IB Mathematics AHL 1.11 The sum of infinite geometric sequences AI HL Paper 1- Exam Style Questions- New Syllabus

(a)
(i) $u_2 = u_1 \times r = 9 \times \left( \frac{2}{3} + \frac{2}{3}i \right) = 6 + 6i$
$\boxed{6 + 6i}$

(ii) $u_3 = u_2 \times r = (6 + 6i) \times \left( \frac{2}{3} + \frac{2}{3}i \right) = 4 + 4i + 4i + 4i^2 = 8i$
$\boxed{8i}$

(b)
To find the ratio $R$ of the moduli, calculate the modulus of $r$:
$|r| = \sqrt{ \left( \frac{2}{3} \right)^2 + \left( \frac{2}{3} \right)^2 }$
$|r| = \sqrt{ \frac{4}{9} + \frac{4}{9} } = \sqrt{ \frac{8}{9} } = \frac{2\sqrt{2}}{3}$
Since $|u_{n+1}| = |u_n \times r| = |u_n| \times |r|$, the sequence is geometric.
Common ratio $R = \boxed{\frac{2\sqrt{2}}{3}}$

(c)
For an infinite geometric series, $S_{\infty} = \frac{u_1}{1 – R}$ [cite: 848]
Here, $u_1 = |9| = 9$ and $R = \frac{2\sqrt{2}}{3}$
$S_{\infty} = \frac{9}{1 – \frac{2\sqrt{2}}{3}} = \frac{27}{3 – 2\sqrt{2}}$
Rationalizing the denominator: $\frac{27(3 + 2\sqrt{2})}{9 – 8}$
$\boxed{81 + 54\sqrt{2}}$ (or $\approx 157.37$)