Foro de estudiantes

Paso base: $$n=0$$

$$\sum_{i=0}^{n}=\sum_{i=0}^{0} x^{i}=x^{0}=1\hspace{0.5cm} \forall x\neq 1$$

$$\frac{1-x^{n+1}}{1-x}=\frac{1-x^{0+1}}{1-x}=\frac{1-x}{1-x}=1\hspace{0.5cm} \forall x\neq 1$$

Se cumple el paso base.

Paso inductivo: Supongamos $$\sum_{i=0}^{k} x^{i}=\frac{1-x^{k+1}}{1-x}$$

$$\sum_{i=0}^{k+1} x^{i}=\sum_{i=0}^{k} x^{i} + x^{k+1}=\frac{1-x^{k+1}}{1-x}+x^{k+1}=\frac{1-x^{k+1}}{1-x}+x^{k+1}\left(\frac{1-x}{1-x}\right)=\frac{1-x^{k+1}}{1-x}+\frac{x^{k+1}-x^{k+2}}{1-x}=\frac{1-x^{k+1}+x^{k+1}-x^{k+2}}{1-x}=\frac{1-x^{k+2}}{1-x}=\frac{1-x^{(k+1)+1}}{1-x}\hspace{0.5cm} \forall x\neq 1$$

Se cumple el paso inductivo.