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S
10 tháng 3 2019
\(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2n}\right)=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2n-1}+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2n}\right)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n-1}+\frac{1}{2n}-\frac{1}{1}-\frac{1}{2}-....-\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\left(\text{đpcm}\right)\)
TD
28 tháng 2 2023
Câu b hướng làm đó là tách con 1/3 và 1/2 ra thành 50 phân số giống nhau. E tách 1/3=50/150 rồi so sánh 1/101, 1/102,...,1/149 với 1/150. Còn vế sau 1/2=50/100 tách tương tự rồi so sánh thôi
AH
Akai Haruma
Giáo viên
28 tháng 2 2023
2a.
$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}$
$< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{50-49}{49.50}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}$
$=1-\frac{1}{50}< 1$ (đpcm)
2 tháng 4 2023
\(=\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{50\cdot\dfrac{49}{2}}\)
\(=\dfrac{1}{2\cdot\dfrac{3}{2}}+\dfrac{1}{3\cdot\dfrac{4}{2}}+...+\dfrac{1}{50\cdot\dfrac{49}{2}}\)
\(=\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{49\cdot50}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
=2*24/50=48/50=24/25
26 tháng 4 2018
\(\left(1+\dfrac{1}{2}\right)+\left(1+\dfrac{1}{2^2}\right)+...+\left(1+\dfrac{1}{2^{50}}\right)\)
= \(\left(1+1+1+...+1\right)+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\right)\)(50 số 1 )
= \(50+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\right)\)
A =\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\)
⇒ 2A = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\)
⇒ 2A - A =\(1-\dfrac{1}{2^{50}}\)
=50+1-\(\dfrac{1}{2^{50}}\)=51-\(\dfrac{1}{2^{50}}>3\)
27 tháng 4 2017
\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}< 1+\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{50^2-1}\)
\(\Leftrightarrow A< 1+\dfrac{1}{3}+\dfrac{1}{8}+...+\dfrac{1}{2499}\)
\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+...++\dfrac{1}{48}-\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(1-\dfrac{1}{51}+\dfrac{1}{2}-\dfrac{1}{50}\right)\)
\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(\dfrac{50}{51}+\dfrac{24}{50}\right)\)
Nhận xét \(\dfrac{50}{51}< 1;\dfrac{24}{50}< 1\Rightarrow A< 1+\dfrac{1}{2}\cdot\left(\dfrac{50}{51}+\dfrac{24}{50}\right)< 1+\dfrac{1}{2}\cdot\left(1+1\right)=2\)
Vậy A<2
BN
27 tháng 4 2017
Nhận xét: \(\dfrac{1}{1^2}=1\)
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
...........
\(\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{50}=2-\dfrac{1}{50}< 2\)
Vậy A < 2
13 tháng 5 2017
Ta có :
\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+......................+\dfrac{1}{50^2}\)
Ta thấy :
\(\dfrac{1}{1^2}=1\)
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
............................
\(\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....................+\dfrac{1}{49.50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...........+\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{50}\)
\(\Rightarrow A< 2-\dfrac{1}{50}< 2\)
\(\Rightarrow A< 2\rightarrowđpcm\)
27 tháng 5 2017
a, \(-1\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{1}{2}+2\dfrac{1}{6}\\ =-\dfrac{5}{3}+\dfrac{3}{4}-\dfrac{1}{2}+\dfrac{13}{6}\\ =\dfrac{-5.4+3.3-1.6+13.2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
b, \(\dfrac{11}{50}\left(-17\dfrac{1}{2}\right)-\dfrac{11}{50}.82\dfrac{1}{2}\\ =\dfrac{11}{50}.\left(-17\dfrac{1}{2}-82\dfrac{1}{2}\right)=\dfrac{11}{50}.\left(-100\right)=-22\)
HD
28 tháng 5 2017
a) \(-1\dfrac{2}{3}\) + \(\dfrac{3}{4}\) \(-\) \(\dfrac{1}{2}\) + \(2\dfrac{1}{6}\)
=\(-\dfrac{5}{3}\) + \(\dfrac{3}{4}\) \(-\) \(\dfrac{1}{2}\) + \(\dfrac{13}{6}\)\()\)
=\(-\) \(\dfrac{20}{12}\) + \(\dfrac{9}{12}\) \(-\) \(\dfrac{6}{12}\) + \(\dfrac{26}{12}\)
= \((\)\(\dfrac{-20}{12}\) + \(\dfrac{26}{12}\) \()\) + \((\) \(\dfrac{9}{12}\) \(-\) \(\dfrac{6}{12}\) \()\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\)
= \(\dfrac{3}{4}\)
b)\(\dfrac{11}{50}\) \((\) \(-17\dfrac{1}{2}\) \()\) \(-\) \(\dfrac{11}{50}\) .\(82\dfrac{1}{2}\)
= \(\dfrac{11}{50}\) . \(-\dfrac{35}{2}\) \(-\) \(\dfrac{11}{50}\) . \(\dfrac{165}{2}\)
= \(\dfrac{11}{50}\). \((\) \(-\dfrac{35}{2}\) \(-\) \(\dfrac{165}{2}\) \()\)
=\(\dfrac{11}{50}\). \(-\)\(100\)
= \(-22\)
Chúc bạn học thật tốt nha ! 
24 tháng 6 2023
1/2^2+1/3^2+...+1/50^2<1/1*2+1/2*3*+...+1/49*50
=1/1-1/2+1/2-1/3+...+1/49-1/50<1
=>S<1+1=2
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2};\dfrac{1}{3^2}< \dfrac{1}{2\cdot3};.....;\dfrac{1}{50^2}< \dfrac{1}{49\cdot50}\\ =>\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+....+\dfrac{1}{49\cdot50}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{49}-\dfrac{1}{50}=\left(1+1\right)-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-.....-\left(\dfrac{1}{49}-\dfrac{1}{49}\right)-\dfrac{1}{50}=2-0-0-0.....-\dfrac{1}{50}\\ =2-\dfrac{1}{50}< 2=>\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 2\)