A=\(\frac{1}{1^2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+.....+\(\frac{1}{50^2}\)
help me please!!!
Chứng tỏ rằng :
A=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\) < \(\frac{1}{3}\)
B=\(\frac{1}{5}+\frac{1}{9}+\frac{1}{10}+\frac{1}{41}+\frac{1}{42}< \frac{1}{2}\)
Please help me if you know!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I need answer!!!!!!!!!!!!!!!!!!
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)
\(2A+A=\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\right)\)
\(3A=1-\frac{1}{64}\)
\(3A=\frac{63}{64}\Rightarrow A=\frac{63}{64}\div3=\frac{21}{64}< \frac{1}{3}\)
B=\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{96\cdot98}+\frac{1}{98\cdot100}\)
Please help me!!!!!!!!
\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}.\frac{49}{100}=\frac{49}{200}\)
Bn Nguyễn Tuấn Minh làm đúng rồi đó bạn
\(B=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{96.98}+\frac{1}{98.100}\)
\(B=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{96.98}+\frac{2}{98.100}\right)\)
\(B=\frac{1}{2}.(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100})\)
\(B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(B=\frac{1}{2}.\frac{49}{100}\)
\(B=\frac{49}{200}\)
chững minh :
a) A= 1+\(\frac{1}{2^3}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{100^2}< 2\)2
b)B=1+\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
giúp với help me
Câu a) Mik chữa lại một chút
Ta có: \(\frac{1}{2^2}< \frac{1}{1\cdot2}\); \(\frac{1}{3^2}< \frac{1}{2\cdot3}\);.......; \(\frac{1}{100^2}< \frac{1}{99\cdot100}\)
Suy ra: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
Suy ra: \(VT< \frac{1}{1}-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy : \(VT+1< 1+1=2\)
1) Tính: \(\frac{3}{2^2}\). \(\frac{8}{3^2}\). \(\frac{15}{4^2}\). ...... .\(\frac{899}{30^2}\)
2) cho A= \(\frac{3}{10}\)+\(\frac{3}{11}\) +\(\frac{3}{12}\) +\(\frac{3}{13}\) +\(\frac{3}{14}\)
Chứng tỏ : 1< A< 2
3) c/m: \(\frac{1}{26}\)+\(\frac{1}{27}\) +\(\frac{1}{28}\) + ...... +\(\frac{1}{50}\) < 1- \(\frac{1}{2}\)+\(\frac{1}{3}\) - \(\frac{1}{4}\) + ......+\(\frac{1}{49}\) -\(\frac{1}{50}\)
Help me, please!!!!
Mình đang cần gấp! Trước thứ hai nha! Thanks!!!
\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot.....\cdot\frac{899}{30^2}\)
\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot.....\cdot\frac{29\cdot31}{30\cdot30}\)
\(=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{3}{4}\cdot\frac{5}{4}\cdot....\cdot\frac{29}{30}\cdot\frac{31}{30}\)
\(=\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{29}{30}\right)\cdot\left(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot....\cdot\frac{31}{30}\right)\)
\(=\frac{1}{30}\cdot\frac{31}{2}\)
\(=\frac{31}{60}\)
b, \(A=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
Ta có:
\(\frac{3}{15}< \frac{3}{10}=\frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{11}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{12}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{13}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{14}< \frac{3}{10}\)
\(\Rightarrow\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}< \frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}\)
\(\Rightarrow\frac{3\cdot5}{15}< A< \frac{3\cdot5}{10}\)
\(\Rightarrow1< A< \frac{15}{10}=\frac{3}{2}\)
Mà \(\frac{3}{2}< 2\)
\(\Rightarrow1< A< 2\)
c ,Ta có
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)+\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
\(Cm:\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{3}{4}\)
Help me! :((
Đặt \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}=A\)
ta có :\(\frac{1}{2^2}=\frac{1}{2\cdot2}=\frac{1}{4}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(...\)
\(\frac{1}{1990^2}=\frac{1}{1990\cdot1990}< \frac{1}{1989\cdot1990}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2\cdot3}+...+\frac{1}{1989\cdot1990}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1989}-\frac{1}{1990}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}=\frac{3}{4}-\frac{1}{1990}< \frac{3}{4}\)
\(\Rightarrow A< \frac{3}{4}\left(ĐPCM\right)\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{3}{4}\)
hk tốt #
Ta có \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{1990^2}< \frac{1}{1989.1990}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1989.1990}\)
\(< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1989}-\frac{1}{1990}\)
\(< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}=\frac{3}{4}-\frac{1}{1990}< \frac{3}{4}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{3}{4}\)
\(\Rightarrow\)Bài toán được chứng minh
Tìm x, biết:
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\) \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(3(3x-\frac{1}{2})^3+\frac{1}{9}=0\) \(\frac{3}{4}-2.|2x-\frac{2}{3}|=2\) \(\frac{2}{3}x-\frac{1}{2}x=\frac{5}{12}\)
Help me, please
ai giải giúp mình bài này với
\(M=\left(1+\frac{1}{1+2}\right)\left(1+\frac{1}{1+2+3}\right)......\left(1+\frac{1}{1+2+3+...+2012}\right)\)
please help me
Tính A = \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.....+\frac{1}{1+2+3+.....+2014}\)
Help me ~~~~
\(\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{2014}-\frac{1}{2015}\)
\(=\frac{1}{2}-\frac{1}{2015}=\frac{2013}{4030}=>A=\frac{2013}{4030}:2=\frac{2013}{2015}\)
tick nhe
1, tim x bết:
,\(\frac{\left(-5^4\right).\left(-15^2\right)-5^4.\left(-3^2.5\right)}{\left(-3^4\right).25^2-\left(-15^2\right).225.5}\)\(:\)\(\frac{x}{5}\)\(=\frac{-1}{6}\)
2, cho A=\(\frac{20}{30}+\frac{20}{70}+\frac{20}{126}+...+\frac{20}{798}\)
B=\(\left(\frac{41}{2}.\frac{42}{2}.\frac{43}{2}...\frac{80}{2}\right):\left(1.3.5...79\right)\)
So sánh A và B.
3, Tính nhanh.
\(\frac{0,875+\frac{1}{2}-7\%-\frac{1}{58}}{\frac{1}{25}-\frac{1}{2}-\frac{2}{7}+\frac{2}{203}}\)\(-125\%\)
HELP ME PLEASE......