\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2};\dfrac{1}{3^2}< \dfrac{1}{2\cdot3};...;\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\\ \Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \RightarrowĐpcm\)
Cho A = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
CMR A < 1
CMR: \(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}< \dfrac{1}{2}\)
A = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+......+\dfrac{1}{2^{100}}\)
B = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{100^2}\)
mk cần gấp
CMR:\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+........+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{2}\)với n thuộc N,n >1
*Rút gọn
1) G=\(\dfrac{2}{3}+\dfrac{2}{3^3}+\dfrac{2}{3^5}+...+\dfrac{2}{3^{99}}\)
2) H=\(\dfrac{1}{2}-\dfrac{1}{2^4}+\dfrac{1}{2^7}-\dfrac{1}{2^{16}}+...-\dfrac{1}{2^{58}}\)
3) E=\(\dfrac{-1}{3}+\left(\dfrac{-1}{3}\right)^2+\left(\dfrac{-1}{3}\right)^3+...+\left(\dfrac{-1}{100}\right)^{100}\)
Cho A = \(\left(\dfrac{1}{2^2}-1\right).\left(\dfrac{1}{3^2}-1\right).\left(\dfrac{1}{4^2}-1\right)....\left(\dfrac{1}{100^2}-1\right)\)
So sánh A với \(\dfrac{-1}{2}\)
Cho A = \(\left(\dfrac{1}{2^2}-1\right).\left(\dfrac{1}{3^2}-1\right).\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)
So sánh A với \(-\dfrac{1}{2}\)
Tính :
a) A= \(\dfrac{1}{2010.2009}-\dfrac{1}{2009.2008}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
b) B= 63.(-124)-124.(-37)+(\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\)).(-66)
Giúp mình với ạ! Mình cần gấp lắm
Tính:
a/\(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{3+\dfrac{3}{2}+\dfrac{3}{3}+\dfrac{3}{4}}{2-\dfrac{2}{2}+\dfrac{2}{3}-\dfrac{2}{4}}\)
b/\(\dfrac{1+\dfrac{1}{4}+\dfrac{1}{1+\dfrac{1}{4}}}{1-\dfrac{1}{4}-\dfrac{1}{1-\dfrac{1}{4}}}\)
c/\(\dfrac{\dfrac{2}{5}-\dfrac{7}{5}}{\dfrac{2}{5}-\dfrac{\dfrac{3}{4}}{\dfrac{3}{4}.\dfrac{3}{7}-1}}-\dfrac{1}{\dfrac{3}{7}\left(\dfrac{3}{4}.\dfrac{3}{7}.\dfrac{2}{5}-\dfrac{2}{5}-\dfrac{3}{4}\right)}\)
d/\(\left(\dfrac{\dfrac{4}{3}}{2+\dfrac{4}{3}}+\dfrac{2-\dfrac{4}{3}}{\dfrac{4}{3}}\right).\left(\dfrac{\dfrac{2}{3}}{4+\dfrac{2}{3}}-\dfrac{4-\dfrac{2}{3}}{\dfrac{2}{3}}\right)\)
Giúp mik với các bạn ơi 1 bài thôi cug đc.
