| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove summation formula |
| Difficulty | Standard +0.3 This is a straightforward proof by induction with a given formula to prove, followed by algebraic manipulation. The induction structure is standard (base case, assume n=k, prove n=k+1), and part (b) requires simple rearrangement using the standard formula for sum of squares. While it involves quartic terms, no novel insight is required—it's a routine Further Maths exercise slightly above average A-level difficulty due to the algebraic manipulation involved. |
| Spec | 4.01a Mathematical induction: construct proofs4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(5 \times 1^4 + 1^2 = \frac{1}{2}(2)^2(2+1)[=6]\) so \(H_1\) is true. | B1 | Checks base case. |
| Assume that \(\displaystyle\sum_{r=1}^{k}\left[(5r^4 + r^2)\right] = \frac{1}{2}k^2(k+1)^2(2k+1)\) | B1 | States inductive hypothesis for some \(k\) including algebraic form. If says for ALL \(k\), then B0. |
| \(\displaystyle\sum_{r=1}^{k+1}\left[(5r^4+r^2)\right] = \frac{1}{2}k^2(k+1)^2(2k+1) + 5(k+1)^4 + (k+1)^2\) | M1 | Considers sum to \(k+1\). |
| \(\frac{1}{2}(k+1)^2\left(2k^3 + k^2 + 10(k+1)^2 + 2\right)\) | M1 | Take out factor of \((k+1)^2\) OR expands the summation expression and the target expression for \(k+1\) and collects like terms for both. |
| \(\frac{1}{2}(k+1)^2(2k^3 + 11k^2 + 20k + 12) = \frac{1}{2}(k+1)^2(k+2)^2(2k+3)\) | A1 | Factorises or having expanded, checks explicitly. At least one intermediate step seen following the award of M1 before reaching the answer. |
| So \(H_{k+1}\) is true. By induction, \(H_n\) is true for all positive integers \(n\). | A1 | States conclusion. Implication must be clearly expressed. |
| Answer | Marks | Guidance |
|---|---|---|
| \(5\displaystyle\sum_{r=1}^{n}r^4 + \frac{1}{6}n(n+1)(2n+1) = \frac{1}{2}n^2(n+1)^2(2n+1)\) | M1 | Uses correct formula for \(\sum r^2\). |
| \([5]\displaystyle\sum_{r=1}^{n}r^4 = \frac{1}{6}n(n+1)(2n+1)(3n(n+1)-1)\) | M1 | Makes \(\sum r^4\) the subject and takes out all linear factors and the remaining term is of correct form. |
| \(\displaystyle\sum_{r=1}^{n}r^4 = \frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)\) | A1 | CAO |
# Question 3:
## Part (a):
$5 \times 1^4 + 1^2 = \frac{1}{2}(2)^2(2+1)[=6]$ so $H_1$ is true. | **B1** | Checks base case.
Assume that $\displaystyle\sum_{r=1}^{k}\left[(5r^4 + r^2)\right] = \frac{1}{2}k^2(k+1)^2(2k+1)$ | **B1** | States inductive hypothesis for some $k$ including algebraic form. If says for ALL $k$, then B0.
$\displaystyle\sum_{r=1}^{k+1}\left[(5r^4+r^2)\right] = \frac{1}{2}k^2(k+1)^2(2k+1) + 5(k+1)^4 + (k+1)^2$ | **M1** | Considers sum to $k+1$.
$\frac{1}{2}(k+1)^2\left(2k^3 + k^2 + 10(k+1)^2 + 2\right)$ | **M1** | Take out factor of $(k+1)^2$ OR expands the summation expression and the target expression for $k+1$ and collects like terms for both.
$\frac{1}{2}(k+1)^2(2k^3 + 11k^2 + 20k + 12) = \frac{1}{2}(k+1)^2(k+2)^2(2k+3)$ | **A1** | Factorises or having expanded, checks explicitly. At least one intermediate step seen following the award of M1 before reaching the answer.
So $H_{k+1}$ is true. By induction, $H_n$ is true for all positive integers $n$. | **A1** | States conclusion. Implication must be clearly expressed.
## Part (b):
$5\displaystyle\sum_{r=1}^{n}r^4 + \frac{1}{6}n(n+1)(2n+1) = \frac{1}{2}n^2(n+1)^2(2n+1)$ | **M1** | Uses correct formula for $\sum r^2$.
$[5]\displaystyle\sum_{r=1}^{n}r^4 = \frac{1}{6}n(n+1)(2n+1)(3n(n+1)-1)$ | **M1** | Makes $\sum r^4$ the subject and takes out all linear factors and the remaining term is of correct form.
$\displaystyle\sum_{r=1}^{n}r^4 = \frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)$ | **A1** | CAO
---3
\begin{enumerate}[label=(\alph*)]
\item Prove by mathematical induction that, for all positive integers $n$,
$$\sum _ { r = 1 } ^ { n } \left( 5 r ^ { 4 } + r ^ { 2 } \right) = \frac { 1 } { 2 } n ^ { 2 } ( n + 1 ) ^ { 2 } ( 2 n + 1 )$$
\item Use the result given in part (a) together with the List of formulae (MF19) to find $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \mathrm { r } ^ { 4 }$ in terms of $n$, fully factorising your answer.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q3 [9]}}