Edexcel FP1 2015 June — Question 3 8 marks

Exam BoardEdexcel
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and Series
TypeSums Between Limits
DifficultyStandard +0.3 This is a straightforward Further Maths FP1 question on series summation. Part (a) requires expanding brackets and applying standard formulae (routine algebraic manipulation), while part (b) uses the 'sums between limits' technique of subtracting two sums. Both parts follow predictable patterns with no novel insight required, making it slightly easier than average even for Further Maths.
Spec4.06a Summation formulae: sum of r, r^2, r^3
3. (a) Using the formulae for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\), show that $$\sum _ { r = 1 } ^ { n } ( r + 1 ) ( r + 4 ) = \frac { n } { 3 } ( n + 4 ) ( n + 5 )$$ for all positive integers \(n\).
(b) Hence show that $$\sum _ { r = n + 1 } ^ { 2 n } ( r + 1 ) ( r + 4 ) = \frac { n } { 3 } ( n + 1 ) ( a n + b )$$ where \(a\) and \(b\) are integers to be found.
Question 3:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\sum_{r=1}^{n}(r+1)(r+4) = \sum_{r=1}^{n} r^2+5r+4\)B1 Expands bracket correctly to \(r^2+5r+4\)
\(= \frac{n}{6}(n+1)(2n+1)+5\cdot\frac{n}{2}(n+1)+4n\)M1 A1 M1: Uses \(\frac{n}{6}(n+1)(2n+1)\) or \(\frac{n}{2}(n+1)\) correctly
\(= \frac{n}{6}\{(n+1)(2n+1)+15(n+1)+24\}\)dM1 dM1: Attempts to remove factor \(\frac{n}{6}\) or \(\frac{n}{3}\) to obtain a quadratic factor; need not be 3 terms
\(= \frac{n}{6}(2n^2+18n+40)\) or \(\frac{n}{3}(n^2+9n+20)\)
\(= \frac{n}{3}(n+4)(n+5)\) given answerA1* (5) Completely correct work including a step with a collected 3 term quadratic prior to bracket, with correct printed answer
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\sum_{r=n+1}^{2n}(r+1)(r+4) = \frac{2n}{3}(2n+4)(2n+5)-\frac{n}{3}(n+4)(n+5)\)M1 Uses \(f(2n)-f(n)\) or \(f(2n)-f(n+1)\) correctly; require all 3 terms in \(2n\) (and \(n+1\) if used)
\(= \frac{n}{3}\{8n^2+36n+40-n^2-9n-20\}\)dM1 dM1: Attempts to remove factor \(\frac{n}{6}\) or \(\frac{n}{3}\) to obtain quadratic factor
\(= \frac{n}{3}\{7n^2+27n+20\} = \frac{n}{3}(n+1)(7n+20)\) or \(a=7, b=20\)A1 (3)(8 marks) A1: Either in expression or as above 
# Question 3:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=1}^{n}(r+1)(r+4) = \sum_{r=1}^{n} r^2+5r+4$ | B1 | Expands bracket correctly to $r^2+5r+4$ |
| $= \frac{n}{6}(n+1)(2n+1)+5\cdot\frac{n}{2}(n+1)+4n$ | M1 A1 | M1: Uses $\frac{n}{6}(n+1)(2n+1)$ or $\frac{n}{2}(n+1)$ correctly |
| $= \frac{n}{6}\{(n+1)(2n+1)+15(n+1)+24\}$ | dM1 | dM1: Attempts to remove factor $\frac{n}{6}$ or $\frac{n}{3}$ to obtain a quadratic factor; need not be 3 terms |
| $= \frac{n}{6}(2n^2+18n+40)$ or $\frac{n}{3}(n^2+9n+20)$ | | |
| $= \frac{n}{3}(n+4)(n+5)$ **given answer** | A1* (5) | Completely correct work including a step with a collected **3 term** quadratic prior to bracket, with correct printed answer |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=n+1}^{2n}(r+1)(r+4) = \frac{2n}{3}(2n+4)(2n+5)-\frac{n}{3}(n+4)(n+5)$ | M1 | Uses $f(2n)-f(n)$ or $f(2n)-f(n+1)$ correctly; require all 3 terms in $2n$ (and $n+1$ if used) |
| $= \frac{n}{3}\{8n^2+36n+40-n^2-9n-20\}$ | dM1 | dM1: Attempts to remove factor $\frac{n}{6}$ or $\frac{n}{3}$ to obtain quadratic factor |
| $= \frac{n}{3}\{7n^2+27n+20\} = \frac{n}{3}(n+1)(7n+20)$ or $a=7, b=20$ | A1 (3)(8 marks) | A1: Either in expression or as above |

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3. (a) Using the formulae for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$, show that

$$\sum _ { r = 1 } ^ { n } ( r + 1 ) ( r + 4 ) = \frac { n } { 3 } ( n + 4 ) ( n + 5 )$$

for all positive integers $n$.\\
(b) Hence show that

$$\sum _ { r = n + 1 } ^ { 2 n } ( r + 1 ) ( r + 4 ) = \frac { n } { 3 } ( n + 1 ) ( a n + b )$$

where $a$ and $b$ are integers to be found.\\

\hfill \mbox{\textit{Edexcel FP1 2015 Q3 [8]}}
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