Question

Tính giá trị biểu thức : 

A = 2/1+2 +  (2+3)/1+2+3 + ....... + (2+3+...+20)/1+2+..+20 

B = (1 - 1/2) (1 - 1/3) ..... (1 - 1/2005)(1 - 1/2006)

Answer 1

\(A=\frac{2}{1+2}+\frac{2+3}{1+2+3}+...+\frac{2+3+...+20}{1+2+3+...+20}\)

\(A=\frac{2}{3}+\frac{5}{6}+...+\frac{209}{210}\)

\(A=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{210}\right)\)

\(A=\left(1+1+....+1\right)\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{210}\right)\)

\(A=19-\left(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{420}\right)\)

\(A=19-\left(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{20.21}\right)\)

\(A=19-2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\)

\(A=19-2\cdot\left(\frac{1}{2}-\frac{1}{21}\right)\)

\(A=19-2\cdot\frac{19}{42}=19-\frac{19}{21}=\frac{380}{21}\)

Vậy A= \(\frac{380}{21}\)

Answer 2

\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2005}\right)\left(1-\frac{1}{2006}\right)\)

\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2004}{2005}\cdot\frac{2005}{2006}\)

\(B=\frac{1\cdot2\cdot...\cdot2004\cdot2005}{2\cdot3\cdot...\cdot2005\cdot2006}\)

\(B=\frac{1}{2006}\)

Vậy \(B=\frac{1}{2006}\)