Question

Demonatrati ca (n+1)+(n+2)+....+(n+n)=n(3n+1)/2 oricare n>=1.

Answer 1

O formula (a lui Gauss): 1+2+3+...+m=m*(m+1)/2, oricare ar fi m numar natural.

Deci (n+1)+(n+2)+...+(n+n)=(1+2+3+...+2n)-(1+2+3+...+n)=2n*(2n+1)/2-n*(n+1)/2=(4n^2+2n-n^2-n)/2=(3n^2+n)/2=n(3n+1)/2.