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BP
30 tháng 8 2023
A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B
\(\Rightarrow\) \(\dfrac{A}{B}\)=2015
DD
1
MV
22 tháng 4 2017
\(M=\dfrac{1+2^2+2^3+...+2^{2012}}{2^{2014}-2}\)
\(=\dfrac{\left(1+2+2^2+2^3+...+2^{2012}\right)-2}{2^{2014}-2}\)
Gọi (1 + 2 + 22 + ... + 22012) - 2 là D
D = (1 + 2 + 22 + ... + 22012) - 2
2D = (2 + 22 + 23 + ... + 22013) - 4
D = (2 + 22 + 23 + ... + 22013) - 4 - (1 + 2 + 22 + ... + 22012) - 2
D = 22013 - 4 - 1 + 2
D = 22013 - 3
M = \(\dfrac{2^{2013}-3}{2^{2014}-2}\)
Tới đây được chưa?
LB
20 tháng 4 2018
\(A=1+\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+...+\dfrac{\dfrac{\left(1+2013\right).2013}{2}}{2013}\)
\(A=1+\dfrac{\dfrac{3.2}{2}}{2}+\dfrac{\dfrac{4.3}{2}}{3}+...+\dfrac{\dfrac{2014.2013}{2}}{2013}\)
\(A=1+\dfrac{3}{2}+\dfrac{2.3}{3}+...+\dfrac{1007.2013}{2013}\)
\(A=1+\dfrac{3}{2}+2+\dfrac{5}{2}...+1007\)
\(2A=2+3+4+5+6+...+2012+2013+2014\)
\(2A=\dfrac{\left(2+2014\right).2013}{2}\)
\(A=\dfrac{2016.2013}{4}=504.2013\)
LB
20 tháng 4 2018
\(B=\dfrac{-2}{1.3}+\dfrac{-2}{2.4}+...+\dfrac{-2}{2012.2014}+\dfrac{-2}{2013.2015}\)
\(-B=\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2012.2014}+\dfrac{2}{2013.2015}\)
\(-B=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2012.2014}\right)\)
\(-B=\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2015-2013}{2013.2015}\right)+\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2014-2012}{2012.2014}\right)\)
\(-B=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2014}\right)\)
\(-B=\left(1-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2014}\right)\)
\(-B=\dfrac{2014}{2015}+\dfrac{2012}{2014.2}=\dfrac{2014^2+1006.2015}{2015.2014}\)
\(B=\dfrac{2014^2+1006.2015}{-2015.2014}\)
11 tháng 1 2018
\(A=\dfrac{-10}{3}+\dfrac{19}{6}\cdot\dfrac{7}{5}-\dfrac{19}{3}\cdot\dfrac{1}{10}+\dfrac{19}{10}\cdot\dfrac{4}{3}\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\dfrac{7}{10}-\dfrac{19}{3}\cdot\dfrac{1}{10}+\dfrac{19}{3}\cdot\dfrac{4}{10}\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\left(\dfrac{7}{10}-\dfrac{1}{10}+\dfrac{4}{10}\right)\)
\(=\dfrac{-10}{3}+\dfrac{19}{3}\cdot\dfrac{10}{10}=\dfrac{-10}{3}+\dfrac{19}{3}\)
\(=\dfrac{9}{3}=3\)
23 tháng 4 2017
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
MV
23 tháng 4 2017
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
Đặt A = 1 + 2 + 22 + 23 + ... + 22012
2A = 2 (1 + 2 + 22 + 23 + ... + 22012)
2A = 2 + 22 + 23 + 24 + ... + 22013
2A - A = (2 + 22 + 23 + 24 + ... + 22013) - (1 + 2 + 22 + 23 + ... + 22012)
=> A = 22013 - 1
Quay lại bài toán, ta có :
\(M=\dfrac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}=\dfrac{2^{2013}-1}{2^{2014}-2}=\dfrac{2^{2013}-1}{2\left(2^{2013}-1\right)}=\dfrac{1}{2}\)