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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 nr=1r(r!)=(n+1)!1, for nZ+.

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 kr=1r(r!)=(k+1)!1)       M1

Note: Assumption of truth must be present.

 

consider n=k+1

k+1r=1r(r!)=kr=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]

Examiners report

[N/A]

Syllabus sections

Topic 1—Number and algebra » AHL 1.15—Proof by induction, contradiction, counterexamples
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Topic 1—Number and algebra

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