\( \sum\limits_{k=0}^{n}{}x^k = \frac{x^{n+1}-1}{x-1} \)
Wenn es für n stimmt betrachte
\( \sum\limits_{k=0}^{n+1}{}x^k = x^{n+1}+ \sum\limits_{k=0}^{n}{}x^k \)
\( x^{n+1}+ \frac{x^{n+1}-1}{x-1} \)
\( = \frac{x^{n+1}(x-1)}{x-1}+ \frac{x^{n+1}-1}{x-1} \)
\( = \frac{x^{n+2}-x^{n+1}}{x-1}+ \frac{x^{n+1}-1}{x-1} \)
\( = \frac{x^{n+2}-x^{n+1}+x^{n+1}-1}{x-1} \)
\( = \frac{x^{n+2}-1}{x-1} \)
Passt !
