| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Sum of powers of roots |
| Difficulty | Challenging +1.2 Part (a) uses the standard identity (α+β+γ)² = α²+β²+γ² + 2(αβ+βγ+γα) with Newton's sums. Part (b) applies the fact that roots satisfy the equation to find α³+β³+γ³. Part (c) requires expanding the sum, using standard summation formulae, and algebraic manipulation—more involved but still follows a clear template using given formulae. This is a typical Further Maths question testing systematic application of root formulae rather than requiring novel insight. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| \((-2)^2 - 2(3)\) | M1 | Uses formula for sum of squares |
| \(-2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\alpha^3 + \beta^3 + \gamma^3 = -2(-2) - 3(-2) - 3(3)\) | M1 | Uses original equation or formula for sum of cubes |
| \(1\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \((\alpha + r)^3 = \alpha^3 + 3\alpha^2 r + 3\alpha r^2 + r^3\) | B1 | Expands |
| \(\sum_{r=1}^{n}\left((\alpha+r)^3 + (\beta+r)^3 + (\gamma+r)^3\right) = \sum_{r=1}^{n}\left(1 + 3(-2)r + 3(-2)r^2 + 3r^3\right)\) | M1 A1 | Collects like terms and uses results from parts (a) and (b) |
| \(n - 6\!\left(\tfrac{1}{2}n(n+1)\right) - 6\!\left(\tfrac{1}{6}n(n+1)(2n+1)\right) + \tfrac{3}{4}n^2(n+1)^2\)<br>\(n - 3n(n+1) - n(n+1)(2n+1) + \tfrac{3}{4}n^2(n+1)^2\) | M1 | Applies formulae from MF19 |
| \(n + \tfrac{1}{4}n(n+1)\!\left(-12 - 4(2n+1) + 3n(n+1)\right)\)<br>\(n + \tfrac{1}{4}n(n+1)\!\left(3n^2 - 5n - 16\right)\) | M1 A1 | Simplifies |
## Question 4(a):
| $(-2)^2 - 2(3)$ | M1 | Uses formula for sum of squares |
|---|---|---|
| $-2$ | A1 | |
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## Question 4(b):
| $\alpha^3 + \beta^3 + \gamma^3 = -2(-2) - 3(-2) - 3(3)$ | M1 | Uses original equation or formula for sum of cubes |
|---|---|---|
| $1$ | A1 | AG |
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## Question 4(c):
| $(\alpha + r)^3 = \alpha^3 + 3\alpha^2 r + 3\alpha r^2 + r^3$ | B1 | Expands |
|---|---|---|
| $\sum_{r=1}^{n}\left((\alpha+r)^3 + (\beta+r)^3 + (\gamma+r)^3\right) = \sum_{r=1}^{n}\left(1 + 3(-2)r + 3(-2)r^2 + 3r^3\right)$ | M1 A1 | Collects like terms and uses results from parts (a) and (b) |
| $n - 6\!\left(\tfrac{1}{2}n(n+1)\right) - 6\!\left(\tfrac{1}{6}n(n+1)(2n+1)\right) + \tfrac{3}{4}n^2(n+1)^2$<br>$n - 3n(n+1) - n(n+1)(2n+1) + \tfrac{3}{4}n^2(n+1)^2$ | M1 | Applies formulae from MF19 |
| $n + \tfrac{1}{4}n(n+1)\!\left(-12 - 4(2n+1) + 3n(n+1)\right)$<br>$n + \tfrac{1}{4}n(n+1)\!\left(3n^2 - 5n - 16\right)$ | M1 A1 | Simplifies |
---4 The cubic equation $x ^ { 3 } + 2 x ^ { 2 } + 3 x + 3 = 0$ has roots $\alpha , \beta , \gamma$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }$.
\item Show that $\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } = 1$.
\item Use standard results from the list of formulae (MF19) to show that
$$\sum _ { r = 1 } ^ { n } \left( ( \alpha + r ) ^ { 3 } + ( \beta + r ) ^ { 3 } + ( \gamma + r ) ^ { 3 } \right) = n + \frac { 1 } { 4 } n ( n + 1 ) \left( a n ^ { 2 } + b n + c \right)$$
where $a$, $b$ and $c$ are constants to be determined.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q4 [10]}}