Question

∑$\frac{2^k+1}{3^k}$=? (de la k=1 pana la n)

Answer 1

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: