Răspuns:
S=∑(2^k+1)/3^k=∑[(2^k)/3^k+1/3^k]=
∑(2/3)^k+∑1/3^k
∑(2/3)^k=2/3+(2/3)^2+(2/3)^3+...+(2/3)^n=suma unei progresii geometroice, cu ratia 2/3=
(2/3)*[(2/3)^n-1]/[1-2/3}=
2/3[(2/3)^n-1]/[1/3]=
2[(2/3)^n-1]
∑(1/3)^k=1/3+(1/3)^2+(1/3)^3+..(1/3)^n=progresie geometrica cu ratia 1/3=
1/3[(1/3)^n-1]/[1-1/3)=1/3*[(1/3)^n-1]/(2/3)=
1/2*[(1/3)^n-1]
S=2[(2/3)^n-1]-1/2[(1/3)^n-1]
Explicație pas cu pas:
