Question

1 . Aratati ce pariate are numarul n( n + 1 ) , unde n este numar natural .

2 . Aratati ce paritate are numarul n( n + 1 ) + 17 , unde n este un numar natural diferit de zero . 

Answer 1

1)n(n+1)
e clar că nr. sunt de paritate diferită
Astfel:
par x impar=par 
                              ⇒  n(n+1)=nr. par
impar x par=par 
2)n(n+1)=nr. par(vezi exer. 1)
par+impar=impar⇒n(n+1)=nr. impar

Answer 2

$1) n\to \ numar \ par \\\\ \boxed{n=2k} \\\\ 2k(2k+1)= 4k^2 +1 -nr. \ impar \\\\\\ n\to numar \ impar \\\\ \boxed{n=2k+1} \\\\ (2k+1)(2k+1+1)= (2k+1)(2k+2) \\\\ =4k^2+4k+2k+2= \\\\ = 4k^2 +6k+2 - nr. \ par\\\\\boxed\boxed{Nr. \ n(n+1) \ are \ parietate \ diferita}}}$

$2)n\to numar \ par \\\\ n(n+1)+17= 4k^2+1+17 = \\\\ = 4k^2+18\to numar \ par \\\\\\ n\to numar \ impar \\\\ n(n+1)+17=4k^2+6k+2+17= \\\\ = 4k^2+6k+19 \to \ numar \ impar \\\\\\ \boxed{\boxed{Nr. \ n(n+1)+17 \ are \ parietate \ diferita}}$