Question 2 - A-Level Maths

Edexcel FP1 2009 January — Question 2 7 marks

Exam Board	Edexcel
Module	FP1 (Further Pure Mathematics 1)
Year	2009
Session	January
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Sequences and Series
Type	Sums Between Limits
Difficulty	Moderate -0.3 This is a straightforward application of standard summation formulae. Part (a) requires substituting known formulae for Σr and Σr² then simplifying algebraically—routine bookwork with no problem-solving. Part (b) uses the difference of two sums, a standard technique. While it requires careful algebra, it involves no novel insight and is typical FP1 fare, making it slightly easier than an average A-level question overall.
Spec	1.04g Sigma notation: for sums of series 4.06a Summation formulae: sum of r, r^2, r^3

2. (a) Show, using the formulae for $\sum r$ and $\sum r ^ { 2 }$, that $$\sum _ { r = 1 } ^ { n } \left( 6 r ^ { 2 } + 4 r - 1 \right) = n ( n + 2 ) ( 2 n + 1 )$$ (b) Hence, or otherwise, find the value of $\sum _ { r = 11 } ^ { 20 } \left( 6 r ^ { 2 } + 4 r - 1 \right)$.

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Question 2:

Part (a)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$6\sum r^2 + 4\sum r - \sum 1 = 6\cdot\frac{n}{6}(n+1)(2n+1) + 4\cdot\frac{n}{2}(n+1) - n$	M1 A1, B1	First M1 for first 2 terms correct; B1 for $-n$
$= \frac{n}{6}(12n^2+18n+6+12n+12-6)$ or $n(n+1)(2n+1)+(2n+1)n$	M1	Attempt to expand and gather terms
$= \frac{n}{6}(12n^2+30n+12) = n(2n^2+5n+2) = n(n+2)(2n+1)$	A1	Final A1 for correct solution only

Part (b)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\sum_{r=1}^{20}(6r^2+4r-1) - \sum_{r=1}^{10}(6r^2+4r-1) = 20\times22\times41 - 10\times12\times21$	M1	Require ($r$ from 1 to 20) subtract ($r$ from 1 to 10) and attempt to substitute
$= 15520$	A1

## Question 2:

### Part (a)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $6\sum r^2 + 4\sum r - \sum 1 = 6\cdot\frac{n}{6}(n+1)(2n+1) + 4\cdot\frac{n}{2}(n+1) - n$ | M1 A1, B1 | First M1 for first 2 terms correct; B1 for $-n$ |
| $= \frac{n}{6}(12n^2+18n+6+12n+12-6)$ or $n(n+1)(2n+1)+(2n+1)n$ | M1 | Attempt to expand and gather terms |
| $= \frac{n}{6}(12n^2+30n+12) = n(2n^2+5n+2) = n(n+2)(2n+1)$ | A1 | Final A1 for correct solution only |

### Part (b)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=1}^{20}(6r^2+4r-1) - \sum_{r=1}^{10}(6r^2+4r-1) = 20\times22\times41 - 10\times12\times21$ | M1 | Require ($r$ from 1 to 20) subtract ($r$ from 1 to 10) and attempt to substitute |
| $= 15520$ | A1 | |

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2. (a) Show, using the formulae for $\sum r$ and $\sum r ^ { 2 }$, that

$$\sum _ { r = 1 } ^ { n } \left( 6 r ^ { 2 } + 4 r - 1 \right) = n ( n + 2 ) ( 2 n + 1 )$$

(b) Hence, or otherwise, find the value of $\sum _ { r = 11 } ^ { 20 } \left( 6 r ^ { 2 } + 4 r - 1 \right)$.\\

\hfill \mbox{\textit{Edexcel FP1 2009 Q2 [7]}}

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