<p>A sequence \(u_1, u_2, u_3, \dots\) is given by</p>
<p style="text-align: center">\[u_1 = 2 \text{ and}\hspace{0.5em} u_n = \frac{3u_{n-1} - 1}{6} \text{ for}\hspace{0.5em} n \ge 2.\]</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Find the exact values of \(u_2\) and \(u_3\).</span></div></li><li><div class="list-item-content"><span class="list-item-text">It is given that&nbsp;\(u_n \to l\) as&nbsp;\(n \to \infty\). Showing your working, find the exact value of \(l\).</span></div></li><li><div class="list-item-paragraph">For this value of \(l\), show that&nbsp;\(u_n = \frac{14}{3}\left(\frac{1}{2}\right)^n + l\) satisfies both the initial condition and the recurrence relation given in the question.</div></li></ol>

A sequence \(u_1, u_2, u_3, \dots\) is given by

\[u_1 = 2 \text{ and } u_n = \frac{3u_{n-1} - 1}{6} \text{ for } n \ge 2.\]

Find the exact values of \(u_2\) and \(u_3\).

It is given that \(u_n \to l\) as \(n \to \infty\). Showing your working, find the exact value of \(l\).

For this value of \(l\), verify that \(u_n = \frac{14}{3}\left(\frac{1}{2}\right)^n + l\) satisfies both the initial condition and the recurrence relation given in the question.

For this value of \(l\), show that \(u_n = \frac{14}{3}\left(\frac{1}{2}\right)^n + l\) satisfies both the initial condition and the recurrence relation given in the question.

Limit of an expression

Recurrence Relations and Summation

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

N2012 P1 Q3 [Modified]

\(u_1 = 2\)

\(u_2 = \dfrac{3(2) - 1}{6} = \dfrac{5}{6}\)

\(u_3 = \dfrac{3\left(\frac{5}{6}\right) - 1}{6} = \dfrac{\frac{5}{2} - 1}{6} = \dfrac{\frac{3}{2}}{6} = \dfrac{1}{4}\)

As \(n \to \infty\), \(u_n \to l\) and \(u_{n-1} \to l\).

Substituting into the recurrence relation:

\(l = \dfrac{3l - 1}{6}\)

\(6l = 3l - 1\)

\(3l = -1\)

\(l = -\dfrac{1}{3}\)

Let \(u_n = \dfrac{14}{3}\left(\dfrac{1}{2}\right)^n + l = \dfrac{14}{3}\left(\dfrac{1}{2}\right)^n - \dfrac{1}{3}\)

**Initial condition:** \(u_1 = \dfrac{14}{3} \cdot \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{7}{3} - \dfrac{1}{3} = \dfrac{6}{3} = 2\) \(\checkmark\)

**Recurrence relation:**

\(\dfrac{3u_{n-1} - 1}{6} = \dfrac{3\left[\dfrac{14}{3}\left(\dfrac{1}{2}\right)^{n-1} - \dfrac{1}{3}\right] - 1}{6}\)

\(= \dfrac{14\left(\dfrac{1}{2}\right)^{n-1} - 1 - 1}{6}\)

\(= \dfrac{14\left(\dfrac{1}{2}\right)^{n-1} - 2}{6}\)

\(= \dfrac{14}{6} \cdot \left(\dfrac{1}{2}\right)^{n-1} - \dfrac{1}{3}\)

\(= \dfrac{7}{3} \cdot \left(\dfrac{1}{2}\right)^{n-1} - \dfrac{1}{3}\)

\(= \dfrac{14}{3} \cdot \left(\dfrac{1}{2}\right)^n - \dfrac{1}{3}\)

\(= u_n\) \(\checkmark\)

Hence verified.

N2012 P1 Q3 [Modified]

N2012 P1 Q3 [Modified]

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