Edexcel FP1 2014 January — Question 6 9 marks

Exam BoardEdexcel
ModuleFP1 (Further Pure Mathematics 1)
Year2014
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and Series
TypeSum of Powers Using Standard Formulae
DifficultyStandard +0.3 This is a straightforward FP1 question requiring algebraic expansion of r(r+1)(r-1) to r³-r, applying two standard formulae, factorizing to find a=2, then solving a quartic equation that simplifies nicely. All steps are routine for Further Maths students with no novel insight required, making it slightly easier than average.
Spec1.04g Sigma notation: for sums of series4.06a Summation formulae: sum of r, r^2, r^3
6. (a) Use the standard results for \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\) and \(\sum _ { r = 1 } ^ { n } r\) to show that for all positive integers \(n\), $$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r - 1 ) = \frac { 1 } { 4 } n ( n + 1 ) ( n - 1 ) ( n + a )$$ where \(a\) is an integer to be determined.
(b) Hence find the value of \(n\), where \(n > 1\), that satisfies $$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r - 1 ) = 10 \sum _ { r = 1 } ^ { n } r ^ { 2 }$$
(a)
AnswerMarks Guidance
\(\sum_{r=1}^{n} r(r+1)(r-1) = \sum_{r=1}^{n} (r^3 - r)\)
\(= \frac{1}{4}n^2(n+1)^2 - \frac{1}{2}n(n+1)\)M1, A1 An attempt to use at least one of the standard formulae correctly; Correct expression
\(= \frac{1}{4}n(n+1)\{n(n+1) - 2\}\)M1 An attempt to factorise out at least \(n(n+1)\)
\(= \frac{1}{4}n(n+1)(n^2 + n - 2)\)
\(= \frac{1}{4}n(n+1)(n-1)(n+2)\)A1 Achieves the correct answer. (Note: \(a = 2\))
(b)
AnswerMarks Guidance
\(\sum_{r=1}^{a} r(r+1)(r-1) = 10\sum_{r=1}^{a} r^2\)
\(\frac{1}{4}n(n+1)(n-1)(n+2) = \frac{10}{6}n(n+1)(2n+1)\)M1 Sets their part (a) \(= \frac{10}{6}n(n+1)(2n+1)\)
\(\frac{1}{4}(n-1)(n+2) = \frac{5}{3}(2n+1)\)
\(3(n^2 + n - 2) = 20(2n+1)\)
\(3n^2 - 37n - 26 = 0\)M1, A1 Manipulates to a "3TQ = 0"; \(3n^2 - 37n - 26 = 0\)
\((3n+2)(n-13) = 0\)M1 A valid method for factorising a 3TQ
\(n = 13\)A1 Only one solution of \(n = 13\) 
**(a)**

| $\sum_{r=1}^{n} r(r+1)(r-1) = \sum_{r=1}^{n} (r^3 - r)$ | | |
| $= \frac{1}{4}n^2(n+1)^2 - \frac{1}{2}n(n+1)$ | M1, A1 | An attempt to use at least one of the standard formulae correctly; Correct expression |
| $= \frac{1}{4}n(n+1)\{n(n+1) - 2\}$ | M1 | An attempt to factorise out at least $n(n+1)$ |
| $= \frac{1}{4}n(n+1)(n^2 + n - 2)$ | | |
| $= \frac{1}{4}n(n+1)(n-1)(n+2)$ | A1 | Achieves the correct answer. (Note: $a = 2$) |

**(b)**

| $\sum_{r=1}^{a} r(r+1)(r-1) = 10\sum_{r=1}^{a} r^2$ | | |
| $\frac{1}{4}n(n+1)(n-1)(n+2) = \frac{10}{6}n(n+1)(2n+1)$ | M1 | Sets their part (a) $= \frac{10}{6}n(n+1)(2n+1)$ |
| $\frac{1}{4}(n-1)(n+2) = \frac{5}{3}(2n+1)$ | | |
| $3(n^2 + n - 2) = 20(2n+1)$ | | |
| $3n^2 - 37n - 26 = 0$ | M1, A1 | Manipulates to a "3TQ = 0"; $3n^2 - 37n - 26 = 0$ |
| $(3n+2)(n-13) = 0$ | M1 | A valid method for factorising a 3TQ |
| $n = 13$ | A1 | Only one solution of $n = 13$ |

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6. (a) Use the standard results for $\sum _ { r = 1 } ^ { n } r ^ { 3 }$ and $\sum _ { r = 1 } ^ { n } r$ to show that for all positive integers $n$,

$$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r - 1 ) = \frac { 1 } { 4 } n ( n + 1 ) ( n - 1 ) ( n + a )$$

where $a$ is an integer to be determined.\\
(b) Hence find the value of $n$, where $n > 1$, that satisfies

$$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r - 1 ) = 10 \sum _ { r = 1 } ^ { n } r ^ { 2 }$$

\hfill \mbox{\textit{Edexcel FP1 2014 Q6 [9]}}
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