Question

5/3•4+5/4•5+5/5•6+...+5/42•43=?
3/1•4+5/4•a+7/a•16+...+1/n(n+1)=?​

Answer 1

Explicație pas cu pas:

$\dfrac{5}{3 \cdot 4} + \dfrac{5}{4 \cdot 5} + \dfrac{5}{5 \cdot 6} + ... + \dfrac{5}{42 \cdot 43} =$

$= 5 \cdot \bigg(\dfrac{4 - 3}{3 \cdot 4} + \dfrac{5 - 4}{4 \cdot 5} + \dfrac{6 - 5}{5 \cdot 6} + ... + \dfrac{43 - 42}{42 \cdot 43} \bigg)$

$= 5 \cdot \bigg(\dfrac{4}{3 \cdot 4} - \dfrac{3}{3 \cdot 4} + \dfrac{5}{4 \cdot 5} - \dfrac{4}{4 \cdot 5} + \dfrac{6}{5 \cdot 6} - \dfrac{5}{5 \cdot 6} + ... + \dfrac{43}{42 \cdot 43} - \dfrac{42}{42 \cdot 43} \bigg)$

$= 5 \cdot \bigg(\dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{6} + ... + \dfrac{1}{42} - \dfrac{1}{43} \bigg)$

$= 5 \cdot \bigg(\dfrac{1}{3} - \dfrac{1}{43} \bigg) = 5 \cdot \dfrac{43 - 3}{3 \cdot 43} = \dfrac{5 \cdot 40}{129}$

$= \dfrac{120}{129}$