Solución al problema 71

Enunciado del problema 71

Demostremos por inducción que $A^n={\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}\right)}^n=\left(\begin{array}{ccc}1& 0& 0\\ n& 1& 0\\ \frac{n^2+n}{2}& n& 1\end{array}\right)$

Caso n=1

$$A=\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}\right)$$

Caso n=2

$$A^2={\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}\right)}^2=\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ 2& 1& 0\\ 3& 2& 1\end{array}\right)$$

Supongamos $A^n=\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}\right)$

$$A^n\cdot A=\left(\begin{array}{ccc}1& 0& 0\\ n& 1& 0\\ \frac{n^2+n}{2}& n& 1\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ n+1& 1& 0\\ \frac{n^2+n}{2}+n+1& n+1& 1\end{array}\right)=$$

$$=\left(\begin{array}{ccc}1& 0& 0\\ n+1& 1& 0\\ \frac{n^2+3n+2}{2}& n+1& 1\end{array}\right)=A^{n+1}$$