}{2^{n}}) làm hộ tớ vs

[yellow]

$\lim _{n \to \infty }(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+...+\frac{2n-1}{2^{n}})$

[chuyensuli]

$\lim _{n \to \infty }(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+...+\frac{2n-1}{2^{n}})$

ok bạn ak

ok bạn ak

giúp tớ cái

[vantho302]

$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{2} + \frac{3}{{{2^2}}} + \frac{5}{{{2^3}}} + ... + \frac{{2n - 1}}{{{2^n}}}} \right)$

Đặt ${S_n} = \frac{1}{2} + \frac{3}{{{2^2}}} + \frac{5}{{{2^3}}} + ... + \frac{{2n - 1}}{{{2^n}}}$
Khi đó ta sẽ thực hiện
$\begin{array}{l}
{S_n} - \frac{1}{2}{S_n} = \frac{1}{2} + \left( {\frac{3}{{{2^2}}} - \frac{1}{{{2^2}}}} \right) + \left( {\frac{5}{{{2^3}}} - \frac{3}{{{2^3}}}} \right) + ... + \left( {\frac{{2n - 1}}{{{2^n}}} - \frac{{2n - 3}}{{{2^n}}}} \right) - \frac{{2n - 1}}{{{2^{n + 1}}}} \\
= \frac{1}{2} + \left( {\frac{1}{2} + \frac{1}{{{2^2}}} + ... + \frac{1}{{{2^{n - 1}}}}} \right) - \frac{{2n - 1}}{{{2^{n + 1}}}} \\
\Leftrightarrow \frac{1}{2}{S_n} = \frac{1}{2} + \left( {\frac{1}{2} + \frac{1}{{{2^2}}} + ... + \frac{1}{{{2^{n - 1}}}}} \right) - \frac{{2n - 1}}{{{2^{n + 1}}}} \\
\Leftrightarrow {S_n} = 1 + 1 + \frac{1}{2} + ...\frac{1}{{{2^{n - 2}}}} - \frac{{2n - 1}}{{{2^{n + 1}}}} \\
\end{array}$
Ta thấy dãy số: $1 + \frac{1}{2} + ...\frac{1}{{{2^{n - 2}}}}$ đây chính là cấp số nhân lùi vô hạn có ${u_1} = 1;q = \frac{1}{2}$
Do đó
$\begin{array}{l}
1 + \frac{1}{2} + ...\frac{1}{{{2^{n - 2}}}} = \frac{{{u_1}}}{{1 - q}} = \frac{1}{{1 - \frac{1}{2}}} = 2 \\
\Rightarrow {S_n} = 1 + 2 - \frac{{2n - 1}}{{{2^n}}} = 3 - \frac{{2n - 1}}{{{2^n}}} \\
\end{array}$
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } {S_n} = \mathop {\lim }\limits_{n \to \infty } \left( {3 - \frac{{2n - 1}}{{{2^n}}}} \right) = 3 - \mathop {\lim }\limits_{n \to \infty } \frac{{2n - 1}}{{{2^n}}} = 3 - \mathop {\lim }\limits_{n \to \infty } \frac{n}{{{2^{n - 1}}}} + \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{2^n}}}$
Mà $\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{2^n}}} = 0$
Ta có: $\left| {\frac{n}{{{2^n}}}} \right| = \frac{n}{{{{\left( {1 + 1} \right)}^n}}} = \frac{n}{{1 + n + \frac{{n\left( {n - 1} \right)}}{2} + ... + 1}} < \frac{n}{{\frac{{n\left( {n - 1} \right)}}{2}}} = \frac{{2n}}{{n\left( {n - 1} \right)}} = \frac{2}{{n - 1}} \to 0$ khi $n \to \infty $
Do đó $\mathop {\lim }\limits_{n \to \infty } \frac{n}{{{2^{n - 1}}}} = 0$
Vậy $\mathop {\lim }\limits_{n \to \infty } {S_n} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{2} + \frac{3}{{{2^2}}} + \frac{5}{{{2^3}}} + ... + \frac{{2n - 1}}{{{2^n}}}} \right) = 3$