Câu hỏi của Ngô Tuấn Anh

Question

$CMR:$$B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}< 1$

Nguyễn Lưu Hà Phương · Accepted Answer

Ta có: $\frac{1}{2^2}< \frac{1}{1.2}$(do 22>1.2) $\frac{1}{3^2}< \frac{1}{2.3}$(do 32>2.3) ........................................... $\frac{1}{8^2}< \frac{1}{7.8}$(do 82>7.8)=> B < $\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}$=> B < $1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}$=> B < $1-\frac{1}{8}< 1$=> B < 1Câu trả lời: Ta có: $\frac{1}{2^2}< \frac{1}{1.2}$(do 22>1.2) $\frac{1}{3^2}< \frac{1}{2.3}$(do 32>2.3) ........................................... $\frac{1}{8^2}< \frac{1}{7.8}$(do 82>7.8)=> B < $\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}$=> B < $1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}$=> B < $1-\frac{1}{8}< 1$=> B < 1

°☆Šuβเη☆°ﾟ · Answer

Ta có $\frac{1}{2^2}< \frac{1}{1.2}$ $\frac{1}{3^2}< \frac{1}{2.3}$ .................... $\frac{1}{8^2}< \frac{1}{7.8}$=> $\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{7.8}$=> $\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}$=> $\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< 1-\frac{1}{8}=\frac{7}{8}< 1\left(đpcm\right)$Câu trả lời: Ta có $\frac{1}{2^2}< \frac{1}{1.2}$ $\frac{1}{3^2}< \frac{1}{2.3}$ .................... $\frac{1}{8^2}< \frac{1}{7.8}$=> $\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{7.8}$=> $\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}$=> $\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< 1-\frac{1}{8}=\frac{7}{8}< 1\left(đpcm\right)$

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