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1/2 + 2/3 + 3/4 + 4/5 + 5/6 + 6/7 + 7/8 + 8/9 + ........+ 95/96 + 96/97 + 97/98 + 98/99 + 99/100 = ?
![](https://rs.olm.vn/images/avt/0.png?1311)
Số các số hạng là:
(2000 - 100) : 1 + 1 = 1901
Tổng là:
(2000 + 100) x 1901 : 2 = 1996050
Đáp số : 1996050
![](https://rs.olm.vn/images/avt/0.png?1311)
=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)
=0+0+...+0
=0
~ mk~
1-2-3+4-5-6-7+8-...+97-98-99+100
= (1-2-3+4) + (5-6-7+8) +...+(97-98-99+100)
= 0+0+...+0
=0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)
\(E=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)
\(=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{99}-1+\dfrac{1}{99}=\dfrac{2}{99}-1=-\dfrac{97}{99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
=100
Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{8}{\dfrac{1}{5}}=40\)
\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(E=1+2-3-4+5+6-7-....+97+98-99\)
\(E=\left(1+2-3-4\right)+\left(5+6-7-8\right)+..+\left(97+98-99-100\right)\)( có \(\frac{100}{2}=25\)nhóm)
\(E=-4+\left(-4\right)+....+\left(-4\right)\)( có \(25\)số )
\(E=\left(-4\right).25=-100\)
\(E=1+2-3-4+.............+98-99-100\)
\(E=1+\left(2-3-4+5\right)+\left(6-7-8+9\right)+...+\left(98-99-100\right)\)
\(E=1+0+0+...+\left(-101\right)\)
\(E=-100\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(2P=2^{101}-2^{100}+2^{98}-2^{97}+...+2^3-2^2\)
=>\(3P=2^{101}-2\)
hay \(P=\dfrac{2^{101}-2}{3}\)
b: \(5Q=5^{101}-5^{100}+5^{99}-5^{98}+...+5^3-5^2+5\)
=>\(6Q=5^{101}+1\)
hay \(Q=\dfrac{5^{101}+1}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
giup oi
\(\Leftrightarrow S=1-2+3-4+5-6+....97-98+99\)
\(\Leftrightarrow S=-1+-1+-1+....+-1+99\)
\(\Leftrightarrow S=-49+99\)
\(\Rightarrow S=50\)
Vậy: \(S=50\)