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1/ Tính:
\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
\(=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{19}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
b) \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{14^2}< \frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{13\cdot15}\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{13\cdot15}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\left(1-\frac{1}{15}\right)< \frac{1}{2}\)
\(\)
a, \(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{n^2}< 1\)
Vì \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(........\)
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....-\frac{1}{n}\)\(=1-\frac{1}{n}=\frac{n-1}{n}< 1\)
a) \(\frac{1}{4}+\frac{3}{4}:x=\frac{5}{8}\)
\(\frac{3}{4}:x=\frac{3}{8}\)
\(x=2\)
vậy x=2
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)
\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2000}{2002}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1000}{2002}\)
\(\frac{1}{x+1}=\frac{1}{2002}\)
\(x+1=2002\)
\(x=2001\)
vậy x=2001
Đăng từ bài thôi bạn à!
a) Áp dụng công thức: \(\frac{1}{a-1}-\frac{1}{a}=\frac{1}{\left(a-1\right)a}>\frac{1}{a.a}=\frac{1}{a^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{4^2}< \frac{1}{3}-\frac{1}{4}\)
..............................
\(\frac{1}{n^2}< \frac{1}{n-1}-\frac{1}{n}\)
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\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}=\frac{1}{n+1}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\) (đpcm)
P = 1/4 + 1/16 + 1/36 + .. + 1/196 = 1/2² + 1/4² + 1/6² +...+ 1/12² + 1/14²
xét tổng quát với số nguyên dương k ta có:
(2k-1)(2k+1) = 4k² - 1 < 4k² = (2k)² => 1/(2k)² < 1/(2k-1)(2k+1)
=> 2/(2k)² < 2 /(2k-1)(2k+1) = 1/(2k-1) - 1/(2k+1) (*)
ad (*) cho k từ 1 đến 7
2/2² < 1/1 - 1/3
2/4² < 1/3 - 1/5
...
2/12² < 1/11 - 1/13
2/14² < 1/13 - 1/15
+ + cộng lại + +
2/2² + 2/4² +...+ 2/14² < 1/1 - 1/15 < 1
=> 2(1/2² + 1/4² +..+ 1/14²) < 1 => P < 1/2 (đpcm)
\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)
= \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
= \(\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}\right)< \frac{1}{4}\left(1+1\right)=\frac{1}{2}\)
#ĐinhBa
