\(D=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{2022\times2023}\)
\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{48\times49\times50}\)
\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{48\times49\times50}\)
\(=\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{48\times49}-\dfrac{1}{49\times50}\)
\(=\dfrac{1}{1\times2}-\dfrac{1}{49\times50}\)
\(=\dfrac{1}{2}-\dfrac{1}{2450}\)
\(=\dfrac{612}{1225}\)
\(\text{#}Toru\)
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{98\times99}+\dfrac{1}{99\times100}\)
\(\dfrac{2}{11\times13}+\dfrac{2}{13\times15}+...+\dfrac{2}{19\times21}+\dfrac{2}{21\times23}\)
\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)
\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)
@seven
a: 1/1*2+1/2*3+...+1/99*100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100
=99/100
b: 2/11*13+2/13*15+...+2/21*23
=1/11-1/13+1/13-1/15+...+1/21-1/23
=1/11-1/23
=12/253
\(\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+.....+\dfrac{2}{99\times100}\)
\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{99.100}\)
\(=2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(=2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2.\left(1-\dfrac{1}{100}\right)=2.\dfrac{99}{100}=\dfrac{99}{50}\)
Cho A=\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{99\times100}\)
So sánh A với 1
Đặt A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
=> A = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=> A = 1 - \(\dfrac{1}{100}\) = \(\dfrac{99}{100}\)
=> 1 = \(\dfrac{100}{100}\)
=> A < 1
A = 11.2+12.3+13.4+...+199.10011.2+12.3+13.4+...+199.100
=> A = 1−12+12−13+13−14+...+199−11001−12+12−13+13−14+...+199−1100
=> A = 1 - 11001100 = 9910099100
=> 1 = 100100100100
=> A < 1
chứng minh
M=\(\dfrac{3}{1^2\times2^2}+\dfrac{5}{2^2\times3^2}+\dfrac{7}{3^2\times4^2}+.......+\dfrac{19}{9^2\times10^2}< 1\)
\(M=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(M=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
\(M=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(M=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)
Tính :
a ) S= 2+4+6+...+2018 ( giải bằng hai cách )
b ) 10 + 102 +103 +...+10100 ( giaỉ bằng hai cách )
c ) \(S=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{100}}\)( giải bằng hai cách )
d ) \(S=\dfrac{1!}{3!}+\dfrac{2!}{4!}+\dfrac{3!}{5!}+....+\dfrac{2018!}{2020!}\)
biết rằng : n! = \(1\times2\times3\times...\times n\)
VD : 1! = 1
2! = \(1\times2\)
3! = \(1\times2\times3\)
4! \(1\times2\times3\times4\)
a: Số số hạng là \(\dfrac{2018-2}{2}+1=1009\left(số\right)\)
Tổng là: \(\dfrac{2018+2}{2}\cdot1009=1009\cdot1010=1019090\)
b: \(10S=10^2+10^3+...+10^{101}\)
\(\Rightarrow9S=10^{101}-10\)
hay \(S=\dfrac{10^{101}-10}{9}\)
c: \(5S=1+\dfrac{1}{5}+...+\dfrac{1}{5^{99}}\)
\(\Leftrightarrow4S=1-\dfrac{1}{5^{100}}\)
hay \(S=\dfrac{1}{4}\left(1-\dfrac{1}{5^{100}}\right)\)
Tính bằng cách hợp lí :
a , \(\dfrac{1}{15}+\dfrac{9}{10}+\dfrac{14}{15}-\dfrac{11}{9}-\dfrac{20}{10}+\dfrac{1}{157}\)
b , \(\dfrac{1}{5}-\dfrac{-1}{3}+\dfrac{-1}{5}-\dfrac{2}{6}\)
c , \(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{2015\times2017}\)
d , \(\dfrac{5}{1\times3}+\dfrac{5}{3\times5}+...+\dfrac{5}{2015\times2017}\)
e , \(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+...+\dfrac{1}{2016\times2017}\)
a: \(=\left(\dfrac{1}{15}+\dfrac{14}{15}\right)+\left(\dfrac{9}{10}-2-\dfrac{11}{9}\right)+\dfrac{1}{157}\)
\(=1+\dfrac{1}{157}+\dfrac{81-180-110}{90}\)
\(=\dfrac{158}{157}+\dfrac{-209}{90}\simeq-1.315\)
b: \(=\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{2}{6}\)
=1/3-1/3
=0
c: \(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2015\cdot2017}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
=2016/2017
Tính giá trị biểu thức:
B=\(\dfrac{5}{1\times2}+\dfrac{13}{2\times3}+\dfrac{25}{3\times4}+\dfrac{41}{4\times5}+...+\dfrac{181}{9\times10}\)
\(B=\dfrac{5}{1.2}+\dfrac{13}{2.3}+\dfrac{25}{3.4}+\dfrac{41}{4.5}+...+\dfrac{181}{9.10}\)
\(=\left(\dfrac{1}{1.2}+\dfrac{4}{1.2}\right)+\left(\dfrac{1}{2.3}+\dfrac{12}{2.3}\right)+\left(\dfrac{1}{3.4}+\dfrac{24}{3.4}\right)+...+\left(\dfrac{1}{9.10}+\dfrac{180}{9.10}\right)\)
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right)+\left(\dfrac{4}{1.2}+\dfrac{12}{2.3}+...+\dfrac{180}{9.10}\right)\)
\(=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)+\left(2+2+...+2\right)\)
\(=1-\dfrac{1}{10}+\left(2.9\right)\)
\(=1-\dfrac{1}{10}+18\)
\(=\dfrac{9}{10}+18\)
\(=18\dfrac{9}{10}\)
Goridano
Tổng của hai số đầu là
\(\dfrac{5}{12}\times2=\dfrac{10}{12}\left(1\right)\)
Tổng của 3 số đầu là:
\(\dfrac{19}{36}\times3=\dfrac{19}{12}\left(2\right)\)
Tổng của 4 số là:
\(\dfrac{143}{249}\times4=\dfrac{143}{60}\)
Từ (1) và (2), ta thấy số thứ 3 là: \(\dfrac{19}{12}-\dfrac{10}{12}=\dfrac{48}{60}=\dfrac{4}{5}\)
Từ (2) và (3), ta thấy số cuối là:
\(\left(\dfrac{3}{4}+\dfrac{4}{5}\right)\div2=\dfrac{31}{40}\)
Số đầu là:
\(\dfrac{31}{40}-\dfrac{1}{10}=\dfrac{20}{40}=\dfrac{1}{2}\)
Theo (1), số thứ 2 là:
\(\dfrac{10}{12}-\dfrac{1}{2}=\dfrac{4}{12}=\dfrac{1}{3}\)
Đáp số : \(\dfrac{1}{2};\dfrac{1}{3};\dfrac{3}{4};\dfrac{4}{5}\)