Đáp án:
\[\lim \frac{{1 + 2 + 4 + .... + {2^{n - 1}}}}{{{4^n} + {{2.3}^n}}} = 0\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1 + 2 + 4 + .... + {2^{n - 1}} = \frac{{{2^n} - 1}}{{2 - 1}} = {2^n} - 1\\
\lim \frac{{1 + 2 + 4 + .... + {2^{n - 1}}}}{{{4^n} + {{2.3}^n}}}\\
= \lim \frac{{{2^n} - 1}}{{{4^n} + {{2.3}^n}}}\\
= \lim \frac{{{{\left( {\frac{1}{2}} \right)}^n} - \frac{1}{{{4^n}}}}}{{1 + 2.{{\left( {\frac{3}{4}} \right)}^n}}}\\
\lim \left[ {{{\left( {\frac{1}{2}} \right)}^n} - {{\left( {\frac{1}{4}} \right)}^n}} \right] = 0;\,\,\,\lim \left[ {1 + 2.{{\left( {\frac{3}{4}} \right)}^n}} \right] = 1\\
\Rightarrow \lim \frac{{{{\left( {\frac{1}{2}} \right)}^n} - \frac{1}{{{4^n}}}}}{{1 + 2.{{\left( {\frac{3}{4}} \right)}^n}}} = 0 \Rightarrow \lim \frac{{1 + 2 + 4 + .... + {2^{n - 1}}}}{{{4^n} + {{2.3}^n}}} = 0
\end{array}\)
