Question

calculati produsul urmator:
$(1-\frac{2}{2*3} )(1-\frac{2}{3*4} )*...*(1-\frac{2}{(n-1)*n} )(1-\frac{2}{n(n+1)} )$

Answer 1

Explicație pas cu pas:

termenul general:

$\boxed {1-\dfrac{2}{n(n+1)} = \dfrac{(n - 1)(n + 2)}{n(n+1)}}$

produsul devine:

$\Big(1-\dfrac{2}{2 \cdot 3} \Big)\Big(1-\dfrac{2}{3 \cdot 4} \Big) \cdot ... \cdot \Big(1-\dfrac{2}{(n-1) \cdot n} \Big)\Big(1-\dfrac{2}{n(n+1)} \Big) =$

$= \dfrac{1 \cdot 4}{2 \cdot 3} \cdot \dfrac{2 \cdot 5}{3 \cdot 4} \cdot \dfrac{3 \cdot 6}{4 \cdot 5} \cdot ... \cdot \dfrac{(n - 2)(n + 1)}{(n - 1)n} \cdot \dfrac{(n - 1)(n + 2)}{n(n+1)}$

$= \bigg(\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{3}{4} \cdot ... \cdot \dfrac{n - 2}{n - 1} \cdot \dfrac{n - 1}{n}\bigg) \cdot \bigg(\dfrac{4}{3} \cdot \dfrac{5}{4} \cdot \dfrac{6}{5} \cdot ... \cdot \dfrac{n + 1}{n} \cdot \dfrac{n + 2}{n+1}\bigg)$

$= \dfrac{1}{n} \cdot \dfrac{n + 2}{3} = \bf {\dfrac{n + 2}{3n}}$