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Fie \( f: R\to R, f(x) = ax^2 + bx + c, a > 0 \) si punctele \( A_n(n, f(n)), n\in N. \) Sa se calculeze \( \lim_{n\to \infty} aria [A_n A_{n + 1}A_{n + 2}] \).

Lena wrote:Fie \( f\ :\ \mathbb{R}\to\mathbb{R} \) , \( f(x)=ax^2+bx+c \) , \( a\ >\ 0 \) si punctele \( A_n\ \left\(\ n\ ,\ f(n)\ \right\) \) , \( n\in\mathbb{N} \) . Sa se calculeze : \( \lim_{n\to\infty}\ \left\[A_nA_{n+1}A_{n+2}\right\] \) .

Solutie : Se stie ca pentru punctele necoliniare : \( \left\|\ \begin{array}{cccc}
M & (\ x_1\ ,\ y_1\ ) \\\\\\
N & (\ x_2\ ,\ y_2\ ) \\\\\\
P & (\ x_3\ ,\ y_3\ )\end{array}\ \right\| \) , aria \( \ \triangle\ MNP \) este data de relatia : \( \fbox{\ \begin{array}{ccc}[MNP] & = & \frac 12 & \cdot & \left|\ \begin{array}{ccc}
x_1\ & y_1\ & 1 \\\\\\
x_2\ & y_2\ & 1 \\\\\\
x_3\ & y_3\ & 1\end{array}\ \right|\end{array}\ } \)

Prin urmare, \( \begin{array}{cccc}\left\[A_nA_{n+1}A_{n+2}\right\] & = & \frac 12 & \cdot & \left|\ \begin{array}{cccc}
n\ & f(n)\ & 1 \\\\\\
n+1\ & f(n+1)\ & 1 \\\\\\
n+2\ & f(n+2)\ & 1\end{array}\ \right| & = & \frac 12 & \cdot & \left|\ \begin{array}{cccc}
-1\ & f(n)-f(n+1)\ & 0 \\\\\\
n+1\ & f(n+1)\ & 1 \\\\\\
1\ & f(n+2)-f(n+1)\ & 0\end{array}\ \right| & = & \frac 12 & \cdot & (-1)^{2+3} & \cdot & \left|\ \begin{array}{cccc}
-1\ & f(n)-f(n+1) \\\\\\\\
1\ & f(n+2)-f(n+1)\end{array}\ \right|\end{array} \)

Asadar, \( \begin{array}{cccc}\left\[A_nA_{n+1}A_{n+2}\right\] & = & \frac 12 & \cdot & \left\[f(n)+f(n+2)-2f(n+1)\right\] & = & \frac a2 & \cdot & \left\[n^2+(n+2)^2-2(n+1)^2\right\] & = & a\ \ \Longrightarrow\ \ \fbox{\ \lim_{n\to\infty}\ \left\[A_nA_{n+1}A_{n+2}\right\]=a\ } \end{array} \)