Edexcel FP1 2013 June — Question 5 10 marks

Exam BoardEdexcel
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeSum from n+1 to 2n or similar range
DifficultyStandard +0.3 This is a straightforward Further Maths question requiring expansion of the product, application of standard summation formulas, and algebraic manipulation. Part (b) uses the standard technique of subtracting sums to handle the shifted range. While it requires careful algebra, it follows a well-established method with no novel insight needed, making it slightly easier than average even for FP1.
Spec4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series
5. (a) Use the standard results for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) to show that $$\sum _ { r = 1 } ^ { n } ( r + 2 ) ( r + 3 ) = \frac { 1 } { 3 } n \left( n ^ { 2 } + 9 n + 26 \right)$$ for all positive integers \(n\).
(b) Hence show that $$\sum _ { r = n + 1 } ^ { 3 n } ( r + 2 ) ( r + 3 ) = \frac { 2 } { 3 } n \left( a n ^ { 2 } + b n + c \right)$$ where \(a\), \(b\) and \(c\) are integers to be found.
Question 5:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\((r+2)(r+3) = r^2 + 5r + 6\)B1
\(\sum(r^2+5r+6) = \tfrac{1}{6}n(n+1)(2n+1) + 5\times\tfrac{1}{2}n(n+1) + 6n\)M1, B1ft M1: use of correct expressions for \(\sum r^2\) and \(\sum r\); B1ft: \(\sum k = nk\)
\(= \tfrac{1}{3}n\!\left[\tfrac{1}{2}(n+1)(2n+1)+\tfrac{15}{2}(n+1)+18\right]\)M1 A1 M1: factors out \(n\) ignoring treatment of constant; A1: correct expression with \(\tfrac{1}{3}n\) or \(\tfrac{1}{6}n\) factored out
\(= \tfrac{1}{3}n\!\left[n^2+9n+26\right]\) *A1* cso Correct completion to printed answer
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\displaystyle\sum_{r=n+1}^{3n} = \tfrac{1}{3}(3n)\!\left((3n)^2+9(3n)+26\right) - \tfrac{1}{3}n\!\left(n^2+9n+26\right)\)M1 A1 M1: \(f(3n) - f(n\) or \(n+1)\) and attempt to use part (a); A1: equivalent correct expression
\(= \tfrac{2}{3}n\!\left(\tfrac{27}{2}n^2+\tfrac{81}{2}n+39-\tfrac{1}{2}n^2-\tfrac{9}{2}n-13\right)\)dM1 Factors out \(\tfrac{2}{3}n\); dependent on previous M1
\(= \tfrac{2}{3}n\!\left(13n^2+36n+26\right)\)A1 Accept correct expression; \(a=13,\ b=36,\ c=26\) 
## Question 5:

### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(r+2)(r+3) = r^2 + 5r + 6$ | B1 | |
| $\sum(r^2+5r+6) = \tfrac{1}{6}n(n+1)(2n+1) + 5\times\tfrac{1}{2}n(n+1) + 6n$ | M1, B1ft | M1: use of correct expressions for $\sum r^2$ **and** $\sum r$; B1ft: $\sum k = nk$ |
| $= \tfrac{1}{3}n\!\left[\tfrac{1}{2}(n+1)(2n+1)+\tfrac{15}{2}(n+1)+18\right]$ | M1 A1 | M1: factors out $n$ ignoring treatment of constant; A1: correct expression with $\tfrac{1}{3}n$ or $\tfrac{1}{6}n$ factored out |
| $= \tfrac{1}{3}n\!\left[n^2+9n+26\right]$ * | A1* cso | Correct completion to printed answer |

### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\displaystyle\sum_{r=n+1}^{3n} = \tfrac{1}{3}(3n)\!\left((3n)^2+9(3n)+26\right) - \tfrac{1}{3}n\!\left(n^2+9n+26\right)$ | M1 A1 | M1: $f(3n) - f(n$ or $n+1)$ and attempt to use part (a); A1: equivalent correct expression |
| $= \tfrac{2}{3}n\!\left(\tfrac{27}{2}n^2+\tfrac{81}{2}n+39-\tfrac{1}{2}n^2-\tfrac{9}{2}n-13\right)$ | dM1 | Factors out $\tfrac{2}{3}n$; dependent on previous M1 |
| $= \tfrac{2}{3}n\!\left(13n^2+36n+26\right)$ | A1 | Accept correct expression; $a=13,\ b=36,\ c=26$ |

---
5. (a) Use the standard results for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ to show that

$$\sum _ { r = 1 } ^ { n } ( r + 2 ) ( r + 3 ) = \frac { 1 } { 3 } n \left( n ^ { 2 } + 9 n + 26 \right)$$

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

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

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

\hfill \mbox{\textit{Edexcel FP1 2013 Q5 [10]}}
This paper (9 questions)
View full paper
Q1 4 Q2 5 Q3 7 Q4 9 Q5 10 Q6 11 Q7 11 Q8 8 Q9 10