SPS SPS FM 2022 October — Question 7 5 marks

Exam BoardSPS
ModuleSPS FM (SPS FM)
Year2022
SessionOctober
Marks5
TopicProof by induction
TypeProve recurrence relation formula
DifficultyStandard +0.3 This is a straightforward proof by induction with a given formula to prove. The inductive step requires substituting the recurrence relation and expanding n³ - n² + 1 + n(3n+1) = (n+1)³ - (n+1)² + 1, which involves routine algebraic manipulation of cubic expressions. While it requires careful algebra, it follows the standard induction template with no conceptual surprises or novel insights needed.
Spec1.01a Proof: structure of mathematical proof and logical steps

7. A sequence of positive integers is defined by $$\begin{aligned} u _ { 1 } & = 1 \\ u _ { n + 1 } & = u _ { n } + n ( 3 n + 1 ) , \quad n \in \mathbb { Z } ^ { + } \end{aligned}$$ Prove by induction that $$u _ { n } = n ^ { 2 } ( n - 1 ) + 1 , \quad n \in \mathbb { Z } ^ { + }$$

7. A sequence of positive integers is defined by

$$\begin{aligned}
u _ { 1 } & = 1 \\
u _ { n + 1 } & = u _ { n } + n ( 3 n + 1 ) , \quad n \in \mathbb { Z } ^ { + }
\end{aligned}$$

Prove by induction that

$$u _ { n } = n ^ { 2 } ( n - 1 ) + 1 , \quad n \in \mathbb { Z } ^ { + }$$

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\hfill \mbox{\textit{SPS SPS FM 2022 Q7 [5]}}