Standard +0.3 This is a standard proof by induction of a summation formula for cubes, which is a textbook exercise in FP1. While it requires understanding of the induction framework and algebraic manipulation, it follows a completely routine template with no novel insight needed. The algebra is straightforward compared to more complex induction proofs, making it slightly easier than average overall but typical for Further Maths content.
But this is the given result with \(k+1\) replacing \(k\). Therefore if true for \(k\) it is true for \(k+1\).
E1
Dependent on A1 and previous E1
Since true for \(n=1\), true for \(n=1,2,3\) and all positive integers.
E1
Dependent on B1 and previous E1 and correct presentation
[7]
# Question 6:
| Answer | Mark | Guidance |
|--------|------|----------|
| When $n=1$, $\frac{1}{4}n^2(n+1)^2 = 1$, so true for $n=1$ | B1 | |
| Assume true for $n=k$: $\sum_{r=1}^{k} r^3 = \frac{1}{4}k^2(k+1)^2$ | E1 | Assume true for $k$ |
| $\Rightarrow \sum_{r=1}^{k+1} r^3 = \frac{1}{4}k^2(k+1)^2 + (k+1)^3$ | M1 | Add $(k+1)$th term to both sides |
| $= \frac{1}{4}(k+1)^2[k^2 + 4(k+1)]$ | M1 | Factor of $\frac{1}{4}(k+1)^2$ |
| $= \frac{1}{4}(k+1)^2[k^2+4k+4]$ | | |
| $= \frac{1}{4}(k+1)^2(k+2)^2 = \frac{1}{4}(k+1)^2((k+1)+1)^2$ | A1 | c.a.o. with correct simplification |
| But this is the given result with $k+1$ replacing $k$. Therefore if true for $k$ it is true for $k+1$. | E1 | Dependent on A1 and previous E1 |
| Since true for $n=1$, true for $n=1,2,3$ and all positive integers. | E1 | Dependent on B1 and previous E1 and correct presentation |
| | **[7]** | |
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