Cho cấp số nhân \(\left( {{u_n}} \right)\)có số hạng đầu \({u_1} = 2\) và công bội \(q = 3\). Số hạng tổng quát của cấp số     nhân là:

Chọn B

\[\lim \left( {\frac{{3n + 2}}{{n + 2}} + {a^2} - 4a} \right) = 0 \Leftrightarrow {a^2} - 4a + 3 = 0 \Leftrightarrow \left[ \begin{array}{l}a = 1\\a = 3\end{array} \right.\]. Vậy \(S = 4\).

Chọn B

\[\mathop {\lim }\limits_{x \to 0} \frac{{2\sqrt {2x + 1} - \sqrt[3]{{{x^2} + x + 8}}}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{2\sqrt {2x + 1} - 2 + 2 - \sqrt[3]{{{x^2} + x + 8}}}}{x}\]\[ = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\sqrt {2x + 1} - 1} \right) + \left( {2 - \sqrt[3]{{{x^2} + x + 8}}} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\sqrt {2x + 1} - 1} \right)}}{x} + \mathop {\lim }\limits_{x \to 0} \frac{{\left( {2 - \sqrt[3]{{{x^2} + x + 8}}} \right)}}{x}\]\[ = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\sqrt {2x + 1} - 1} \right)\left( {\sqrt {2x + 1} + 1} \right)}}{{x\left( {\sqrt {2x + 1} + 1} \right)}} + \mathop {\lim }\limits_{x \to 0} \frac{{\left( {2 - \sqrt[3]{{{x^2} + x + 8}}} \right)\left( {4 + 2\sqrt[3]{{{x^2} + x + 8}} + \sqrt[3]{{{{\left( {{x^2} + x + 8} \right)}^2}}}} \right)}}{{x\left( {4 + 2\sqrt[3]{{{x^2} + x + 8}} + \sqrt[3]{{{{\left( {{x^2} + x + 8} \right)}^2}}}} \right)}}\]\[ = \mathop {\lim }\limits_{x \to 0} \frac{{4x}}{{x\left( {\sqrt {2x + 1} + 1} \right)}} + \mathop {\lim }\limits_{x \to 0} \frac{{8 - \left( {{x^2} + x + 8} \right)}}{{x\left( {4 + 2\sqrt[3]{{{x^2} + x + 8}} + \sqrt[3]{{{{\left( {{x^2} + x + 8} \right)}^2}}}} \right)}}\]\[ = \mathop {\lim }\limits_{x \to 0} \frac{4}{{\left( {\sqrt {2x + 1} + 1} \right)}} + \mathop {\lim }\limits_{x \to 0} \frac{{ - 1 - x}}{{\left( {4 + 2\sqrt[3]{{{x^2} + x + 8}} + \sqrt[3]{{{{\left( {{x^2} + x + 8} \right)}^2}}}} \right)}} = 2 - \frac{1}{{12}} = \frac{{23}}{{12}}\]

Vậy \(a = 23,\,b = 12\). \(a - 2b = - 1\).