Question

aratati ca C de n luate cate 0+C de n luate cate 1+C de n luate cate 2+...+C de n luate cate n=2 la n.
C=combinari.

Answer 1

$(1+x)^n = C_{n}^0+C_{n}^1x+C_{n}^2x^2+C_{n}^3x^3+...+C_{n}^nx^n \\\\ \text{(Formula lui Newton de dezvoltare a binomului)}\\ \\ \text{Facem x = 1:} \\ \\ (1+1)^n = C_{n}^0+C_{n}^1\cdot 1+C_{n}^2\cdot 1^2+C_{n}^3\cdot 1^3+...+C_{n}^n\cdot 1^n \\\\\boxed{2^n =C_{n}^0+C_{n}^1+C_{n}^2+C_{n}^3+...+C_{n}^n}~~~~~\text{q.e.d}$