Question

Tính B=\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{n+1}\right)\) với \(n\in N\)

Answer 1

\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{n+1}\right)\\ =\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}....\dfrac{n}{n+1}\\ =\dfrac{1.2.3...n}{2.3...\left(n+1\right)}\\ =\dfrac{1}{n+1}\)

Answer 2

\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{n}{n+1}\)

\(=\dfrac{1}{n+1}\)