| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Sum of Powers Using Standard Formulae |
| Difficulty | Moderate -0.3 This is a straightforward algebraic manipulation question requiring expansion of (2r-1)², application of standard summation formulae, and simplification. While it involves Further Maths content, it's a routine 'show that' proof with clear steps and no novel insight required, making it slightly easier than an average A-level question. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.04g Sigma notation: for sums of series4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^{n} 3(4r^2 - 4r + 1) = 12\sum_{r=1}^{n} r^2 - 12\sum_{r=1}^{n} r + \sum_{r=1}^{n} 3\) | M1 | Expanding given expression to give a 3 term quadratic and attempt to substitute |
| \(= \frac{12}{6}n(n+1)(2n+1) - \frac{12}{2}n(n+1), +3n\) | A1, B1 | First A for first two terms correct or equivalent; B for \(+3n\) appearing |
| \(= n[2(n+1)(2n+1) - 6(n+1) + 3]\) | M1 | Factorising by \(n\) |
| \(= n[4n^2 - 1] = n(2n+1)(2n-1)\) | A1 cso | Completely correct solution |
| [5] |
$\sum_{r=1}^{n} 3(4r^2 - 4r + 1) = 12\sum_{r=1}^{n} r^2 - 12\sum_{r=1}^{n} r + \sum_{r=1}^{n} 3$ | M1 | Expanding given expression to give a 3 term quadratic and attempt to substitute
$= \frac{12}{6}n(n+1)(2n+1) - \frac{12}{2}n(n+1), +3n$ | A1, B1 | First A for first two terms correct or equivalent; B for $+3n$ appearing
$= n[2(n+1)(2n+1) - 6(n+1) + 3]$ | M1 | Factorising by $n$
$= n[4n^2 - 1] = n(2n+1)(2n-1)$ | A1 cso | Completely correct solution
| [5] |
---\begin{enumerate}
\item Show, using the formulae for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$, that
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } 3 ( 2 r - 1 ) ^ { 2 } = n ( 2 n + 1 ) ( 2 n - 1 ) , \text { for all positive integers } n .$$
\hfill \mbox{\textit{Edexcel FP1 2013 Q1 [5]}}