Mình tính đc kết quả 22 nhưng hơi khó hiểu mong mọi người giải dùm

C = \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)...\left(1-\frac{1}{210}\right)=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{209}{210}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{418}{420}\)

= \(\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{19.22}{20.21}=\frac{2.2\left(2.3.4...19\right)\left(5.6...22\right)}{\left(2.3.4..20\right)\left(3.4.5..21\right)}=\frac{4.22}{19.3.4}=\frac{22}{57}\)

\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{210}\right)\)

\(B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{209}{210}\)

\(B=\frac{1}{3}.\frac{1}{3}.\frac{9}{2}.\frac{14}{15}...\frac{209}{210}\)

\(B=\frac{1}{6}.\frac{9}{2}.\frac{14}{15}...\frac{209}{210}\)

\(B=\frac{1}{2}.\frac{1}{1}.\frac{7}{5}...\frac{209}{210}\)

\(B=\frac{7}{10}...\frac{209}{210}\)

\(B=\frac{62}{210}\)

Lời giải:

$\frac{A}{2}=\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-...-\frac{1}{420}$

$\frac{A}{2}=\frac{1}{2}-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\right)$

Xét:

$\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}$

$=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}$

$=\frac{1}{2}-\frac{1}{21}$

Do đó:

$\frac{A}{2}=\frac{1}{2}-(\frac{1}{2}-\frac{1}{21})=\frac{1}{21}$

$\Rightarrow A=\frac{2}{21}$

C=2/3.5/6.9/10...209/210

C=4/6.10/12.18/20...418/420 là do nhân với 2

C=1.4/2.3.2.5/3.4.3.6/4.5...19.22/20.21

C=1.2.3....19/2.3.4...20.4.5.6...22/3.4.5...21

C=1/20.22/3

C=11/30

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