Q = 1+ 2/2.3 + 2/ 3.4 + 2/4.5 +... + 2/50/51
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A=1.2+2.3+...+49.50
=>3A=1.2.3+2.3.3+...+49.50.3
=>3A=1.2.(3-0)+2.3.(4-1)+....+49.50.(51-48)
=>3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50
=>3A=49.50.51
=>A=49.25.51=62475
=>3A=
Đặt A=1.2+2.3+3.4+4.5+...+49.50
3A=1.2.3+2.3.3+3.4.3+4.5.3+...+49.50.3
3A=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+49.50.(51-48)
3A=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+49.50.51-48.49.50
3A=(1.2.3+2.3.4+3.4.5+4.5.6+...+49.50.51)-(0.1.2+1.2.3+2.3.4+3.4.5+...+48.49.50)
3A=49.50.51-0.1.2
3A=49.50.51
A=49.50.17
A=41650
Cho A=1/1.2 + 1/2.3 + + 1/ 3.4+...+1/49.50 ; B = 1.2+2.3+3.4+4.5+5.6+...+49.50
Tính 50 mủ 2 A – B/17
\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}=\frac{1}{5}\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}=\frac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\frac{1}{5}\)
ĐKXĐ: \(x\ne0;x\ne-1\)
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2008}{2010}\)
\(\Leftrightarrow2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2008}{2010}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2008}{2010}\)(Biết công thức này chứ?)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2008}{2010}\)
\(\Leftrightarrow1-\dfrac{2}{x+1}=\dfrac{2008}{2010}\)
\(\Leftrightarrow\dfrac{x-1}{x+1}=\dfrac{2008}{2010}\Leftrightarrow2010x-2010=2008x+2008\Leftrightarrow x=2009\left(tm\right)\)
Vậy x = 2009
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{49.50}\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(A=2.\frac{12}{25}=\frac{2.12}{25}=\frac{24}{25}\)
\(Q=1+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{50\cdot51}\)
\(Q=1+2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{50\cdot51}\right)\)
\(Q=1+2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(Q=1+2\cdot\left(\frac{1}{2}-\frac{1}{51}\right)\)
\(Q=1+\frac{49}{51}\)
\(Q=\frac{100}{51}\)