18N.1.AHL.TZ0.H_6

pestleMathematicsAAHLPaper 118N· ahl-1-15-proof-by-induction-contradiction-counterexamplessource ↗

Use mathematical induction to prove that $\sum\limits_{r = 1}^n {r\left( {r{\text{!}}} \right)} = \left( {n + 1} \right){\text{!}} - 1$ , for $n \in {\mathbb{Z}^ + }$ .

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

consider $n = 1$ . $1\left( {1{\text{!}}} \right) = 1$ and $2{\text{!}} - 1 = 1$ therefore true for $n = 1$ R1

Note: There must be evidence that $n = 1$ has been substituted into both expressions, or an expression such LHS=RHS=1 is used. “therefore true for $n = 1$ ” or an equivalent statement must be seen.

assume true for $n = k$ , (so that $\sum\limits_{r = 1}^k {r\left( {r{\text{!}}} \right)} = \left( {k + 1} \right){\text{!}} - 1$ ) M1

Note: Assumption of truth must be present.

consider $n = k + 1$

$\sum\limits_{r = 1}^{k + 1} {r\left( {r{\text{!}}} \right)} = \sum\limits_{r = 1}^k {r\left( {r{\text{!}}} \right)} + \left( {k + 1} \right)\left( {k + 1} \right){\text{!}}$ (M1)

${\text{ = }}\left( {k + 1} \right){\text{!}} - 1 + \left( {k + 1} \right)\left( {k + 1} \right){\text{!}}$ A1

${\text{ = }}\left( {k + 2} \right)\left( {k + 1} \right){\text{!}} - 1$ M1

Note: M1 is for factorising $\left( {k + 1} \right){\text{!}}$

${\text{ = }}\left( {k + 2} \right){\text{!}} - 1$

$= \left( {\left( {k + 1} \right) + 1} \right){\text{!}} - 1$

so if true for $n = k$ , then also true for $n = k + 1$ , and as true for $n = 1$ then true for all $n$ $\left( { \in {\mathbb{Z}^ + }} \right)$ R1

Note: Only award final R1 if all three method marks have been awarded.
Award R0 if the proof is developed from both LHS and RHS.

[6 marks]

Examiners’ report

[N/A]