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=> (x+2020)/5=(x+2020)/6=(x+2020)/3+(x+2020)/2
=>(x+2020)(1/5+1/6)=(x+2020)(1/3+1/2)
Với x+2020=0=>x=-2020
Với x+2020 khác 0=>1/5+1/6=1/3+1/2 ,vô lí
Vậy x=-2020
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Đặt A = \(\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+....+\frac{1}{5^{2015}}\)
5A = \(1+\frac{1}{5^1}+\frac{1}{5^2}+....+\frac{1}{5^{2014}}\)
4A = 5A - A = \(1-\frac{1}{5^{2015}}\)
=> A = \(\frac{1-\frac{1}{5^{2015}}}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^{^2}+...+\left(\frac{1}{5}\right)^{2015}\)
\(A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2015}}\)
\(5A=5\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2015}}\right)\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2014}}\)
\(\Rightarrow5A-A=\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2014}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2015}}\right)\)
\(\Rightarrow4A=1-\frac{1}{5^{2015}}\)
\(\Rightarrow A=\frac{1-\frac{1}{5^{2015}}}{4}\)
Vì \(1-\frac{1}{5^{2015}}
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N=(xy2z3+x2y3z4)+(x3y4z5+x4y5z6)+...+(x2013y2014z2015+x2014y2015z2016)
Xét dạng tổng quát của các nhóm:
x2n-1y2nz2n+1=(-1).1.(-1)=1
x2ny2n+1z2n+2=1.(-1).1=-1
Do đó (x2n-1y2nz2n+1+x2ny2n+1z2n+2)=1+(-1)=0
=>N=0+0+...+0
=0