<p>A sequence of negative numbers is defined by ${{x}_{n+1}}=-\sqrt{8+3{{x}_{n}}}$ for $n=1$, $2$, $3$, …</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that $-\frac{8}{3}<{{x}_{n}}<0$ for $n=1$, $2$, $3$, …</div></li><li><div class="list-item-paragraph">Find the exact value of $l$, given that when $n\to \infty $, ${{x}_{n}}\to l$.</div></li><li><div class="list-item-paragraph">Show that ${{x}_{n+1}}^{2}-{{l}^{2}}=3\left( {{x}_{n}}-l \right)$.</div></li><li><div class="list-item-paragraph">Show algebraically that if ${{x}_{n}} < l$ for some value of $n$, then ${{x}_{n+1}}>l$.</div></li></ol>

A sequence of negative numbers is defined by ${{x}_{n+1}}=-\sqrt{8+3{{x}_{n}}}$ for $n=1$, $2$, $3$, …

Show that $-\frac{8}{3}<{{x}_{n}}<0$ for $n=1$, $2$, $3$, …

Find the exact value of $l$, given that when $n\to \infty $, ${{x}_{n}}\to l$.

Show that ${{x}_{n+1}}^{2}-{{l}^{2}}=3\left( {{x}_{n}}-l \right)$.

Show algebraically that if ${{x}_{n}} < l$ for some value of $n$, then ${{x}_{n+1}}>l$.

Default solution

$\{x_n\}$ are negative numbers $\therefore x_n < 0$

For $x_n$ to be real and negative, $8 + 3x_n > 0$

$n \to \infty$, $x_n \to l$, $x_{n+1} \to l$

\[x_{n+1} = -\sqrt{8 + 3x_n} \Rightarrow l = -\sqrt{8 + 3l}\]

\[l^2 - 3l - 8 = 0 \Rightarrow l^2 = 3l + 8\]

\[l = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-8)}}{2(1)} = \dfrac{3 \pm \sqrt{41}}{2}\]

\[l = \dfrac{3 - \sqrt{41}}{2} \text{ since } l < 0\]

\[x_{n+1}^2 - l^2 = \left(-\sqrt{8 + 3x_n}\right)^2 - l^2\]

Since $x_{n+1} = -\sqrt{8 + 3x_n}$ and from (ii), $l^2 - 3l - 8 = 0 \Rightarrow l^2 = 3l + 8$

From (iii), $3(x_n - l) = x_{n+1}^2 - l^2$

Note: $\{x_n\}$ contains all negative terms $\therefore x_{n+1}$ is negative, $l$ is negative

Worked Example 5

Worked Example 5

Video Solution:

Default solution