mateforum.ro

Pentru \( a > 0 \) definim sirul \( (x_n)_{n\geq 1 } \) prin relatia de recurenta:
\(
x_{n + 1} = a + ax_1x_2x_3 \cdots x_{n - 1} + (x_1x_2 \cdots x_{n - 1})^2 \text{ ,} \forall n\geq 2
\)
cu \( x_1 = 1 \) si \( x_2 = 1 + a \).

\( a) \) Sa se arate ca \( x_{n + 1} = x_n^2 - ax_n + a \), \( \forall n\geq 2 \).

\( b) \) Sa se arate ca
\(
\lim_{n\to \infty} \sum_{k = 1}^{n} \frac {a^k}{x_k} = a(a + 1)
\).

Concursul interjudetean Papiu, Tg Mures, 2008 (cls 12)

a) ok by induction

Can someone post the solution of b) please ?

thanks

a) P(n): \( x_n=x_1x_2 \cdots x_{n-1}+a \)

P(2) - true

Suppose P(k) true, will prove that \( P(k) \to P(k+1) \).
We know that
\( x_k=x_1x_2 \cdots x_{k-1}+a \).

\( x_{k+1}=a+ \prod_{i=1}^{k-1}x_i (a + \prod_{i=1}^{k-1}x_i) \)
\( x_{k+1}=a+ \prod_{i=1}^{k-1}x_i \cdot x_k \)
\( x_{k+1}=a+ \prod_{i=1}^k x_i \)

So \( P(k) \to P(k+1) \) and P(2) true \( \Rightarrow \) P(n) true \( \forall n\in \mathbb{N}, n \geq 2 \).


Therefore we have
\( (x_1x_2 \cdots x_{n-1} + x_n)(x_1x_2 \cdots x_{n-1} +a - x_n)=0 \)
\( ax_1x_2 \cdots x_n-1 + (x_1x_2 \cdots x_n-1)^2 = x_n^2 - ax_n \)
\( x_{n+1}=x_n^2-ax_n+a \).

For b/

with the relation in a/

a^(k+1)/(x(k+1)-a)= a^k/(x(k)-a) - a^k/x(k)