Question

24. a) Så se arate 1/k*(k+2)=1/2*(1/k-1/(k+2)) , unde k€N*

Answer 1

$\dfrac{1}{k(k+2)} = \dfrac{1}{2}\cdot \dfrac{(k+2)-k}{k(k+2)} = \\ \\ = \dfrac{1}{2}\cdot\Big[\dfrac{k+2}{k(k+2)}-\dfrac{k}{k(k+2)}\Big] = \dfrac{1}{2}\cdot \Big(\dfrac{1}{k}-\dfrac{1}{k+2}\Big)$