Đáp án:
\[S = \frac{1}{2}\]
Giải thích các bước giải:
Tổng của CSN lùi vô hạn có |q|<1 là:
\(D = q + {q^2} + {q^3} + ... + {q^n} = \frac{{{q^{n + 1}} - q}}{{q - 1}} = \frac{q}{{1 - q}}\,\,\,\,\,\left( {\lim {q^{n + 1}} = 0} \right)\)
Ta có:
\(\begin{array}{l}
S = \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{4} - \frac{1}{9}} \right) + .... + \left( {\frac{1}{{{2^n}}} - \frac{1}{{{3^n}}}} \right) + ....\\
= \left( {\frac{1}{2} + \frac{1}{4} + .... + \frac{1}{{{2^n}}}} \right) - \left( {\frac{1}{3} + \frac{1}{9} + .... + \frac{1}{{{3^n}}}} \right)\\
A = \frac{1}{2} + \frac{1}{4} + .... + \frac{1}{{{2^n}}} = \frac{1}{2} + {\left( {\frac{1}{2}} \right)^2} + .... + {\left( {\frac{1}{2}} \right)^n} = \frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}} = 1\\
B = \frac{1}{3} + \frac{1}{9} + .... + \frac{1}{{{3^n}}} = \frac{1}{3} + {\left( {\frac{1}{3}} \right)^2} + .... + {\left( {\frac{1}{3}} \right)^n} = \frac{{\frac{1}{3}}}{{1 - \frac{1}{3}}} = \frac{1}{2}\\
\Rightarrow S = A - B = \frac{1}{2}
\end{array}\)
