SPS SPS FM 2020 June — Question 10 6 marks

Exam BoardSPS
ModuleSPS FM (SPS FM)
Year2020
SessionJune
Marks6
TopicProof by induction
TypeProve summation formula
DifficultyStandard +0.3 This is a straightforward proof by induction with a summation formula. The base case is trivial, and the inductive step requires standard algebraic manipulation to show the formula holds for n+1. The factorization is relatively clean and the question follows a standard template with no novel insights required, making it slightly easier than average.
Spec4.01a Mathematical induction: construct proofs

10. Prove by induction that for \(n \in \mathbb { Z } ^ { + }\) $$2 \times 4 + 4 \times 5 + 6 \times 6 + \ldots + 2 n ( n + 3 ) = \frac { 2 } { 3 } n ( n + 1 ) ( n + 5 )$$

10.

Prove by induction that for $n \in \mathbb { Z } ^ { + }$

$$2 \times 4 + 4 \times 5 + 6 \times 6 + \ldots + 2 n ( n + 3 ) = \frac { 2 } { 3 } n ( n + 1 ) ( n + 5 )$$

\hfill \mbox{\textit{SPS SPS FM 2020 Q10 [6]}}