Răspuns:
Explicație pas cu pas:
Observam ca:
[tex]A^2 = \left[\begin{array}{ccc}1&1&2\\0&1&1\\0&0&1\end{array}\right] * \left[\begin{array}{ccc}1&1&2\\0&1&1\\0&0&1\end{array}\right] = \left[\begin{array}{ccc}1&2&5\\0&1&2\\0&0&1\end{array}\right][/tex]
[tex]A^3 = \left[\begin{array}{ccc}1&1&2\\0&1&1\\0&0&1\end{array}\right] * \left[\begin{array}{ccc}1&2&5\\0&1&2\\0&0&1\end{array}\right] = \left[\begin{array}{ccc}1&3&9\\0&1&3\\0&0&1\end{array}\right][/tex]
[tex]A^4 = \left[\begin{array}{ccc}1&1&2\\0&1&1\\0&0&1\end{array}\right] * \left[\begin{array}{ccc}1&3&9\\0&1&3\\0&0&1\end{array}\right] = \left[\begin{array}{ccc}1&4&14\\0&1&4\\0&0&1\end{array}\right][/tex]
Notam prin Aₙ matricea:
[tex]A_{n}= \left[\begin{array}{ccc}1&n&\frac{n*(n+3)}{2} \\0&1&n\\0&0&1\end{array}\right][/tex]
Si demonstram prin inductie matematica relatia: Aₙ = Aⁿ, pentru orice n numar natural diferit de 0
[tex]A_{1}= \left[\begin{array}{ccc}1&1&\frac{1*(1+3)}{2} \\0&1&1\\0&0&1\end{array}\right] = \left[\begin{array}{ccc}1&1&\frac{4}{2} \\0&1&1\\0&0&1\end{array}\right] =\left[\begin{array}{ccc}1&1&2 \\0&1&1\\0&0&1\end{array}\right] = A^1[/tex]
Presupunem ca Aₙ = Aⁿ si demonstram ca Aₙ₊₁ = Aⁿ⁺¹
[tex]A^{n+1} = A^n*A = A_{n}*A= \left[\begin{array}{ccc}1&n&\frac{n*(n+3)}{2} \\0&1&n\\0&0&1\end{array}\right]*\left[\begin{array}{ccc}1&1&2\\0&1&1\\0&0&1\end{array}\right] =[/tex]
[tex]= \left[\begin{array}{ccc}1+0+0&1+n+0&2+n+\frac{n*(n+3)}{2}\\0+0+0&0+1+0&0+1+n\\0+0+0&0+0+0&0+0+1\end{array}\right] =[/tex]
[tex]= \left[\begin{array}{ccc}1&(n+1)&\frac{4+2n+n^2+3n}{2}\\0&1&(n+1)\\0&0&1\end{array}\right] = \left[\begin{array}{ccc}1&(n+1)&\frac{n^2+5n+4}{2}\\0&1&(n+1)\\0&0&1\end{array}\right] =[/tex]
[tex]\left[\begin{array}{ccc}1&(n+1)&\frac{(n+1)(n+4)}{2}\\0&1&(n+1)\\0&0&1\end{array}\right] =\left[\begin{array}{ccc}1&(n+1)&\frac{(n+1)[(n+1)+3]}{2}\\0&1&(n+1)\\0&0&1\end{array}\right]=A_{n+1}[/tex]
