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\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(=\dfrac{1}{1}\cdot\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{1}{7}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=\dfrac{1}{1}-\dfrac{1}{7}=\dfrac{7}{7}-\dfrac{1}{7}=\dfrac{6}{7}\)
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\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{20}.\left(1+2+....+20\right)\)
\(=1+\frac{1}{2}\times\frac{2.3}{2}+\frac{1}{3}\times\frac{3.4}{2}+...+\frac{1}{20}\times\frac{20.21}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)
\(=\frac{\left(2+21\right).20:2}{2}=\frac{230}{2}=115\)
Số cuối là
\(\frac{1}{10}.\left(1+2+3+...+10\right)\) hay \(\frac{1}{20}.\left(1+2+3+...+20\right)\) ??
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a= 1/2+1/2^2+.............1/2^10
2a=1+1/2+1/2^2+............+1/^9
2a-a=1-1/2^10
=1-1/2004
=2003/2004
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\(A = \frac{1}{1.2} + \frac{1}{2.3}+..+ \frac{1}{9.10}\)
\(= 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...-\frac{1}{10}\)
\(= 1 -\frac{1}{10}\)
\(=\frac{9}{10}\)
A = 1/1.2 + 1/2.3 + 1/3.4 + .. + 1/9.10
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10
A = 1 - 1/10
A = 9/10
Đặt A = \(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
Nhân cả hai với 2 ta được :
2A = \(2\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
=> 2A = \(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
Trừ 2A cho A , ta được :
2A - A = \(\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
=> A = \(1-\frac{1}{2^{10}}\)
A=1/2+1/2^2+1/2^3+.............+1/2^10
2A=1+1/2+1/2^2+.............+1/2^9
2a-a=(1/2+1/2^2+1/2^3.....+1/2^10)-(1+1/2+1/2^2............+1/2^9)
=>a=1-1/2^10
=>a=1-1/1024
=>A=1023/1024