CAIE FP1 2011 June — Question 1 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeAlternating series summation
DifficultyStandard +0.3 This question requires knowing the standard sum of squares formula and applying algebraic manipulation to find an alternating series. The first part is direct application of a formula (sum of even squares = 4 times sum of first n squares), and the second part requires recognizing that the alternating sum can be found by subtracting the even squares from all squares. While it involves multiple steps, both are standard techniques with no novel insight required, making it slightly easier than average for Further Maths.
Spec4.06a Summation formulae: sum of r, r^2, r^3
1 Find \(2 ^ { 2 } + 4 ^ { 2 } + \ldots + ( 2 n ) ^ { 2 }\). Hence find \(1 ^ { 2 } - 2 ^ { 2 } + 3 ^ { 2 } - 4 ^ { 2 } + \ldots - ( 2 n ) ^ { 2 }\), simplifying your answer.
Question 1:
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(2^2 + 4^2 + \ldots + (2n)^2 = \frac{4n(n+1)(2n+1)}{6}\)M1A1 Finds four times sum of first \(n\) squares
\(1^2 - 2^2 + 3^2 - 4^2 + \ldots - (2n)^2 = \frac{2n(2n+1)(4n+1)}{6} - \frac{8n(n+1)(2n+1)}{6}\)M1A1 Subtracts eight times sum of first \(n\) squares from sum of first \(2n\) squares
\(= \frac{n(2n+1)}{3}(4n+1-4n-4) = -n(2n+1)\)A1 Simplifies
Or: \(\frac{4n(n+1)(2n+1)}{6} - \frac{4n(n+1)}{2} + n - \frac{4n(n+1)(2n+1)}{6} = -2n^2 - n\)(M1A1)(A1) Alternative method 
# Question 1:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $2^2 + 4^2 + \ldots + (2n)^2 = \frac{4n(n+1)(2n+1)}{6}$ | M1A1 | Finds four times sum of first $n$ squares |
| $1^2 - 2^2 + 3^2 - 4^2 + \ldots - (2n)^2 = \frac{2n(2n+1)(4n+1)}{6} - \frac{8n(n+1)(2n+1)}{6}$ | M1A1 | Subtracts eight times sum of first $n$ squares from sum of first $2n$ squares |
| $= \frac{n(2n+1)}{3}(4n+1-4n-4) = -n(2n+1)$ | A1 | Simplifies |
| Or: $\frac{4n(n+1)(2n+1)}{6} - \frac{4n(n+1)}{2} + n - \frac{4n(n+1)(2n+1)}{6} = -2n^2 - n$ | (M1A1)(A1) | Alternative method |

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1 Find $2 ^ { 2 } + 4 ^ { 2 } + \ldots + ( 2 n ) ^ { 2 }$.

Hence find $1 ^ { 2 } - 2 ^ { 2 } + 3 ^ { 2 } - 4 ^ { 2 } + \ldots - ( 2 n ) ^ { 2 }$, simplifying your answer.

\hfill \mbox{\textit{CAIE FP1 2011 Q1 [5]}}
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Q1 5 Q2 5 Q3 6 Q4 8 Q5 8 Q6 8 Q7 11 Q8 11 Q9 11 Q10 13 Q11 EITHER Q11 OR