| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Proving standard summation formulae |
| Difficulty | Moderate -0.3 This is a standard, highly structured proof of the sum of cubes formula using method of differences. Each part explicitly guides the student through the steps: (i) telescoping sum (routine), (ii) binomial expansion verification (straightforward algebra), (iii) equating and rearranging to isolate the desired sum. While it requires careful algebraic manipulation across multiple parts, it's a textbook example with no novel insight required—easier than average for FP1. |
| Spec | 1.04g Sigma notation: for sums of series4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Show that terms cancel in pairs. Obtain given answer correctly | M1 A1 | 2 |
| (ii) Attempt to expand and simplify. Obtain given answer correctly | M1 A1 | 2 |
| (iii) Correct \(\sum r\) stated \(\sum 1 = n\) | B1 B1 | |
| Consider sum of 4 separate terms on RHS. Required sum is LHS – 3 terms | M1* *DM1 | |
| \((n+1)^4 - 1 - n(n+1)(2n+1) - 2n(n+1) - n\) | A1 | |
| \(4\sum_{r=1}^{n} r^3 = n^2(n+1)^2\) | A1 | 6 |
| 10 |
**(i)** Show that terms cancel in pairs. Obtain given answer correctly | M1 A1 | 2 |
**(ii)** Attempt to expand and simplify. Obtain given answer correctly | M1 A1 | 2 |
**(iii)** Correct $\sum r$ stated $\sum 1 = n$ | B1 B1 | |
Consider sum of 4 separate terms on RHS. Required sum is LHS – 3 terms | M1* *DM1 | |
$(n+1)^4 - 1 - n(n+1)(2n+1) - 2n(n+1) - n$ | A1 | |
$4\sum_{r=1}^{n} r^3 = n^2(n+1)^2$ | A1 | 6 | Obtain given answer correctly
| | | **10** |7 (i) Use the method of differences to show that
$$\sum _ { r = 1 } ^ { n } \left\{ ( r + 1 ) ^ { 4 } - r ^ { 4 } \right\} = ( n + 1 ) ^ { 4 } - 1$$
(ii) Show that $( r + 1 ) ^ { 4 } - r ^ { 4 } \equiv 4 r ^ { 3 } + 6 r ^ { 2 } + 4 r + 1$.\\
(iii) Hence show that
$$4 \sum _ { r = 1 } ^ { n } r ^ { 3 } = n ^ { 2 } ( n + 1 ) ^ { 2 }$$
\hfill \mbox{\textit{OCR FP1 2009 Q7 [10]}}