OCR FP1 2006 January — Question 2 5 marks

Exam BoardOCR
ModuleFP1 (Further Pure Mathematics 1)
Year2006
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve summation formula
DifficultyStandard +0.3 This is a standard textbook proof by induction of a well-known summation formula. The algebraic manipulation in the inductive step is straightforward (factoring a cubic), and this exact question appears in most Further Pure textbooks as a foundational example. While it requires understanding of induction structure, it demands no novel insight and is slightly easier than average due to its routine nature.
Spec4.01a Mathematical induction: construct proofs

2 Prove by induction that, for \(n \geqslant 1 , \sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )\).

AnswerMarks Guidance
\(1^2 = \frac{1}{6} \times 1 \times 2 \times 3\)B1 Show result true for \(n = 1\) or \(2\)
\(\frac{1}{6}n(n+1)(2n+1) + (n+1)^2\)M1 Add next term to given sum formula, any letter OK
\(\frac{1}{6}(n+1)(n+2)(2(n+1)+1)\)DM1 A1 Attempt to factorise or expand and simplify; Correct expression obtained
(Specific statement of induction conclusion, with no errors seen)A1 5 marks total
$1^2 = \frac{1}{6} \times 1 \times 2 \times 3$ | B1 | Show result true for $n = 1$ or $2$

$\frac{1}{6}n(n+1)(2n+1) + (n+1)^2$ | M1 | Add next term to given sum formula, any letter OK

$\frac{1}{6}(n+1)(n+2)(2(n+1)+1)$ | DM1 A1 | Attempt to factorise or expand and simplify; Correct expression obtained

(Specific statement of induction conclusion, with no errors seen) | A1 | 5 marks total
2 Prove by induction that, for $n \geqslant 1 , \sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$.

\hfill \mbox{\textit{OCR FP1 2006 Q2 [5]}}