| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Proving standard summation formulae |
| Difficulty | Standard +0.8 This is a Further Maths question requiring the method of differences to derive a standard summation formula. Part (a) is routine algebra, part (b) is straightforward expansion, but part (c) requires understanding telescoping series and algebraic manipulation to isolate the sum—a technique beyond standard A-level that requires methodological insight rather than just recall. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks |
|---|---|
| \((2r+1)^3 = (2r)^3 + 3(2r)^2 + 3(2r) + 1\) | M1 |
| \(A = 8, B = 12, C = 6\) | A1 |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \((2r-1)^3 = (2r)^3 - 3(2r)^2 + 3(2r) - 1\) | M1 | |
| \((2r+1)^3 - (2r-1)^3 = 24r^2 + 2\) | A1cso | \((*)\) |
| (2) |
| Answer | Marks |
|---|---|
| \(r=1: \quad 3^3 - 1^3 = 24 \times 1^2 + 2\) | M1 A1 |
| \(r=2: \quad 5^3 - 3^3 = 24 \times 2^2 + 2\) | |
| \(\vdots\) | |
| \(r=n: \quad (2n+1)^3 - (2n-1)^3 = 24 \times n^2 + 2\) | |
| Summing: \((2n+1)^3 - 1 = 24\sum r^2 + \left(\sum 2\right)\) | M1 |
| \(\left(\sum 2\right) = 2n\) | B1 |
| Proceeding to \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\) | A1cso |
| (5) 9 | 1st M1 require coefficients of 1,3,3,1 or equivalent. 1st M1 require 1,-3,3,-1 or equivalent. 1st M1 for attempt with at least 1,2 and \(n\) if summing expression incorrect, RHS of display not required at this stage. 1st A1 for 1,2 and n correct. 2nd M1 require cancelling and use of \(24r^2 + 2\). Award B1 for correct \(kn\) for their approach. 2nd A1 is for correct solution only. |
**Part (a):**
| $(2r+1)^3 = (2r)^3 + 3(2r)^2 + 3(2r) + 1$ | M1 | |
| $A = 8, B = 12, C = 6$ | A1 | |
| | (2) | |
**Part (b):**
| $(2r-1)^3 = (2r)^3 - 3(2r)^2 + 3(2r) - 1$ | M1 | |
| $(2r+1)^3 - (2r-1)^3 = 24r^2 + 2$ | A1cso | $(*)$ |
| | (2) | |
**Part (c):**
| $r=1: \quad 3^3 - 1^3 = 24 \times 1^2 + 2$ | M1 A1 | |
| $r=2: \quad 5^3 - 3^3 = 24 \times 2^2 + 2$ | | |
| $\vdots$ | | |
| $r=n: \quad (2n+1)^3 - (2n-1)^3 = 24 \times n^2 + 2$ | | |
| Summing: $(2n+1)^3 - 1 = 24\sum r^2 + \left(\sum 2\right)$ | M1 | |
| $\left(\sum 2\right) = 2n$ | B1 | |
| Proceeding to $\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)$ | A1cso | |
| | (5) 9 | 1st M1 require coefficients of 1,3,3,1 or equivalent. 1st M1 require 1,-3,3,-1 or equivalent. 1st M1 for attempt with at least 1,2 and $n$ if summing expression incorrect, RHS of display not required at this stage. 1st A1 for 1,2 and n correct. 2nd M1 require cancelling and use of $24r^2 + 2$. Award B1 for correct $kn$ for their approach. 2nd A1 is for correct solution only. |4. Given that
$$( 2 r + 1 ) ^ { 3 } = A r ^ { 3 } + B r ^ { 2 } + C r + 1 ,$$
\begin{enumerate}[label=(\alph*)]
\item find the values of the constants $A , B$ and $C$.
\item Show that
$$( 2 r + 1 ) ^ { 3 } - ( 2 r - 1 ) ^ { 3 } = 24 r ^ { 2 } + 2$$
\item Using the result in part (b) and the method of differences, show that
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2011 Q4 [9]}}