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\(A=\frac{1}{2^2}+\cdot\cdot\cdot+\frac{1}{2018^2}\)<\(\frac{1}{1\cdot2}+\cdot\cdot\cdot+\frac{1}{2017\cdot2018}\)
\(\Rightarrow A\)<\(1-\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{2017}-\frac{1}{2018}\)
\(\Rightarrow A\)<\(1-\frac{1}{2018}\)<\(1\)
a, Ta có: \(4\equiv1\left(mod3\right)\)
\(\Rightarrow4^{2018}\equiv1\left(mod3\right)\)
\(\Rightarrow4^{2018}-1⋮3\)
b, Ta có: \(5\equiv1\left(mod4\right)\)
\(\Rightarrow5^{2019}\equiv1\left(mod4\right)\)
\(\Rightarrow5^{2019}-1⋮4\)
c, \(4\equiv-1\left(mod5\right)\)
\(\Rightarrow4^{2019}\equiv-1\left(mod5\right)\)
\(\Rightarrow4^{2019}+1⋮5\)
d, \(5\equiv-1\left(mod6\right)\)
\(\Rightarrow5^{2017}\equiv-1\left(mod6\right)\)
\(\Rightarrow5^{2017}+1⋮6\)
1. Vì \(4\) chia \(3\) dư \(1\)
\(\Rightarrow4^{2018}\) chia \(3\) dư \(1^{2018}=1.\)
\(\Rightarrow4^{2018}-1\) chia hết cho \(3.\)
\(B=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+.....+\frac{1}{5^{2018}}+\frac{1}{5^{2019}}\)
\(\Rightarrow5B=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+.......+\frac{1}{5^{2017}}+\frac{1}{5^{2018}}\)
\(\Rightarrow5B-B=1-\frac{1}{5^{2019}}\)
\(\Rightarrow4B=1-\frac{1}{5^{2019}}\)
\(\Rightarrow B=\frac{1-\frac{1}{5^{2019}}}{4}< \frac{1}{4}\left(đpcm\right)\)