The positive numbers ${{x}_{n}}$ satisfy the equation ${{x}_{n+1}}=\sqrt{{{x}_{n}}+3}$, for&nbsp

Question

The positive numbers ${{x}_{n}}$ satisfy the equation ${{x}_{n+1}}=\sqrt{{{x}_{n}}+3}$, for $n=1$, $2$, $3$, … As $n\to \infty $, ${{x}_{n}}\to \alpha $. Find the exact value of $\alpha $.[2] Prove that ${{\left( {{x}_{n+1}} \right)}^{2}}-{{\alpha }^{2}}={{x}_{n}}-\alpha $.[2] Hence, or otherwise, show that if ${{x}_{n}}>\alpha $ for a certain value of $n$, then ${{x}_{n+2}}>\alpha $.[3]

Accepted Answer

(i) $\alpha=\frac{1+\sqrt{13}}{2}$ (ii) $(x_{n+1})^2-\alpha^2=x_n-\alpha$ (iii) $x_{n+2}>\alpha$

2007 VJC P1 Q6

2007 VJC P1 Q6

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