Cho S= $\frac{1}{2²}$ +$\frac{1}{3²}$+$\frac{1}{4²}$+..+$\frac{1}{10²}$ CM S> $\frac{9}{22}$ 06/11/2021 Bởi Harper Cho S= $\frac{1}{2²}$ +$\frac{1}{3²}$+$\frac{1}{4²}$+..+$\frac{1}{10²}$ CM S> $\frac{9}{22}$
Đáp án: $\begin{array}{l}Do:{2^2} < 2.3\\ \Rightarrow \frac{1}{{{2^2}}} > \frac{1}{{2.3}}\\TT:\frac{1}{{{3^2}}} > \frac{1}{{3.4}};\frac{1}{{{4^2}}} > \frac{1}{{4.5}};…\frac{1}{{{{10}^2}}} > \frac{1}{{10.11}}\\ \Rightarrow \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + … + \frac{1}{{{{10}^2}}}\\ > \frac{1}{{2.3}} + \frac{1}{{3.4}} + \frac{1}{{4.5}} + … + \frac{1}{{10.11}}\\ \Rightarrow S > \frac{1}{2} – \frac{1}{3} + \frac{1}{3} – \frac{1}{4} + \frac{1}{4} – \frac{1}{5} + … + \frac{1}{{10}} – \frac{1}{{11}}\\ \Rightarrow S > \frac{1}{2} – \frac{1}{{11}}\\ \Rightarrow S > \frac{{11 – 2}}{{22}}\\ \Rightarrow S > \frac{9}{{22}}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
Do:{2^2} < 2.3\\
\Rightarrow \frac{1}{{{2^2}}} > \frac{1}{{2.3}}\\
TT:\frac{1}{{{3^2}}} > \frac{1}{{3.4}};\frac{1}{{{4^2}}} > \frac{1}{{4.5}};…\frac{1}{{{{10}^2}}} > \frac{1}{{10.11}}\\
\Rightarrow \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + … + \frac{1}{{{{10}^2}}}\\
> \frac{1}{{2.3}} + \frac{1}{{3.4}} + \frac{1}{{4.5}} + … + \frac{1}{{10.11}}\\
\Rightarrow S > \frac{1}{2} – \frac{1}{3} + \frac{1}{3} – \frac{1}{4} + \frac{1}{4} – \frac{1}{5} + … + \frac{1}{{10}} – \frac{1}{{11}}\\
\Rightarrow S > \frac{1}{2} – \frac{1}{{11}}\\
\Rightarrow S > \frac{{11 – 2}}{{22}}\\
\Rightarrow S > \frac{9}{{22}}
\end{array}$