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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  r = 1 n r ( 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  r = 1 k r ( r ! ) = ( k + 1 ) ! 1 )       M1

Note: Assumption of truth must be present.

 

consider  n = k + 1

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