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\(A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2014}}\\ 3A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2013}}\\ 3A-A=\left(3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2013}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2014}}\right)\\ 2A=3-\dfrac{1}{3^{2014}}\\ A=\left(3-\dfrac{1}{3^{2014}}\right):2\\ A=3:2-\dfrac{1}{3^{2014}}:2\\ A=\dfrac{3}{2}-\dfrac{1}{3^{2014}\cdot2}< \dfrac{3}{2}\)
Vậy \(A< \dfrac{3}{2}\)
b: \(B=2013+\dfrac{2013}{3}+\dfrac{2013}{6}+\dfrac{2013}{10}+\dfrac{2013}{15}\)
\(=2013\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}\right)\)
\(=4026\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)\)
\(=4026\cdot\dfrac{5}{6}=3355\)
Câu 2:
\(A=2014+\dfrac{2014}{1+2}+\dfrac{2014}{1+2+3}+...+\dfrac{2014}{1+2+3+...+2013}\)
\(=2014\left(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+2013}\right)\)
\(=2014\left(1+\dfrac{1}{2\left(2+1\right)}.2+\dfrac{1}{3\left(3+1\right)}.2+...+\dfrac{1}{2013\left(2013+1\right)}.2\right)\)
\(=2014\left(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2013.2014}\right)\)
\(=4028\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2013.2014}\right)\)
Bạn tự tính nốt nhé
1)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2012^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2011\cdot2012}\left(1\right)\)\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2011\cdot2012}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\\ =\dfrac{1}{1}-\dfrac{1}{2012}< 1\left(2\right)\)
Từ (1) và (2) ta có: A < 1
2)
\(A=2014+\dfrac{2014}{1+2}+\dfrac{2014}{1+2+3}+...+\dfrac{2014}{1+2+3+...+2013}\\ =2014\cdot\left(\dfrac{1}{1}+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+2013}\right)\\ =2014\cdot\left(\dfrac{1}{\left(1\cdot2\right):2}+\dfrac{1}{\left(2\cdot3\right):2}+\dfrac{1}{\left(3\cdot4\right):2}+...+\dfrac{1}{\left(2013\cdot2014\right):2}\right)\\ =2014\cdot\left(\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{2013\cdot2014}\right)\\ =2014\cdot2\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2013\cdot2014}\right)\\ =4028\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\right)\\ =4028\cdot\left(1-\dfrac{1}{2014}\right)\\ =4028\cdot\dfrac{2013}{2014}\\ =4026\)
3)
Để A là số nguyên thì \(6n+42⋮6n\Rightarrow42⋮6n\Rightarrow6n\inƯ\left(42\right)\)
\(Ư\left(42\right)=\left\{1;2;3;6;7;14;21;42\right\}\)
| 6n | 1 | 2 | 3 | 6 | 7 | 14 | 21 | 42 |
| n | \(\dfrac{1}{6}\) | \(\dfrac{1}{3}\) | \(\dfrac{1}{2}\) | 1 | \(\dfrac{7}{6}\) | \(\dfrac{7}{3}\) | \(\dfrac{7}{2}\) | 7 |
Vì n là số tự nhiên nên n = 1 hoặc n = 7
4)
\(A=\dfrac{17^{18}+1}{17^{19}+1}< \dfrac{17^{18}+1+16}{17^{19}+1+16}=\dfrac{17^{18}+17}{17^{19}+17}=\dfrac{17\cdot\left(17^{17}+1\right)}{17\cdot\left(17^{18}+1\right)}=\dfrac{17^{17}+1}{17^{18}+1}=B\)
Vậy A<B
A= \(\dfrac{-3}{5}-\dfrac{-4}{5}+\dfrac{-9}{10}\)
A = \(\dfrac{-7}{10}\)
\(A=\dfrac{-10}{3}+\dfrac{19}{6}\cdot\dfrac{7}{5}-\dfrac{19}{3}\cdot\dfrac{1}{10}+\dfrac{19}{10}\cdot\dfrac{4}{3}\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\dfrac{7}{10}-\dfrac{19}{3}\cdot\dfrac{1}{10}+\dfrac{19}{3}\cdot\dfrac{4}{10}\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\left(\dfrac{7}{10}-\dfrac{1}{10}+\dfrac{4}{10}\right)\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\dfrac{10}{10}=\dfrac{-10}{3}+\dfrac{19}{3}\)
\(=\dfrac{9}{3}=3\)
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2015}}\)
\(\Rightarrow5A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{2014}}\)
\(\Rightarrow5A-A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{2014}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow4A=1-\dfrac{1}{5^{2015}}\)
\(\Rightarrow A=\dfrac{1}{4}-\dfrac{1}{5^{2015}.4}< \dfrac{1}{4}\)
\(\Rightarrowđpcm\)
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\)
\(\Rightarrow5A=5\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow5A=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}+\dfrac{1}{5^{2014}}\)
\(\Rightarrow5A-A=\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}+\dfrac{1}{5^{2014}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow4A=1-\dfrac{1}{5^{2015}}\)
\(\Rightarrow A=\dfrac{1}{4}-\dfrac{1}{5^{2015}.4}\)
\(\Rightarrow A< \dfrac{1}{4}\)
\(A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}\)
\(3A=3\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}\right)\)
\(3A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2013}}\)
\(3A-A=\left(3+1+...+\dfrac{1}{3^{2013}}\right)-\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{2014}}\right)\)
\(2A=3-\dfrac{1}{3^{2014}}\Rightarrow A=\dfrac{3}{2}-\dfrac{\dfrac{1}{3^{2014}}}{2}< \dfrac{3}{2}\)
Vậy \(A< \dfrac{3}{2}\)
A=1+13+132+133+...+132014A=1+13+132+133+...+132014
3A=3(1+13+132+133+...+132014)3A=3(1+13+132+133+...+132014)
3A=3+1+13+...+1320133A=3+1+13+...+132013
3A−A=(3+1+...+132013)−(1+13+...+132014)3A−A=(3+1+...+132013)−(1+13+...+132014)
2A=3−132014⇒A=32−1320142<32