Date | November 2018 | Marks available | 6 | Reference code | 18N.1.AHL.TZ0.H_6 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Question number | H_6 | Adapted from | N/A |
Question
Use mathematical induction to prove that n∑r=1r(r!)=(n+1)!−1, for n∈Z+.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
consider n=1. 1(1!)=1 and 2!−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 k∑r=1r(r!)=(k+1)!−1) M1
Note: Assumption of truth must be present.
consider n=k+1
k+1∑r=1r(r!)=k∑r=1r(r!)+(k+1)(k+1)! (M1)
= (k+1)!−1+(k+1)(k+1)! A1
= (k+2)(k+1)!−1 M1
Note: M1 is for factorising (k+1)!
= (k+2)!−1
=((k+1)+1)!−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 (∈Z+) 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]