=2×(16+112+120+130+...+1

=2×(12×3+13×4+14×5+...+L2×(2×31​+3×41​+4×51​+...+9×101​)

=2×(12−13+13−14+14−15+...+19−1

=45

Lời giải:

$\frac{A}{2}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}$
$=\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}+\frac{7-6}{6\times 7}+\frac{9-8}{8\times 9}+\frac{10-9}{9\times 10}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}$

$=1-\frac{1}{9}=\frac{8}{9}$

$\Rightarrow A=2\times \frac{8}{9}=\frac{16}{9}$

\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+\dfrac{1}{55}\)

\(A=2\times\dfrac{1}{2}\times\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+\dfrac{1}{55}\right)\)

\(A=2\times\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\right)\)

\(A=2\times\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{9\times10}+\dfrac{1}{10\times11}\right)\)

\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)\)

\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{11}\right)\)

\(A=2\times\dfrac{9}{22}\)

\(A=\dfrac{9}{11}\)

\(a=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{45}\)

\(a=\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+\frac{1}{3.5}+...+\frac{1}{5.9}\)

\(a=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)

\(a=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(a=2\left(\frac{1}{2}-\frac{1}{10}\right)\)

=> \(a=2.\frac{2}{5}\)

=> \(a=\frac{4}{5}\)

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{45}\)

\(\Rightarrow\frac{1}{2}A=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\right)\cdot\frac{1}{2}\)

\(=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)

\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(\Rightarrow A=\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)

\(a=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+....+\frac{1}{45}\)

\(a=\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+\frac{1}{3.5}+...+\frac{1}{5.9}\)

\(\frac{1}{2}.a=\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+\frac{1}{3.5}+...+\frac{1}{5.9}\right)\)

\(\frac{1}{2}.a=\frac{1}{2.1.3}+\frac{1}{2.2.3}+\frac{1}{2.2.5}+\frac{1}{2.3.5}+...+\frac{1}{2.5.9}\)

\(\frac{1}{2}.a=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\)

\(\frac{1}{2}.a=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(\frac{1}{2}.a=\frac{1}{2}-\frac{1}{10}\)

\(\frac{1}{2}.a=\frac{2}{5}\)

\(a=\frac{2}{5}:\frac{1}{2}=\frac{2}{5}.2\)

=> \(a=\frac{4}{5}\)

= 7536573657865734657365873464876

\(=1-2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{90}\right)=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.10}\right)=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.......-\frac{1}{10}\right)=1-2\left(\frac{1}{2}-\frac{1}{10}\right)=1-\frac{2.4}{10}=1-\frac{4}{5}=\frac{1}{5}\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{45}\)

\(=2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\right)\)

\(=2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\right)\)

\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=2\times\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=\frac{4}{5}\)

\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{36}\) + \(\dfrac{1}{45}\)

= \(\dfrac{2}{4}\) + \(\dfrac{2}{6}\) + \(\dfrac{2}{12}\) + \(\dfrac{2}{20}\) + \(\dfrac{2}{30}\) + ... + \(\dfrac{2}{72}\) + \(\dfrac{2}{90}\)

= \(\dfrac{2}{2.2}\) + \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + \(\dfrac{2}{5.6}\) + ... + \(\dfrac{2}{8.9}\) + \(\dfrac{2}{9.10}\)

= 2 (\(\dfrac{1}{2.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + ... + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\))