\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
So sánh B với 1
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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
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 < \(\dfrac{1}{2}\)
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
=>\(\dfrac{11}{20}>\dfrac{11}{21}\)
=>\(-\dfrac{11}{20}< -\dfrac{11}{21}\)
=>\(B< -\dfrac{11}{21}\)
AH
Akai Haruma
Giáo viên
1 tháng 8 2018
Lời giải:
\(A=\frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+...+(\frac{1}{2})^{98}+(\frac{1}{2})^{99}\)
\(\Rightarrow 2A=1+\frac{1}{2}+(\frac{1}{2})^2+...+(\frac{1}{2})^{97}+(\frac{1}{2})^{98}\)
Trừ theo vế:
\(2A-A=1-(\frac{1}{2})^{99}\)
\(A=1-(\frac{1}{2})^{99}< 1\)
Ta có đpcm.
9 tháng 4 2017
Vì có 98 số hạng -> tích là số chẵn -> làm như dạng bt ! ( đổi ngc lại thành 1 - nhìn cho quen mắt )
27 tháng 7 2018
ta có : \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\Rightarrow\dfrac{1}{2}A=A-\dfrac{1}{2}A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^{100}\) \(\Rightarrow A=2.\left(\dfrac{1}{2}A\right)=1-2\left(\dfrac{1}{2}\right)^{100}< 1\left(đpcm\right)\)
AT
27 tháng 7 2018
\(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow2A=2\cdot\left[\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\right]\)
\(2A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)
\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}-\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^3-...-\left(\dfrac{1}{2}\right)^{99}\)
\(A=1-\left(\dfrac{1}{2}\right)^{99}< 1\left(đpcm\right)\)
MH
25 tháng 1 2022
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
25 tháng 1 2022
b, \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)=\dfrac{99.98...1}{100.99...2}=\dfrac{1}{100}\)
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-\(\dfrac{1}{2^{50}}>3\)
ta có : \(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow\dfrac{1}{2}B=\dfrac{1}{2}\left(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\right)=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\) \(\Rightarrow B-\dfrac{1}{2}B=\dfrac{1}{2}B=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^{100}\)
\(\Rightarrow B=2.\dfrac{1}{2}B=1\left(\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^{100}\right)=1-\left(\dfrac{1}{2}\right)^{99}< 1\)
vậy \(B< 1\)
2B= 1+ 1/2+ (1/2)2+ ....+(1/2)98
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B= 1/2+ (1/2)2+ ....+(1/2)99
B= 1- (1/2)99 <1
=>B <1