Prove by mathematical induction \(\sum\limits_{r = 1}^n {r(r!) = (n + 1)! - 1} \), \(n \in {\mathbb{Z}^ + }\).

let \(n = 1\)

LHS \( = 1 \times 1! = 1\)

RHS \( = (1 + 1)! - 1 = 2 - 1 = 1\)

hence true for \(n = 1\)     R1

assume true for \(n = k\)

\(\sum\limits_{r = 1}^k {r(r!) = (k + 1)! - 1} \)     M1

\(\sum\limits_{r = 1}^{k + 1} {r(r!) = (k + 1)! - 1}  + (k + 1) \times (k + 1)!\)     M1A1

\( = (k + 1)!(1 + k + 1) - 1\)

\( = (k + 1)!(k + 2) - 1\)     A1

\( = (k + 2)! - 1\)     A1

hence if true for \(n = k\), true for \(n = k + 1\)     R1

since the result is true for \(n = 1\) and \({\text{P}}(k) \Rightarrow {\text{P}}(k + 1)\) the result is proved by mathematical induction \(\forall n \in {\mathbb{Z}^ + }\)     R1

[8 marks]

This question was done poorly on a number of levels. Many students knew the structure of induction but did not show that they understood what they were doing. The general notation was poor for both the induction itself and the sigma notation.

In noting the case for \(n = 1\) too many stated the equation rather than using the LHS and RHS separately and concluding with a statement. There were also too many who did not state the conclusion for this case.

Many did not state the assumption for \(n = k\) as an assumption.

Most stated the equation for \(n = k + 1\) and worked with the equation. Also common was the lack of sigma and inappropriate use of n and k in the statement. There were some very nice solutions however.

The final conclusion was often not complete or not considered which would lead to the conclusion that the student did not really understand what induction is about.