1/(1.3)+1/(2.4)+1/(3.5)+1/(4.6)+...+1/(2021.2023)
(1+1/1.3)(1+1/2.4)(1+1/3.5).....(1+1/2021.2023)
1/1.3-1/2.4+1/3.5+1/4.6+...+1/97.99-1/98.100 = ?
1/1.3-1/2.4+1/3.5-1/4.6+...+1/97.99-1/98.100 = ?
=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
C=(1+1/1.3)(1+1/2.4)(1+1/3.5)(1+1/4.6)....(1+1/98.100)
\(C=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.\dfrac{25}{4.6}....\dfrac{9801}{9800}=\)
\(=\dfrac{2^2.3^2.4^2.5^2.....99^2}{1.2.3^2.4^2.5^2....98^2.99.100}=\dfrac{2.99}{100}=\dfrac{198}{100}=1,98\)
Cho A = 1/1.3 + 1/2.4 + 1/3.5 + 1/3.5 + 1/4.6 + ... + 1/98.100 .Chứng tỏ A < 3/4
(1+1/1.3)(1+1/2.4)(1+1/3.5)(1+1/4.6)...(1+1/2013.2015
\(\left(1+\frac{1}{1.3}\right).....\left(1+\frac{1}{2013.2015}\right)=\frac{2^2}{1.3}.....\frac{2014^2}{2013.2015}=\)\(\frac{2.3.....2014}{1.2.....2013}.\frac{2.3.....2014}{3.4.....2015}=2014.\frac{2}{2015}=\frac{4028}{2015}\)
1/1.3 - 1/2.4 + 1/3.5 - 1/4.6+.....+1/97.99-1/98.100
\(\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}\)
\(=1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}-\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}-\frac{1}{98}+\frac{1}{100}\)
\(=1-\frac{1}{2}-\frac{1}{99}-\frac{1}{98}\)
\(=\frac{2327}{4851}\)
Đặt A=1/1.3 - 1/2.4 +1/3.5 -1/4.6 +.....+1/97.99 -1/98.100
4A= 4/1.3 -4/2.4 +4/3.5 -4/4.6 +.....+4/97.99 -4/98.100
=(4/1.3 +4/3.5 +...+4/97.99) - (4/2.4 +4/4.6 +...+4/98.100)
=(1/1 -1/3+1/3-1/5+...+1/97-1/99)-(1/2 -1/4 -....1/98-1/100)
=(1/1-1/99)-(1/2-1/100)
4A=98/99 - 99/100
A= (98/99-99/100) :4
ngắn gọn nhất nè
1/1.3-1/2.4+1/3.5-1/4.6+...+1/97.99-1/98.100
= 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
tính C=1.3+2.4+3.5+4.6+.....+(n-1).(n+1)
Tính
C=1/1.2 - 1/1.3 + 1/2.4 - 1/3.5 + 1/4.6- 1/5.7 +...+ 1/18.20 - 1/19.20
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