Question

 1+3+5+...+(2n-1)? Cum se rezolva pas cu pas?

Answer 1

Se cunoaste formula  $1+2+...+n=\dfrac{n(n+1)}{2}.$

Prin urmare, avem  $1+2+...+2n=\dfrac{2n(2n+1)}{2}=n(2n+1)$ .

Observam ca :

$1+3+...+(2n-1)=[1+2+...+2n]-[2+4+...+2n]\\ \\ =[1+2+...+2n]-[2(1+2+...+n)]\\ \\ =n(2n+1)-2\dfrac{n(n+1)}{2}\\ \\ =n(2n+1)-n(n+1) \\ \\ =n(2n+1-n-1)\\ \\ =n^2$