| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Session | Specimen |
| Marks | 4 |
| Topic | Sequences and Series |
| Type | Sum of Powers Using Standard Formulae |
| Difficulty | Moderate -0.8 This is a straightforward application of standard summation formulae (sum of r³, r², and constants) requiring algebraic manipulation to reach the given form. It's a 'show that' question with the answer provided, making it easier than average as students only need to verify rather than derive. The algebra is routine for Further Maths students, though factoring to the final form requires some care. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
$S = 4\sum_{r=1}^{n} r^3 + 2\sum_{r=1}^{n} r^2 + 5n = n^2(n+1)^2 + \frac{n}{3}(n+1)(2n+1) + 5n$
Use of at least one standard $\Sigma$ result correctly: M1
$\sum_{r=1}^{n} 1 = n$: B1
$S = \frac{n}{3}\{3n^3 + 6n^2 + 3n + 2n^2 + 3n + 1 + 15\} = \frac{n}{3}\{3n^3 + 8n^2 + 6n + 16\} = \frac{n}{3}(n^2+2)(3n+8)$
For serious factorisation attempt: M1
For given answer legitimately obtained: A1
**Total: 4 marks**1 Using standard results given in MF20, show that
$$\sum _ { r = 1 } ^ { n } \left( 4 r ^ { 3 } + 2 r ^ { 2 } + 5 \right) = \frac { 1 } { 3 } n \left( n ^ { 2 } + 2 \right) ( 3 n + 8 )$$
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 Q1 [4]}}