Standard +0.3 This is a standard proof by induction with a summation formula. While it's Further Maths content, the structure is routine: verify base case, assume for n=k, prove for n=k+1 by algebraic manipulation. The algebra is straightforward (expanding r(r+3) and factoring cubics), making this slightly easier than average overall but typical for Further Pure induction questions.
Answer: If expression is true for \(n = k\), it's also true for \(n = k+1\). As it's true for \(n = 1\), by mathematical induction, it's true for all positive integers \(n\).
E1
Award for a perfect solution including the last line.
Answer: When $n = 1$, LHS $= 1 \times 4 = 4$ and RHS $= \frac{1}{3} \times 1 \times 2 \times 6 = 4$. Therefore, expression is valid for $n = 1$. | B1 |
Answer: Assume result is true for $n = k$, i.e. $\sum_{r=1}^{k} r(r+3) = \frac{1}{3}k(k+1)(k+5)$ | M1 |
Answer: Consider $n = k+1$: $\sum_{r=1}^{k+1} r(r+3) = \frac{1}{3}k(k+1)(k+5) + (k+1)(k+4)$ | M1 |
Answer: $= \frac{1}{3}(k+1)[k(k+5) + 3(k+4)]$ | A1 |
Answer: $= \frac{1}{3}(k+1)[k^2 + 8k + 12]$ | A1 |
Answer: $= \frac{1}{3}(k+1)(k+2)(k+6)$ | A1 |
Answer: $= \frac{1}{3}(k+1)((k+1)+1)((k+1)+5)$ | A1 |
Answer: If expression is true for $n = k$, it's also true for $n = k+1$. As it's true for $n = 1$, by mathematical induction, it's true for all positive integers $n$. | E1 | Award for a perfect solution including the last line.
Prove, by mathematical induction, that $\sum_{r=1}^{n} r(r+3) = \frac{1}{3}n(n+1)(n+5)$
for all positive integers $n$. [6]
\hfill \mbox{\textit{WJEC Further Unit 1 2018 Q2 [6]}}