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Ta có : \(P=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2019}-\dfrac{1}{2020}=1-\dfrac{1}{2020}=\dfrac{2019}{2020}\)
mà \(2019< 2020\)nên P < 1 ( đpcm )
\(P=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2019.2021}\)
\(P=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021}\)
\(P=1-\dfrac{1}{2021}\)
\(P=\dfrac{2020}{2021}\)
Vì \(\dfrac{2020}{2021}< 1\) ⇒ \(P< 1\) ( điều phải chứng minh )
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2019.2021}\)
= \(2.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2019.2021}\right)\)
= \(1.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2019.2021}\right)\)
= \(1.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2019}-\dfrac{1}{2021}\right)\)
= \(1.\left(1-\dfrac{1}{2021}\right)\)
= \(1.\dfrac{2020}{2021}\)
= \(\dfrac{2020}{2021}\)
Đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2019\cdot2021}\)
\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{2019\cdot2021}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2019}-\frac{1}{2021}\)
\(2A=1-\frac{1}{2021}=\frac{2020}{2021}\)
\(A=\frac{2020}{2021}:2=\frac{2020\cdot2}{2021}=\frac{4040}{2021}\)
bn
Tran Le Khanh Linh lm sai r nếu chia 2 thì 2021.2 chứ ko phải 2020.2
A=2 + 2^2+ 2^3 + 2^4 + 2^5 + 2^6 + 2^6 + 2^7 + 2^8 + 2^9
=>A=(2+2^2+2^3)+....+(2^7+2^8+2^9)
A=(2.1+2.2+2.2^2)+.......+(2^7.1+2^7.2+2^7.2^2)
A=2(1+2+2^2)+....+2^7(1+2+2^2)
A=2.7+....+2^7.7
A=(2+...+2^7).7 chia hết cho 7
vậy A chia hết cho 7
tick nha
B=1-1/4+1/4-1/7+1/7-1/10+...+1/27-1/30. =>B=1-1/30=29/30.
=1/1-1/4+1/4-1/7+1/7-1/10+...+1/27-1/30
=1/1-(1/4-1/4)-(1/7-1/7)-(1/10-1/10)-...-(1/27-1/27)-1/30
=1-0-0-0-...-0-1/30
=1-1/30
=1/29
Bạn tự kết luận nhé..!
=2.(2\1.3+2\3.5+...+2\9.11)
=2.(1-1\11)
làm tắt bạn tự hiểu nhé
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2021}\)
\(A=1-\frac{1}{2021}\)
\(A=\frac{2020}{2021}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2019.2021}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)