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A=20182+20162+20142+...+42 +22-(20172 +20152+20132+...+ 32 + 1)
A=(2018²-2017²)+(20162-20152)+(2014²-2013²)+...+(2² −1²)
A=2018+2017+2016+2015+2014+2013+...+2+1
\(A=\dfrac{2018\left(2018+1\right)}{2}=\text{2 037 171}\)
PT đã cho tương đương với:
\(\left(\frac{x}{2017}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+3}{2014}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}=\frac{x+2017}{2015}+\frac{x+2017}{2014}\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2017}+\frac{1}{2016}\right)=\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2014}\right)\)
\(\Leftrightarrow x+2017=0\Leftrightarrow x=-2017\)
Ta có:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2014}+\frac{1}{2015}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)=\frac{1}{1008}+\frac{1}{1009}+....+\frac{1}{2015}\)
Mà \(P=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)
\(\Leftrightarrow S-P=0\) \(\Rightarrow\left(S-P\right)^{2016}=0\)
