\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
Chứng minh rằng : B<1
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Những câu hỏi liên quan
19 tháng 4 2016
Ta có : \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}\)
Mà \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{8^2}<\frac{1}{7.8}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=1-\frac{1}{8}<1\)
Vậy B < 1
14 tháng 8 2018
Ta có
B = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+...+\frac{1}{8^2}\) \(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)= \(1-\frac{1}{8}< 1\)
AK
12 tháng 4 2018
Ta có : \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{8^2}< \frac{1}{7.8}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\)
\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow B< 1-\frac{1}{8}\)
\(\Rightarrow B< \frac{7}{8}\)
\(\Rightarrow B< \frac{8}{8}=1\)
Vậy \(B< 1\left(Đpcm\right)\)
Chúc bạn học tốt !!!
12 tháng 4 2018
nhan xet1/2^2<1/1.2=1/1-1/2
1/3^2<1/2.3=1/2-1/3
1/4^2<1/3.4=1/3-1/4
..................................
1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/8<
1/1-1/8=8/8-1/8=7/8<1 vay B<1
29 tháng 4 2019
B < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
B < \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
B < \(1-\frac{1}{8}\)mà 1 - 1/8 < 1
=> B < 1 ( dpcm )
Vậy ...
29 tháng 4 2019
\(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}< 1-\frac{1}{8}=\frac{7}{8}< 1\)
Vậy B<1
Hok tốt
G
13 tháng 3 2018
\(B=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=1-\frac{1}{8}< 1\)
13 tháng 3 2018
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}\)
vì \(\frac{1}{2^2}>\frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)
\(...\)
\(\frac{1}{8^2}< \frac{1}{7\cdot8}\)
nên \(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{7\cdot8}\) (1)
\(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{7\cdot8}\)
\(B=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+...+\frac{8-7}{7\cdot8}\)
\(B=\left(\frac{2}{1\cdot2}-\frac{1}{1\cdot2}\right)+\left(\frac{3}{2\cdot3}-\frac{2}{2\cdot3}\right)+...+\left(\frac{8}{7\cdot8}-\frac{1}{7\cdot8}\right)\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
\(B=1-\frac{1}{8}\)
\(B=\frac{7}{8}< 1\) (2)
(1)(2) \(\Rightarrow A< B< 1\)
\(\Rightarrow A< 1\) (đpct)
10 tháng 5 2016
Ta có:
1/2^2 < 1/1.2
1/3^2 < 1/2.3
1/4^2 < 1/3.4
..........
1/8^2 < 1/7.8
=> B = 1/2^2 + 1/3^2 + 1/4^2 + ... +1/8^2 < 1/1.2 + 1/2.3 + 1/3.4 +....+ 1/7.8 = 1 - 1/2 +1/2 - 1/3 + 1/3 - 1/4 +...+1/7 - 1/8 = 1 - 1/8 < 1
=> B < 1 (ĐPCM)
Ta có
\(\frac{1}{2^2}< \frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)
\(........\)
\(\frac{1}{8^2}< \frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
Mà \(\frac{3}{8}< 1\)
\(\Rightarrow B< 1\)
Đặt A =\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{7}-\frac{1}{8}\)
\(A=1-\frac{1}{8}< 1\)
\(\Leftrightarrow B< A< 1\)