so sánh A=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{6^2}\)+...\(\dfrac{1}{\left(2n\right)^2}\)với \(\dfrac{1}{2}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
H9
HT.Phong (9A5)
CTVHS
20 tháng 9 2023
\(B=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{100^2}\right)\)
\(B=\left(\dfrac{2^2}{2^2}-\dfrac{1}{2^2}\right)\cdot\left(\dfrac{3^2}{3^2}-\dfrac{1}{3^2}\right)....\left(\dfrac{100^2}{100^2}-\dfrac{1}{100^2}\right)\)
\(B=\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}....\cdot\dfrac{100^2-1}{100^2}\)
\(B=\dfrac{\left(2+1\right)\left(2-1\right)}{2^2}\cdot\dfrac{\left(3+1\right)\left(3-1\right)}{3^2}\cdot...\cdot\dfrac{\left(100+1\right)\left(100-1\right)}{100^2}\)
\(B=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{99\cdot101}{100^2}\)
\(B=\dfrac{1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot101}{2^2\cdot3^2\cdot4^2\cdot5^2\cdot....\cdot100^2}\)
\(B=\dfrac{1\cdot101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}\)
\(B=\dfrac{101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}\)
Mà: \(\dfrac{1}{2}=\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot100}\)
Ta có: \(101< 3\cdot4\cdot5\cdot...\cdot100\)
\(\Rightarrow\dfrac{101}{2\cdot3\cdot4\cdot5\cdot...\cdot100}< \dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot100}\)
\(\Rightarrow B< \dfrac{1}{2}\)
17 tháng 8 2021
\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)
\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)
\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)
\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)
Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B
31 tháng 12 2023
a: \(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\)
\(=-\dfrac{1}{10}\)
9<10
=>1/9>1/10
=>\(-\dfrac{1}{9}< -\dfrac{1}{10}\)
=>\(A>-\dfrac{1}{9}\)
b: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{11}{10}\)
\(=\dfrac{-1}{10}\cdot\dfrac{11}{2}=\dfrac{-11}{20}\)
20<21
=>\(\dfrac{11}{20}>\dfrac{11}{21}\)
=>\(-\dfrac{11}{20}< -\dfrac{11}{21}\)
=>\(B< -\dfrac{11}{21}\)
H9
HT.Phong (9A5)
CTVHS
19 tháng 10 2023
Ta có:
\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{10}-1\right)\)
\(A=-\dfrac{1}{2}\cdot-\dfrac{2}{3}-\dfrac{3}{4}\cdot...\cdot-\dfrac{9}{10}\)
\(A=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-9}{2\cdot3\cdot4\cdot...\cdot10}\)
\(A=-\dfrac{1}{10}\)
Mà: \(10>9\)
\(\Rightarrow\dfrac{1}{10}< \dfrac{1}{9}\)
\(\Rightarrow-\dfrac{1}{10}>-\dfrac{1}{9}\)
\(\Rightarrow A>-\dfrac{1}{9}\)
22 tháng 11 2018
a) Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{\left(2n\right)^2}\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)
Ta có:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{n}\)
\(\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{n}+1\)
\(\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 2-\dfrac{1}{n}\)
\(\Rightarrow\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)< \dfrac{1}{2^2}\left(2-\dfrac{1}{2}\right)\)
\(\Rightarrow A< \dfrac{1}{2^2}.2-\dfrac{1}{2^2}.\dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}-\dfrac{1}{2^3}< \dfrac{1}{2}\)
Vậy \(A< \dfrac{1}{2}\left(Đpcm\right)\)
b) Đặt \(B=\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{\left(2n+1\right)^2}\)
Ta có:
\(B< \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(B< \dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(B< \dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}\left(\dfrac{2n+1}{2n+1}-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}.\dfrac{2n}{2n+1}\)
\(B< \dfrac{2n}{4n+2}\)
\(B< \dfrac{2n}{2\left(2n+1\right)}\)
\(B< \dfrac{n}{2n+1}\)
22 tháng 2 2022
a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)
b: 
HN
12 tháng 3 2017
a/ \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\)
Vậy A < 1
b/ Dựa vô câu a mà làm câu b nhé
\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{\left(2n\right)^2}=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)
\(< \dfrac{1}{4}\left(1+1-\dfrac{1}{n}\right)=\dfrac{1}{2}-\dfrac{1}{4n}< \dfrac{1}{2}\)
Vậy \(B< \dfrac{1}{2}\)
H9
HT.Phong (9A5)
CTVHS
19 tháng 9 2023
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{2020^2}-1\right)\)
\(B=\left(\dfrac{1}{2^2}-\dfrac{2^2}{2^2}\right)\left(\dfrac{1}{3^2}-\dfrac{3^2}{3^2}\right)....\left(\dfrac{1}{2020^2}-\dfrac{2020^2}{2020^2}\right)\)
\(B=\left(\dfrac{1-2^2}{2^2}\right)\left(\dfrac{1-3^2}{3^2}\right)...\left(\dfrac{1-2020^2}{2020^2}\right)\)
\(B=\dfrac{\left(1-2\right)\left(1+2\right)}{2^2}\cdot\dfrac{\left(1-3\right)\left(1+3\right)}{3^2}....\cdot\dfrac{\left(2020-1\right)\left(2020+1\right)}{2020^2}\)
\(B=\dfrac{-1\cdot3}{2^2}\cdot\dfrac{-2\cdot4}{3^2}\cdot\dfrac{-3\cdot5}{4^2}\cdot....\cdot\dfrac{-2019\cdot2021}{2020}\)
\(B=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-2019}{2\cdot3\cdot4\cdot....\cdot2020}\)
\(B=\dfrac{-1\cdot-1\cdot-1\cdot....\cdot-1}{1}\)
\(B=-1\) (2019 số -1)
Mà: \(-1< \dfrac{1}{2}\)
\(\Rightarrow B< \dfrac{1}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 9 2023
\(\dfrac{1}{2^2}\); \(\dfrac{1}{3^2}\);...;\(\dfrac{1}{2020^2}\) < 1 ⇒ 0 > \(\dfrac{1}{2^2}\) - 1 > \(\dfrac{1}{3^2}\) - 1 >..> \(\dfrac{1}{2020^2}\) - 1
Xét dãy số 2; 3; 4;...; 2020 dãy số này có số số hạng là:
(2020 - 2):1 + 1 = 2019 (số hạng)
Vậy B là tích của 2019 số âm nên B < 0 ⇒ B < \(\dfrac{1}{2}\)
5 tháng 11 2021
\(a,P=\dfrac{1}{\left(2+1\right)\left(2+1-1\right):2}+\dfrac{1}{\left(3+1\right)\left(3+1-1\right):2}+...+\dfrac{1}{\left(2017+1\right)\left(2017+1-1\right):2}\\ P=\dfrac{1}{2\cdot3:2}+\dfrac{1}{3\cdot4:2}+...+\dfrac{1}{2017\cdot2018:2}\\ P=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)=2\cdot\dfrac{504}{1009}=\dfrac{1008}{1009}\)
\(b,\) Ta có \(\dfrac{1}{4^2}< \dfrac{1}{2\cdot4};\dfrac{1}{6^2}< \dfrac{1}{4\cdot6};...;\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{\left(2n-2\right)2n}\)
\(\Leftrightarrow VT< \dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{\left(2n-2\right)2n}\\ \Leftrightarrow VT< \dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{\left(2n-2\right)2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{2n}\right)< \dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}\)
Ta có :22A=1+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{n^2}\)
22A-A=1-\(\dfrac{1}{\left(2n\right)^2}\)
3A=\(\dfrac{\left(2n\right)^2-1}{\left(2n\right)^2}\) <\(\dfrac{n^2}{\left(2n\right)^2}\)=\(\dfrac{1}{2}\)
3A<\(\dfrac{1}{2}\) suy ra A<\(\dfrac{1}{2}\)
A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\)+.......+\(\dfrac{1}{\left(2.n\right)^2}\)
A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{\left(2.2\right)^2}\)+ \(\dfrac{1}{\left(2.3\right)^2}\) +....+\(\dfrac{1}{\left(2.n\right)^2}\)
A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^2.2^2}\) + \(\dfrac{1}{2^2.3^2}\)+......+ \(\dfrac{1}{2^2.n^2}\)
A = \(\dfrac{1}{2^2}\) \(\times\) ( 1 + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+.......+ \(\dfrac{1}{n^2}\))
22 \(\times\) A = 1 + \(\dfrac{1}{2^2}\)+ \(\dfrac{1}{3^2}\)+......+\(\dfrac{1}{n^2}\)
4A = 1 + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) +......+ \(\dfrac{1}{n^2}\)
4A = 1 + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ...+\(\dfrac{1}{n.n}\)
1 = 1
\(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\)
\(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\)
...................
\(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right).n}\)
Cộng vế với vế ta có:
4A = 1+\(\dfrac{1}{2.2}\)+\(\dfrac{1}{3.3}\)+....+\(\dfrac{1}{n.n}\) <1+ \(\dfrac{1}{1.2}\)+ \(\dfrac{1}{2.3}\)+ ......+ \(\dfrac{1}{\left(n-1\right).n}\)
4A < 1+ \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+....+\(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\) = 2 - \(\dfrac{1}{n}\)
A < ( 2 - \(\dfrac{1}{n}\)): 4
A < 2 : 4 - \(\dfrac{1}{n}\) : 4
A < \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\)
Vậy A < \(\dfrac{1}{2}\)