| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2014 |
| Session | June |
| Marks | 4 |
| Topic | Sequences and series, recurrence and convergence |
| Type | Sum from n+1 to 2n or similar range |
| Difficulty | Standard +0.8 This question requires recognizing that the sum can be rewritten by substituting k = N+r to transform it into a standard sum of squares formula, then manipulating the result algebraically to match the given form. While it uses a provided formula, the transformation and algebraic manipulation to find the specific value of 'a' requires non-trivial problem-solving insight beyond routine application. |
| Spec | 1.04g Sigma notation: for sums of series4.06a Summation formulae: sum of r, r^2, r^3 |
(i) $S = N^2 + (N+1)^2 + (N+2)^2 + ... + (2N-2)^2 + (2N-1)^2 + (2N)^2$ **B1** [1]
(ii) $= \sum_{r=1}^{2N} r^2 - \sum_{r=1}^{N-1} r^2$ **M1**
$= \frac{2N}{6}(2N+1)(4N+1) - \frac{(N-1)}{6}(N)(2N-1)$ **A1 A1**
$= \frac{N}{6}\left(16N^2 + 12N + 2 - [2N^2 - 3N + 1]\right) = \frac{N}{6}\left(14N^2 + 15N + 1\right)$
$= \frac{N}{6}(N+1)(14N+1)$ **A1** [4]
**ALTERNATIVE**
$S = \sum_{r=0}^{N}\left(N^2 + 2Nr + r^2\right)$ Splitting into three parts **M1**
$= (N+1)N^2 + 2N \cdot \frac{N}{2}(N+1) + \frac{N}{6}(N+1)(2N+1)$ **B1 M1** Use of both these standard results
$= \frac{(N+1)}{6}\left(6N^2 + 6N^2 + 2N^2 + N\right)$
$= \frac{N}{6}(N+1)(14N+1)$ **A1**1 The series $S$ is given by $S = \sum _ { r = 0 } ^ { N } ( N + r ) ^ { 2 }$.\\
(i) Write out the first three terms and the last three terms of the series for $S$.\\
(ii) Use the standard result $\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$ to show that $S = \frac { 1 } { 6 } N ( N + 1 ) ( a N + 1 )$ for some positive integer $a$ to be determined.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2014 Q1 [4]}}