| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove summation with exponentials |
| Difficulty | Challenging +1.2 Part (i) is a standard geometric series induction proof requiring routine application of the inductive hypothesis. Part (ii) requires combining the result with de Moivre's theorem, taking an infinite limit, and extracting the imaginary part—this involves multiple non-trivial steps and complex number manipulation beyond typical A-level, placing it moderately above average difficulty for Further Maths. |
| Spec | 1.04j Sum to infinity: convergent geometric series |r|<14.01a Mathematical induction: construct proofs4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 = \frac{z^1 - 1}{z-1}\), so true when \(n=1\) | B1 | Shows base case |
| Assume that \(1 + z + \ldots + z^{k-1} = \frac{z^k - 1}{z-1}\) | B1 | States inductive hypothesis |
| \(1 + z + \ldots + z^{k-1} + z^k = \frac{z^k-1}{z-1} + z^k = \frac{z^k - 1 + z^k(z-1)}{z-1} = \frac{z^{k+1}-1}{z-1}\), so true when \(n = k+1\) | M1 A1 | Combines fractions |
| \(H_k \rightarrow H_{k+1}\). Hence, by induction, true for all positive integers | A1 | States conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Since \( | z | < 1\), \(\sum_{m=0}^{\infty} z^m = \frac{-1}{z-1}\) |
| \(\sum_{m=1}^{\infty} 2^{-m}\sin m\theta = \text{Im}\!\left(\sum_{m=0}^{\infty} z^m\right) = \text{Im}\!\left(\frac{-1}{\frac{1}{2}\cos\theta + \mathrm{i}\frac{1}{2}\sin\theta - 1}\right)\) | M1 A1 | Uses de Moivre's theorem |
| \(\text{Im}\!\left(\frac{-\!\left(\frac{1}{2}\cos\theta - 1 - \mathrm{i}\frac{1}{2}\sin\theta\right)}{\frac{1}{4}\cos^2\theta - \cos\theta + 1 + \frac{1}{4}\sin^2\theta}\right)\) | M1 | Multiply numerator and denominator by conjugate |
| \(\dfrac{\frac{1}{2}\sin\theta}{\frac{5}{4} - \cos\theta} = \dfrac{2\sin\theta}{5 - 4\cos\theta}\) | A1 | States imaginary part, AG |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 = \frac{z^1 - 1}{z-1}$, so true when $n=1$ | B1 | Shows base case |
| Assume that $1 + z + \ldots + z^{k-1} = \frac{z^k - 1}{z-1}$ | B1 | States inductive hypothesis |
| $1 + z + \ldots + z^{k-1} + z^k = \frac{z^k-1}{z-1} + z^k = \frac{z^k - 1 + z^k(z-1)}{z-1} = \frac{z^{k+1}-1}{z-1}$, so true when $n = k+1$ | M1 A1 | Combines fractions |
| $H_k \rightarrow H_{k+1}$. Hence, by induction, true for all positive integers | A1 | States conclusion |
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Since $|z| < 1$, $\sum_{m=0}^{\infty} z^m = \frac{-1}{z-1}$ | B1 | States $|z|<1$ and uses formula for sum to infinity of geometric progression |
| $\sum_{m=1}^{\infty} 2^{-m}\sin m\theta = \text{Im}\!\left(\sum_{m=0}^{\infty} z^m\right) = \text{Im}\!\left(\frac{-1}{\frac{1}{2}\cos\theta + \mathrm{i}\frac{1}{2}\sin\theta - 1}\right)$ | M1 A1 | Uses de Moivre's theorem |
| $\text{Im}\!\left(\frac{-\!\left(\frac{1}{2}\cos\theta - 1 - \mathrm{i}\frac{1}{2}\sin\theta\right)}{\frac{1}{4}\cos^2\theta - \cos\theta + 1 + \frac{1}{4}\sin^2\theta}\right)$ | M1 | Multiply numerator and denominator by conjugate |
| $\dfrac{\frac{1}{2}\sin\theta}{\frac{5}{4} - \cos\theta} = \dfrac{2\sin\theta}{5 - 4\cos\theta}$ | A1 | States imaginary part, AG |8 (i) Prove by mathematical induction that, for $z \neq 1$ and all positive integers $n$,
$$1 + z + z ^ { 2 } + \ldots + z ^ { n - 1 } = \frac { z ^ { n } - 1 } { z - 1 }$$
(ii) By letting $z = \frac { 1 } { 2 } ( \cos \theta + \mathrm { i } \sin \theta )$, use de Moivre's theorem to deduce that
$$\sum _ { m = 1 } ^ { \infty } \left( \frac { 1 } { 2 } \right) ^ { m } \sin m \theta = \frac { 2 \sin \theta } { 5 - 4 \cos \theta }$$
\hfill \mbox{\textit{CAIE FP1 2019 Q8 [10]}}