OCR Further Pure Core 2 2018 March — Question 3 3 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2018
SessionMarch
Marks3
TopicSequences and series, recurrence and convergence
TypeSum from n+1 to 2n or similar range
DifficultyStandard +0.3 This is a straightforward application of a given formula requiring the algebraic manipulation ∑(r=121 to 300) r² = ∑(r=1 to 300) r² - ∑(r=1 to 120) r². The formula is provided, so students only need to substitute values and perform arithmetic—no derivation or novel insight required. Slightly above average difficulty due to the manipulation step and arithmetic with large numbers, but still a routine textbook exercise.
Spec4.06a Summation formulae: sum of r, r^2, r^3
In this question you must show detailed reasoning. Use the formula \(\sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1)\) to evaluate \(121^2 + 122^2 + 123^2 + \ldots + 300^2\). [3]
AnswerMarks Guidance
\(\frac{1}{6} \times 300 \times (300 + 1)(2 \times 300 + 1)\)M1 Any correct numerical use of formula
\(\frac{1}{6} \times 120 \times (120 + 1)(2 \times 120 + 1)\)M1 Must be 120, not 121 (unless corrected by \(121^2\) later).
\((= 9045050 - 583220) = 8461830\)A1 [3] 
$\frac{1}{6} \times 300 \times (300 + 1)(2 \times 300 + 1)$ | M1 | Any correct numerical use of formula | Implied by 9045050 seen

$\frac{1}{6} \times 120 \times (120 + 1)(2 \times 120 + 1)$ | M1 | Must be 120, not 121 (unless corrected by $121^2$ later). | Implied by 583220 seen

$(= 9045050 - 583220) = 8461830$ | A1 [3]

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In this question you must show detailed reasoning.

Use the formula $\sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1)$ to evaluate $121^2 + 122^2 + 123^2 + \ldots + 300^2$. [3]

\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q3 [3]}}
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