AQA FP2 2007 June — Question 1 7 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2007
SessionJune
Marks7
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 method of differences question with clear guidance. Part (a) is routine algebraic verification requiring expansion of (r+1)³ terms. Part (b) applies the standard telescoping technique with given endpoints. While it requires careful bookkeeping with the sum limits (50 to 99), the method is explicitly stated and the algebraic manipulation is mechanical for Further Maths students.
Spec4.06b Method of differences: telescoping series

1
  1. Given that \(\mathrm { f } ( r ) = ( r - 1 ) r ^ { 2 }\), show that $$\mathrm { f } ( r + 1 ) - \mathrm { f } ( r ) = r ( 3 r + 1 )$$
  2. Use the method of differences to find the value of $$\sum _ { r = 50 } ^ { 99 } r ( 3 r + 1 )$$ (4 marks)

Question 1:
Part (a)
AnswerMarks Guidance
WorkingMarks Guidance
\(f(r+1) - f(r) = r(r+1)^2 - (r-1)r^2\)M1
\(= r(r^2 + 2r + 1 - r^2 + r)\)A1 any expanded form
\(= r(3r+1)\)A1 AG
Part (b)
AnswerMarks Guidance
WorkingMarks Guidance
\(r=50\): \(f(51)-f(50)\); \(r=51\): \(f(52)-f(51)\); \(r=99\): \(f(100)-f(99)\)M1A1 PI, clearly shown. Accept \(\sum_1^{99} - \sum_1^{49}\)
\(\sum_{r=50}^{99} r(3r+1) = f(100) - f(50)\)m1 clear cancellation
\(= 867500\)A1F cao
## Question 1:

### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $f(r+1) - f(r) = r(r+1)^2 - (r-1)r^2$ | M1 | |
| $= r(r^2 + 2r + 1 - r^2 + r)$ | A1 | any expanded form |
| $= r(3r+1)$ | A1 | AG |

### Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $r=50$: $f(51)-f(50)$; $r=51$: $f(52)-f(51)$; $r=99$: $f(100)-f(99)$ | M1A1 | PI, clearly shown. Accept $\sum_1^{99} - \sum_1^{49}$ |
| $\sum_{r=50}^{99} r(3r+1) = f(100) - f(50)$ | m1 | clear cancellation |
| $= 867500$ | A1F | cao |

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1
\begin{enumerate}[label=(\alph*)]
\item Given that $\mathrm { f } ( r ) = ( r - 1 ) r ^ { 2 }$, show that

$$\mathrm { f } ( r + 1 ) - \mathrm { f } ( r ) = r ( 3 r + 1 )$$
\item Use the method of differences to find the value of

$$\sum _ { r = 50 } ^ { 99 } r ( 3 r + 1 )$$

(4 marks)
\end{enumerate}

\hfill \mbox{\textit{AQA FP2 2007 Q1 [7]}}