Toán Cho M=1/5^2+1/6^2+1/7^2+1/8^2+…+1/20^2.C/m M<1/5 mao lin sắp nộp rồi 07/12/2021 By Natalia Cho M=1/5^2+1/6^2+1/7^2+1/8^2+…+1/20^2.C/m M<1/5 mao lin sắp nộp rồi
Đáp án: $\begin{array}{l}M = \frac{1}{{{5^2}}} + \frac{1}{{{6^2}}} + \frac{1}{{{7^2}}} + … + \frac{1}{{{{20}^2}}}\\Do:\frac{1}{{{5^2}}} < \frac{1}{{4.5}};\frac{1}{{{6^2}}} < \frac{1}{{5.6}};\frac{1}{{{7^2}}} < \frac{1}{{6.7}};…;\frac{1}{{{{20}^2}}} < \frac{1}{{19.20}}\\ \Rightarrow \frac{1}{{{5^2}}} + \frac{1}{{{6^2}}} + \frac{1}{{{7^2}}} + … + \frac{1}{{{{20}^2}}} < \frac{1}{{4.5}} + \frac{1}{{5.6}} + \frac{1}{{6.7}} + … + \frac{1}{{19.20}}\\ \Rightarrow M < \frac{1}{4} – \frac{1}{5} + \frac{1}{5} – \frac{1}{6} + … + \frac{1}{{19}} – \frac{1}{{20}}\\ \Rightarrow M < \frac{1}{4} – \frac{1}{{20}}\\ \Rightarrow M < \frac{1}{5}\end{array}$ Trả lời
Đáp án:
$\begin{array}{l}
M = \frac{1}{{{5^2}}} + \frac{1}{{{6^2}}} + \frac{1}{{{7^2}}} + … + \frac{1}{{{{20}^2}}}\\
Do:\frac{1}{{{5^2}}} < \frac{1}{{4.5}};\frac{1}{{{6^2}}} < \frac{1}{{5.6}};\frac{1}{{{7^2}}} < \frac{1}{{6.7}};…;\frac{1}{{{{20}^2}}} < \frac{1}{{19.20}}\\
\Rightarrow \frac{1}{{{5^2}}} + \frac{1}{{{6^2}}} + \frac{1}{{{7^2}}} + … + \frac{1}{{{{20}^2}}} < \frac{1}{{4.5}} + \frac{1}{{5.6}} + \frac{1}{{6.7}} + … + \frac{1}{{19.20}}\\
\Rightarrow M < \frac{1}{4} – \frac{1}{5} + \frac{1}{5} – \frac{1}{6} + … + \frac{1}{{19}} – \frac{1}{{20}}\\
\Rightarrow M < \frac{1}{4} – \frac{1}{{20}}\\
\Rightarrow M < \frac{1}{5}
\end{array}$