| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2007 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Factorial or product method of differences |
| Difficulty | Standard +0.8 This is a Further Maths question requiring recognition of a telescoping series pattern through factorial manipulation, followed by summation using method of differences. Part (i) requires algebraic manipulation of factorials, part (ii) needs insight to apply telescoping, and part (iii) tests understanding of divergence. While systematic, it demands more sophistication than standard A-level series questions and is appropriately challenging for FP1. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((r+1)^2r!\) | M1 | Factor of \(r!\) or \((r+1)!\) seen |
| A1 | Factor of \((r+1)\) found | |
| A1 | Obtain given answer correctly | |
| (ii) | M1 | Express terms as differences using (i) |
| A1 | At least 1st two and last term correct | |
| \((n+2)! - 2!\) | A1 | Show that pairs of terms cancel |
| (iii) | B1ft | Obtain correct answer in any form |
(i) $(r+1)^2r!$ | M1 | Factor of $r!$ or $(r+1)!$ seen
| A1 | Factor of $(r+1)$ found
| A1 | Obtain given answer correctly
(ii) | M1 | Express terms as differences using (i)
| A1 | At least 1st two and last term correct
$(n+2)! - 2!$ | A1 | Show that pairs of terms cancel
(iii) | B1ft | Obtain correct answer in any form | (1 mark, 8 total)8 (i) Show that $( r + 2 ) ! - ( r + 1 ) ! = ( r + 1 ) ^ { 2 } \times r !$.\\
(ii) Hence find an expression, in terms of $n$, for
$$2 ^ { 2 } \times 1 ! + 3 ^ { 2 } \times 2 ! + 4 ^ { 2 } \times 3 ! + \ldots + ( n + 1 ) ^ { 2 } \times n ! .$$
(iii) State, giving a brief reason, whether the series
$$2 ^ { 2 } \times 1 ! + 3 ^ { 2 } \times 2 ! + 4 ^ { 2 } \times 3 ! + \ldots$$
converges.
\hfill \mbox{\textit{OCR FP1 2007 Q8 [8]}}