CAIE FP1 2008 June — Question 7 8 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2008
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve summation formula
DifficultyStandard +0.8 This is a two-part Further Maths induction question requiring a standard proof followed by algebraic manipulation using given formulae to derive a new result. While the induction itself is routine, the second part requires careful algebraic rearrangement and factorization to identify the quadratic Q(n), making it moderately challenging but still within typical FP1 scope.
Spec4.01a Mathematical induction: construct proofs4.06a Summation formulae: sum of r, r^2, r^3

7 Prove by induction that $$\sum _ { r = 1 } ^ { n } \left( 3 r ^ { 5 } + r ^ { 3 } \right) = \frac { 1 } { 2 } n ^ { 3 } ( n + 1 ) ^ { 3 }$$ for all \(n \geqslant 1\). Use this result together with the List of Formulae (MF10) to prove that $$\sum _ { r = 1 } ^ { n } r ^ { 5 } = \frac { 1 } { 12 } n ^ { 2 } ( n + 1 ) ^ { 2 } \mathrm { Q } ( n )$$ where \(\mathrm { Q } ( n )\) is a quadratic function of \(n\) which is to be determined.

AnswerMarks
Verifies \(H_1\) to be trueB1
\(H_k: \sum_{r=1}^{k}(3r^5 + r^3) = (1/2)k^3(k + 1)^3\)B1
\(H_k \Rightarrow \sum_{r=1}^{k+1}(3r^5 + r^3) = (1/2)k^3(k + 1)^3 + 3(k + 1)^5 + (k + 1)^3\)M1
\(= ... = (1/2)(k + 1)^2(k + 2)^3\)A1
Thus \(H_k \Rightarrow H_{k+1}\) and concludesA1
\(3\sum_{r=1}^{n}r^5 + (1/4)n^2(n + 1)^2 = (1/2)n^3(n + 1)^3\)M1
\(\Rightarrow ... \Rightarrow \sum_{r=1}^{n}r^5 = (1/12)n^2(n + 1)^2(2n^2 + 2n - 1)\)M1A1
Verifies $H_1$ to be true | B1 |
$H_k: \sum_{r=1}^{k}(3r^5 + r^3) = (1/2)k^3(k + 1)^3$ | B1 |
$H_k \Rightarrow \sum_{r=1}^{k+1}(3r^5 + r^3) = (1/2)k^3(k + 1)^3 + 3(k + 1)^5 + (k + 1)^3$ | M1 |
$= ... = (1/2)(k + 1)^2(k + 2)^3$ | A1 |
Thus $H_k \Rightarrow H_{k+1}$ and concludes | A1 |
$3\sum_{r=1}^{n}r^5 + (1/4)n^2(n + 1)^2 = (1/2)n^3(n + 1)^3$ | M1 |
$\Rightarrow ... \Rightarrow \sum_{r=1}^{n}r^5 = (1/12)n^2(n + 1)^2(2n^2 + 2n - 1)$ | M1A1 |
7 Prove by induction that

$$\sum _ { r = 1 } ^ { n } \left( 3 r ^ { 5 } + r ^ { 3 } \right) = \frac { 1 } { 2 } n ^ { 3 } ( n + 1 ) ^ { 3 }$$

for all $n \geqslant 1$.

Use this result together with the List of Formulae (MF10) to prove that

$$\sum _ { r = 1 } ^ { n } r ^ { 5 } = \frac { 1 } { 12 } n ^ { 2 } ( n + 1 ) ^ { 2 } \mathrm { Q } ( n )$$

where $\mathrm { Q } ( n )$ is a quadratic function of $n$ which is to be determined.

\hfill \mbox{\textit{CAIE FP1 2008 Q7 [8]}}