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a) \(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+.......+\frac{6}{44.47}+\frac{6}{47.50}\)
\(=2\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+......+\frac{3}{44.47}+\frac{3}{47.50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{44}-\frac{1}{47}+\frac{1}{47}-\frac{1}{50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=1-\frac{1}{25}\)
\(=\frac{24}{25}\)
đặt \(A=\frac{1}{9.11}+\frac{1}{11.13}+........+\frac{1}{41.43}+\frac{1}{43.45}\)
\(2A=\frac{2}{9.11}+\frac{2}{11.13}+.......+\frac{2}{41.43}+\frac{2}{43.45}\)
\(2A=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+......+\frac{1}{41}-\frac{1}{43}+\frac{1}{43}-\frac{1}{45}\)
\(2A=\frac{1}{9}-\frac{1}{45}\)
\(2A=\frac{4}{45}\)
\(A=\frac{4}{45}\div2\)
\(A=\frac{2}{45}\)
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\(A=\frac{5}{2.5}+\frac{5}{5.8}+\frac{5}{8.11}+...+\frac{5}{47.50}\)
\(=\frac{5}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{47.50}\right)\)
\(=\frac{5}{3}\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{50-47}{47.50}\right)\)
\(=\frac{5}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{47}-\frac{1}{50}\right)\)
\(=\frac{5}{3}\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{4}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 3S = 3/2.5 + 3/5.8 + ... + 3/47.50
3S = 1/2 - 1/5 + 1/5 - 1/8 + ... +1/47 - 1/50
3S = 1/2 - 1/50
3S = 12/25
=> S = 12/25 : 3 = 4/25
k, đây là dạng toán sai phân hữu hạn.
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số hạng tổng quát là 1/[n.(n+3)] = (1/3).[(n+3)-n]/[n.(n+3)] = (1/3). [1/n - 1/(n+3)]
=>
A = (1/3).[(1/2 - 1/5) + (1/5 - 1/8) + (1/8 - 1/11) +...+(1/44 - 1/47) + (1/47 - 1/50)]
= (1/3).[1/2 - 1/50]
= (1/3). (24/50) = (1/3).(12/25) = 4/25
vậy A = 4/25
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good luck!
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= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + .... + 1/29 - 1/32
= 1/2 - 1/32
= ..... ( tự bấm máy tính nhé )
Ta có: \(E=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{29}-\frac{1}{32}\)
\(\Rightarrow E=\frac{1}{2}-\frac{1}{32}\)
\(\Rightarrow E=\frac{16-1}{32}=\frac{15}{32}\)
Vậy \(E=\frac{15}{32}\)
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3x/2.5 + 3x/5.8 + 3x/8.11 + 3x/11.14 = 1/21
=> x . ( 3/2.5 + 3/5.8 + 3/8.11 + 3/11.14 ) = 1/21
=> x . ( 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 ) = 1/21
x . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 ) = 1/21
x . ( 1/2 - 1/14 ) = 1/21
x . 3/7 = 1/21
x = 1/21 : 3/7
=> x = 1/9
\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)
<=> \(x\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right)=\frac{1}{21}\)
<=> \(x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
<=> \(x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
<=> \(x\cdot\frac{3}{7}=\frac{1}{21}\)
<=> \(x=\frac{1}{9}\)
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\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{1}{3}\cdot\frac{99}{202}=\frac{33}{202}\)
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Q=(-1)+(-3)+(-5)+...+(-99)
Dãy số trên là dãy số cách đều -4 đơn vị và có 51 số hạng.
\(\Rightarrow\) Q = [ -99 + ( -1) . 51 : 2 = -2550
Vậy Q= -2500
S= \(\dfrac{1}{2.5}\) + \(\dfrac{1}{5.8}\) + \(\dfrac{1}{8.10}\) + ... + \(\dfrac{1}{47.50}\)
S= \(\dfrac{1}{3}\) . ( \(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\) + ... + \(\dfrac{3}{47.50}\) )
S= \(\dfrac{1}{3}\) . ( \(\dfrac{1}{2}\) - \(\dfrac{1}{50}\) )
S = \(\dfrac{1}{3}\) . \(\dfrac{12}{25}\)
S= \(\dfrac{4}{25}\)
Vậy S = \(\dfrac{4}{25}\)
Lời giải:
$A=2.5+5.8+8.11+...+44.47+47.50$
$9A=2.5.1+2.5.8+5.8(11-2)+8.11(14-5)+....+44.47(50-41)+47.50.(53-44)$
$=2.5.1+2.5.8+5.8.11+8.11.14+...+44.47.50+47.50.53-(2.5.8+5.8.11+....+41.44.47+44.47.50)$
4=2.5.1+47.50.53=124560$
$\Rightarrow A=13840$