Voi nota combinari de n luate cate k cu Cn^k sper ca e ok asa
C10^0+3*C10^1+...+3*C10^10=3(C10^0+C10^1+...+C10^10)-2(deoarece 2*C10^0)=2
Dar,din formula binomului lui Newton avem:
(a+b)^n=a^n*b^0*Cn^0+a^n-1*b^1+Cn^1+...+a^0*Cn^n*b^n
De unde rezulta a=1,b=1.,n=10
Deci avem 3*(a+b)^n-2 adica
3*2^10-2=3*1024-2=3072-2=3070