\(\frac{2017}{1.2}+\frac{2017}{3.4}+\frac{2017}{4.5}+.....+\frac{2017}{99.100}=?\)
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MT
9 tháng 4 2020
Đặt S = ( 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/2017.2018 )
Đặt A = ( 1/1.2 + 1/3.4 + ... + 1/2017.2018)
= 1 - 1/2 + 1/3 - 1/4 + ... + 1/2017 - 1/2018
= ( 1 + 1/3 + ... + 1/2017 ) - ( 1/2 + 1/4 + ... + 1/2018 )
= ( 1 + 1/2 + ... + 1/2018 ) - 2 ( 1/2 + 1/4 + ... + 1/2018) )
= ( 1 + 1/2 + ... + 1/2018 ) - ( 1 + 1/2 + ... + 1/1009 )
= 1/1010 + 1/1011 + ... + 1/2018
=> A - ( 1/1010 + 1/1011 + ... + 1/2017 ) = 1/2018
=> S = 1/2018
Vậy S = 1/2018
23 tháng 4 2017
Ta có: \(\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}\)
\(=1+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)\)
\(=\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2018}\)
\(=2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)\)
Giờ ta thế vào bài toán ban đầu được
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)
\(=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}\)
\(=\frac{2017}{2018}\)
18 tháng 8 2019
a) \(A=\frac{2+2^2+...+2^{2017}}{1-2^{2017}}\)
Đặt \(B=2+2^2+...+2^{2017}\)
\(\Rightarrow2B=2^2+2^3+...+2^{2018}\)
\(\Rightarrow2B-B=\left(2^2+2^3+...+2^{2018}\right)-\left(2+...+2^{2017}\right)\)
\(\Rightarrow B=2^{2018}-2\)
\(\Rightarrow A=\frac{2^{2018}-2}{1-2^{2017}}\)
\(\Rightarrow A=\frac{-2.\left(1-2^{2017}\right)}{1-2^{2017}}\)
\(\Rightarrow A=-2\)
18 tháng 8 2019
b)Đề phải là CM: \(A< \frac{2017}{2016^2}\)
\(A=\frac{1}{2017}+\frac{2}{2017^2}+...+\frac{22017}{2017^{2017}}+\frac{2018}{2017^{2018}}\)
\(\Rightarrow2017A=1+\frac{2}{2017}+...+\frac{22017}{2017^{2016}}+\frac{2018}{2017^{2017}}\)
\(\Rightarrow2017A-A=\left(1+...+\frac{2018}{2017^{2017}}\right)-\left(\frac{1}{2017}+...+\frac{2017}{2017^{2017}}+\frac{2018}{2017^{2018}}\right)\)
\(\Rightarrow2016A=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}-\frac{2018}{2017^{2018}}\)
Đặt \(\Rightarrow S=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}\)
\(\Rightarrow2017S=2017+1+\frac{1}{2017}+...+\frac{1}{2017^{2016}}\)
\(\Rightarrow2017S-S=\left(2017+1+...+\frac{1}{2017^{2016}}\right)-\left(1+...+\frac{1}{2017^{2017}}\right)\)
\(\Rightarrow2016S=2017-\frac{1}{2017^{2017}}< 2017\)
\(\Rightarrow2016S< 2017\)
\(\Rightarrow S< \frac{2017}{2016}\)
\(\Rightarrow2016A< \frac{2017}{2016}\)
\(\Rightarrow A< \frac{2017}{2016^2}\left(đpcm\right)\)
BT
22 tháng 8 2016
Có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{a.b}=\frac{-\left(a+b\right)}{c\left(a+b+c\right)}\)
\(\Leftrightarrow\left(a+b\right)c\left(a+b+c\right)=-\left(a+b\right)ab\)
\(\Leftrightarrow\left(a+b\right)\left(ca+cb+c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(c+a\right)\left(c+b\right)=0.\)
Vậy: hoặc a + b = 0 hoặc c + a = 0 hoặc c + b =0.
Vai trò của a, b, c như nhau nên giả sử \(a+b=0\Leftrightarrow a=-b.\)
Khi đó: \(\frac{1}{a^{2007}}+\frac{1}{b^{2007}}+\frac{1}{c^{2007}}=\frac{1}{a^{2007}}+\frac{1}{\left(-a\right)^{2007}}+\frac{1}{c^{2007}}=\frac{1}{c^{2007}}.\)
\(\frac{1}{a^{2007}+b^{2007}+c^{2007}}=\frac{1}{a^{2007}+\left(-a\right)^{2007}+c^{2007}}=\frac{1}{c^{2007}}.\)
Vậy: \(\frac{1}{a^{2007}}+\frac{1}{b^{2007}}+\frac{1}{c^{2007}}=\frac{1}{a^{2007}+b^{2007}+c^{2007}}.\)(đpcm).
\(=2017.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2017.\left(1-\frac{1}{100}\right)\)
\(=2017.\frac{99}{100}\)
\(=\frac{199693}{100}\)
\(\frac{2017}{1.2}+\frac{2017}{3.4}+\frac{2017}{4.5}+...+\frac{2017}{99.100}\) \(=2017.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\) \(=2017.\left(1-\frac{1}{100}\right)\) \(=2017.\frac{99}{100}\) \(=\frac{199693}{100}\)