Question

Se considera sirul de numere reale (xn)n>=1 si Sn=x1+x2+...+xn, n€N\{0} .Stiind ca 2Sn=3^n-1,sa se demonstreze ca (xn)n>=1 este progresie geometrica.

Answer 1

[tex]2S_n=3^n-1=\ \textgreater \ S_n= \frac{3^n-1}{2}\\ S_{n-1} = \frac{3^{n-1}-1}{2}\\ x_n=S_n-S_{n-1}=\frac{3^n-1}{2}-\frac{3^{n-1}-1}{2}= \frac{3^{n-1}(3-1)}{2}=3^{n-1}\\ x_n=3^{n-1}=\ \textgreater \ (x_n)\ progresie\ geometrica[/tex]