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\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(S=1-\frac{1}{2018}\)
\(S=\frac{2018}{2018}-\frac{1}{2018}\)
\(S=\frac{2017}{2018}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}.\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}=\frac{2017}{2018}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
P=1x2+2x3+3x4+...+2017x2018
3P = 1x2x3 + 2x3x3 + 3x4x3 + ... + 2017x2018x3
3P = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + ... +2017x2018x(2019-2016)
3P = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + ... + 2017x2018x2019 - 2016x2017x2018
3P = 2017x2018x2019
P = 2017x2018x2019 : 3
P = 2739315938
P = 1x2+2x3+3x4+...+2017x2018
3xP = 1x2x3+2x3x3+3x4x3+...+2017x2018x3
3xP = 1x2x3+2x3x(4-1)+3x4x(5-2)+...+2017x2018x(2019-2016)
3xP = 1x2x3+2x3x4-2x3x1+3x4x5-3x4x2+...+2017x2018x2019-2017x2018x2016
3xP = 2017x2018x2019
3xP = 8217947814
P = 8217947814 : 3
P = 2739315938
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A = 1.2 + 2.3 + 3.4 + ... + 2017.2018
⇒ 3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2017.218.(2019 - 2016)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2017.2018.2019 - 2016.2017.2018
= 2017.2018.2019
= 2017.2018.2019
B = 2018³/3 ⇒ 3B = 2018³
Ta có:
2017.2019 = (2018 - 1).(2018 + 1)
= 2018² - 1²
= 2018.2018 - 1 < 2018.2018
⇒ 2017.2018.2019 < 2018.2018.2018
⇒ 3A < 3B
⇒ A < B
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\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)
\(1-\frac{1}{x+1}=\frac{99}{100}\)
=> \(\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
=> x+1 = 100
=> x = 100 - 1
=> x = 99
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1/1.2 +1/2.3 +1/3.4 +....+1/99.100
=1-1/2+1/2-1/3+1/3-14+.....+1/99-1/100
=1-1/100
=99/100
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\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2009}-\dfrac{1}{2010}\\ =1-\dfrac{1}{2010}=\dfrac{2009}{2010}\)
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x= -2018/2017
Bài làm:
Ta có: \(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2017.2018}=-1\)
\(\Leftrightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2018}\right)=-1\)
\(\Leftrightarrow x.\frac{2017}{2018}=-1\)
\(\Rightarrow x=-\frac{2018}{2017}\)