Cho A=(1+1/1.3).(1+1/2.4).(1+1/3.5).(1+1/4.6)...(1+1/2020.2022) Chứng tỏ A=2021/1011
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KG
1
29 tháng 3 2023
a: \(A=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2022\cdot2024}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2022}-\dfrac{1}{2024}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{1011}{2024}=\dfrac{1011}{4848}< \dfrac{1}{4}\)
b: \(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2013\cdot2015}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2014}{2015}=\dfrac{1007}{2015}< \dfrac{1}{2}\)
NT
0
TQ
1
18 tháng 3 2023
\(P=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+...+\dfrac{1}{2021.2023}\)
Ta sẽ "tách" P làm 2 phần:
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2021.2023}\)
\(B=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{2020.2022}\)
Do đó \(P=A+B\)
Ta có \(A=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2023-2021}{2021.2023}\right)\)
\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)
\(A=\dfrac{1}{2}\left(1-\dfrac{1}{2023}\right)\)
\(A=\dfrac{1011}{2023}\)
Mặt khác, \(B=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{2020.2022}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2020.2022}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+\dfrac{8-6}{6.8}+...+\dfrac{2022-2020}{2020.2022}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(B=\dfrac{505}{2022}\)
Từ đó \(P=A+B=\dfrac{1011}{2023}+\dfrac{505}{2022}=\dfrac{3065857}{4090506}\)
TL
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 9 2023
P = (1 + \(\dfrac{1}{1.3}\)).(1 + \(\dfrac{1}{2.4}\)).(1 + \(\dfrac{1}{3.5}\))....(1 + \(\dfrac{1}{2020.2022}\))
P = \(\dfrac{1.3+1}{1.3}\). \(\dfrac{2.4+1}{2.4}\).\(\dfrac{3.5+1}{3.5}\)....\(\dfrac{2020.2022+1}{2020.2022}\)
P=\(\dfrac{\left(2-1\right)\left(2+1\right)+1}{1.3}\).\(\dfrac{\left(3-1\right)\left(3+1\right)+1}{2.4}\)...\(\dfrac{\left(2021+1\right).\left(2022-1\right)+1}{2020.2022}\)
P = \(\dfrac{2.2}{1.3}\).\(\dfrac{3.3}{2.4}\).\(\dfrac{4.4}{3.5}\)....\(\dfrac{2021.2021}{2020.2022}\)
P = \(\dfrac{2.2021}{2022}\)
P = \(\dfrac{2021}{1011}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 9 2023
P = (1+\(\dfrac{1}{1.3}\)).(1+\(\dfrac{1}{2.4}\)).(1 + \(\dfrac{1}{3.5}\))...(1+\(\dfrac{1}{2020.2022}\))
P =\(\dfrac{1.3+1}{1.3}\).\(\dfrac{2.4+1}{2.4}\).\(\dfrac{3.5+1}{3.5}\)...\(\dfrac{2020.2022+1}{2020.2022}\)
P = \(\dfrac{(2-1)(2+1)+1}{1.3}\).\(\dfrac{(3-1)(3+1)+1}{2.4}\)...\(\dfrac{(2021-1)(2021+1)}{2020.2022}\)
P = \(\dfrac{2.2}{1.3}\).\(\dfrac{3.3}{2.4}\).\(\dfrac{4.4}{3.5}\)...\(\dfrac{2021.2021}{2020.2022}\)
P = \(\dfrac{2021}{1011}\)
VT
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
19 tháng 10 2023
\(C=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.\dfrac{25}{4.6}....\dfrac{9801}{9800}=\)
\(=\dfrac{2^2.3^2.4^2.5^2.....99^2}{1.2.3^2.4^2.5^2....98^2.99.100}=\dfrac{2.99}{100}=\dfrac{198}{100}=1,98\)
3 tháng 8 2015
=1-1/3-1/2+1/4+1/3-1/5-1/4+1/6+...+1/97-1/99-1/98+1/100
=1-1/2-1/99-1/98=2327/4851
\(A=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)\left(1+\frac{1}{4\cdot6}\right)...\left(1+\frac{1}{2020\cdot2022}\right)\)
\(A=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot\frac{25}{4\cdot6}\cdot....\cdot\frac{4084441}{2021\cdot2022}\)
\(A=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\frac{4\cdot4}{3\cdot5}\cdot\frac{5\cdot5}{4\cdot6}\cdot...\cdot\frac{2021\cdot2021}{2020\cdot2022}\)
\(A\frac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot2021\cdot2021}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot....\cdot2020\cdot2022}\)
\(A=\frac{\left(2\cdot3\cdot4\cdot5\cdot...\cdot2021\right)\left(2\cdot3\cdot4\cdot5\cdot...\cdot2021\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot2020\right)\left(3\cdot4\cdot5\cdot6\cdot...\cdot2022\right)}\)
\(A=\frac{2021\cdot2}{2022}\)
\(A=\frac{4042}{2022}\)
\(A=\frac{2021}{1011}\left(đpcm\right)\)