So sánh P:1/3^2-1/3^4+1/3^6-1/3^8+....+1/3^2018-1/3^2020 và 0,1 Bài này hơi khó giúp mk với

Đáp án:

\[P < 0,1\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
P = \frac{1}{{{3^2}}} - \frac{1}{{{3^4}}} + \frac{1}{{{3^6}}} - \frac{1}{{{3^8}}} + .... + \frac{1}{{{3^{2018}}}} - \frac{1}{{{3^{2020}}}}\\
 \Rightarrow {3^2}.P = 1 - \frac{1}{{{3^2}}} + \frac{1}{{{3^4}}} - \frac{1}{{{3^6}}} + ..... + \frac{1}{{{3^{2016}}}} - \frac{1}{{{3^{2018}}}}\\
 \Leftrightarrow {3^2}P + P = \left( {\frac{1}{{{3^2}}} - \frac{1}{{{3^4}}} + \frac{1}{{{3^6}}} - \frac{1}{{{3^8}}} + .... + \frac{1}{{{3^{2018}}}} - \frac{1}{{{3^{2020}}}}} \right)\\
 + \left( {1 - \frac{1}{{{3^2}}} + \frac{1}{{{3^4}}} - \frac{1}{{{3^6}}} + ..... + \frac{1}{{{3^{2016}}}} - \frac{1}{{{3^{2018}}}}} \right)\\
 \Leftrightarrow 10P = 1 - \frac{1}{{{3^{2020}}}}\\
 \Rightarrow P = 0,1 - \frac{1}{{{{10.3}^{2020}}}} < 0,1
\end{array}\)