The positive numbers ${{x}_{n}}$ satisfy the relation ${{x}_{n+1}}=\sqrt{{{x}_{n}}+5}$ for $n=1$, $2

Question

The positive numbers ${{x}_{n}}$ satisfy the relation ${{x}_{n+1}}=\sqrt{{{x}_{n}}+5}$ for $n=1$, $2$, $3$, … As $n	o \infty $, ${{x}_{n}}	o l$. Find (in either order) the value of $l$ to $3$ decimal places and the exact value of $l$. Prove that ${{\left( {{x}_{n+1}} ight)}^{2}}-{{l}^{2}}={{x}_{n}}-l$. Hence show that, if ${{x}_{n}}>l$, then ${{x}_{n}}>{{x}_{n+1}}>l$.

Accepted Answer

$$l = \dfrac{1 + \sqrt{21}}{2}$$

2007 Specimen Paper P1 Q8

2007 Specimen Paper P1 Q8

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