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Date May 2022 Marks available 8 Reference code 22M.3.AHL.TZ1.1
Level Additional Higher Level Paper Paper 3 Time zone Time zone 1
Command term Prove Question number 1 Adapted from N/A

Question

This question asks you to explore some properties of polygonal numbers and to determine and prove interesting results involving these numbers.


A polygonal number is an integer which can be represented as a series of dots arranged in the shape of a regular polygon. Triangular numbers, square numbers and pentagonal numbers are examples of polygonal numbers.

For example, a triangular number is a number that can be arranged in the shape of an equilateral triangle. The first five triangular numbers are 1, 3, 6, 10 and 15.

The following table illustrates the first five triangular, square and pentagonal numbers respectively. In each case the first polygonal number is one represented by a single dot.

For an r-sided regular polygon, where r+, r3, the nth polygonal number Prn is given by

Prn=r-2n2-r-4n2, where n+.

Hence, for square numbers, P4n=4-2n2-4-4n2=n2.

The nth pentagonal number can be represented by the arithmetic series

P5n=1+4+7++3n-2.

For triangular numbers, verify that P3n=nn+12.

[2]
a.i.

The number 351 is a triangular number. Determine which one it is.

[2]
a.ii.

Show that P3n+P3n+1n+12.

[2]
b.i.

State, in words, what the identity given in part (b)(i) shows for two consecutive triangular numbers.

[1]
b.ii.

For n=4, sketch a diagram clearly showing your answer to part (b)(ii).

[1]
b.iii.

Show that 8P3n+1 is the square of an odd number for all n+.

[3]
c.

Hence show that P5n=n3n-12 for n+.

[3]
d.

By using a suitable table of values or otherwise, determine the smallest positive integer, greater than 1, that is both a triangular number and a pentagonal number.

[5]
e.

A polygonal number, Prn, can be represented by the series

Σm=1n1+m-1r-2 where r+, r3.

Use mathematical induction to prove that Prn=r-2n2-r-4n2 where n+.

[8]
f.

Markscheme

P3n=3-2n2-3-4n2  OR  P3n=n2--n2        A1

P3n=n2+n2        A1

 

Note: Award A0A1 if P3n=n2+n2 only is seen.

Do not award any marks for numerical verification.

 

so for triangular numbers, P3n=nn+12        AG

 

[2 marks]

a.i.

METHOD 1

uses a table of values to find a positive integer that satisfies P3n=351        (M1)

for example, a list showing at least 3 consecutive terms 325, 351, 378

 

Note: Award (M1) for use of a GDC’s numerical solve or graph feature.

 

n=26  (26th triangular number)        A1

 

Note: Award A0 for n=27,26. Award A0 if additional solutions besides n=26 are given.

 

METHOD 2

attempts to solve nn+12=351 n2+n-702=0 for n        (M1)

n=-1±12-41-7022  OR  n-26n+27=0

n=26  (26th triangular number)        A1

 

Note: Award A0 for n=27,26. Award A0 if additional solutions besides n=26 are given.

 

[2 marks]

a.ii.

attempts to form an expression for P3n+P3n+1 in terms of n        M1

 

EITHER

P3n+P3n+1nn+12+n+1n+22

n+12n+22 2n+1n+12         A1

 

OR

P3n+P3n+1n22+n2+n+122+n+12

n2+n2+n2+2n+1+n+12  n2+2n+1         A1

 

THEN

n+12         AG

 

[2 marks]

b.i.

the sum of the nth and n+1th triangular numbers

is the n+1th square number         A1

 

[1 mark]

b.ii.

      A1

 

Note: Accept equivalent single diagrams, such as the one above, where the 4th and 5th triangular numbers and the 5th square number are clearly shown.
Award A1 for a diagram that show P34 (a triangle with 10 dots) and P35 (a triangle with 15 dots) and P45 (a square with 25 dots).

 

[1 mark]

b.iii.

METHOD 1

8P3n+1=8nn+12+1 =4nn+1+1          A1

attempts to expand their expression for 8P3n+1          (M1)

=4n2+4n+1

=2n+12          A1

and 2n+1 is odd          AG

 

METHOD 2

8P3n+1=8n+12-P3n+1+1=8n+12-n+1n+22+1  A1

attempts to expand their expression for 8P3n+1          (M1)

8n2+2n+1-4n2+3n+2+1 =4n2+4n+1

=2n+12          A1

and 2n+1 is odd          AG

 

Method 3

8P3n+1=8nn+12+1 =An+B2 (where A,B+)          A1

attempts to expand their expression for 8P3n+1          (M1)

4n2+4n+1 =A2n2+2ABn+B2

now equates coefficients and obtains B=1 and A=2

=2n+12          A1

and 2n+1 is odd          AG

 

[3 marks]

c.

EITHER

u1=1 and d=3          (A1)

substitutes their u1 and their d into P5n=n22u1+n-1d          M1

P5n=n22+3n-1 =n22+3n-3          A1

 

OR

u1=1 and un=3n-2          (A1)

substitutes their u1 and their un into P5n=n2u1+un          M1

P5n=n21+3n-2          A1

 

OR

P5n=31-2+32-2+33-2+3n-2

P5n=31+32+33++3n-2n =31+2+3++n-2n        (A1)

substitutes nn+12 into their expression for P5n          M1

P5n=3nn+12-2n

P5n=n23n+1-4          A1

 

OR

attempts to find the arithmetic mean of n terms          (M1)

=1+3n-22          A1

multiplies the above expression by the number of terms n

P5n=n21+3n-2          A1

 

THEN

so P5n=n3n-12          AG

 

[3 marks]

d.

METHOD 1

forms a table of P3n values that includes some values for n>5         (M1)

forms a table of P5m values that includes some values for m>5         (M1)

 

Note: Award (M1) if at least one P3n value is correct. Award (M1) if at least one P5m value is correct. Accept as above for n2+n values and 3m2-m values.

 

n=20 for triangular numbers          (A1)

m=12 for pentagonal numbers          (A1)

 

Note: Award (A1) if n=20 is seen in or out of a table. Award (A1) if m=12 is seen in or out of a table. Condone the use of the same parameter for triangular numbers and pentagonal numbers, for example, n=20 for triangular numbers and n=12 for pentagonal numbers.

 

210 (is a triangular number and a pentagonal number)          A1

 

Note: Award all five marks for 210 seen anywhere with or without working shown.

 

METHOD 2

EITHER

attempts to express P3n=P5m as a quadratic in n         (M1)

n2+n+m-3m2=0 (or equivalent)

attempts to solve their quadratic in n         (M1)

n=-1±12m2-4m+12=-1±12-4m-3m22

 

OR

attempts to express P3n=P5m as a quadratic in m         (M1)

3m2-m-n2+n=0 (or equivalent)

attempts to solve their quadratic in m         (M1)

m=1±12n2-12n+16=1±-12+12n2+n6

 

THEN

n=20 for triangular numbers          (A1)

m=12 for pentagonal numbers          (A1)

210 (is a triangular number and a pentagonal number)          A1

 

METHOD 3

nn+12=m3m-12

let n=m+k n>m and so 3m2-m=m+km+k+1        M1

2m2-2k+1m-k2+k=0          A1

attempts to find the discriminant of their quadratic

and recognises that this must be a perfect square        M1

Δ=4k+12+8k2+k

N2=4k+12+8k2+k =4k+13k+1

determines that k=8 leading to 2m2-18m-72=0m=-3,12 and so m=12          A1

210 (is a triangular number and a pentagonal number)          A1

 

 

METHOD 4

nn+12=m3m-12

let m=n-k m<n and so n2+n=n-k3n-k-1       M1

2n2-23k+1n+3k2+k=0          A1

attempts to find the discriminant of their quadratic

and recognises that this must be a perfect square        M1

Δ=43k+12-83k2+k

N2=43k+12-83k2+k =4k+13k+1

determines that k=8 leading to 2n2-50n+200=0n=5,20 and so n=20          A1

210 (is a triangular number and a pentagonal number)          A1

 

[5 marks]

e.

Note: Award a maximum of R1M0M0A1M1A1A1R0 for a ‘correct’ proof using n and n+1.

 

consider n=1: Pr1=1+1-1r-2=1 and Pr1=r-212-r-412=1

so true for n=1              R1 

 

Note: Accept Pr1=1 and Pr1=r-212-r-412=1.
Do not accept one-sided considerations such as 'Pr1=1 and so true for n=1'.
Subsequent marks after this R1 are independent of this mark can be awarded.

 

Assume true for n=k, ie. Prk=r-2k2-r-4k2          M1

 

Note: Award M0 for statements such as “let n=k ”. The assumption of truth must be clear.
Subsequent marks after this M1 are independent of this mark and can be awarded.

 

Consider n=k+1:

(Prk+1 can be represented by the sum

Σm=1k+11+m-1r-2=Σm=1k1+m-1r-2+1+kr-2 and so

Prk+1=r-2k2-r-4k2+1+kr-2  Prk+1=Prk+1+kr-2         M1

=r-2k2-r-4k+2+2kr-22         A1

 =r-2k2+2k-r-4k+22

=r-2k2+2k+1-r-2-r-4k+22         M1

=r-2k+12-r-4k-r-42          (A1)

 =r-2k+12-r-4k+12         A1

hence true for n=1 and n=k true n=k+1 true         R1

therefore true for all n+

 

Note: Only award the final R1 if the first five marks have been awarded. Award marks as appropriate for solutions that expand both the LHS and (given) RHS of the equation.

 

[8 marks]

f.

Examiners report

Part (a) (i) was generally well done. Unfortunately, some candidates adopted numerical verification. Part (a) (ii) was generally well done with the majority of successful candidates using their GDC judiciously and disregarding = −27 as a possible solution. A few candidates interpreted the question as needing to deal with P3(351).

Although part (b) (i) was generally well done, a significant number of candidates laboured unnecessarily to show the required result. Many candidates set their LHS to equal the RHS throughout the solution. Part (b) (ii) was generally not well done with many candidates unable to articulate clearly in words and symbols what the given identity shows for the sum of two consecutive triangular numbers. In part (b) (iii), most candidates were unable to produce a clear diagram illustrating the identity stated in part (b) (i). 

Part (c) was reasonably well done. Most candidates were able to show algebraically that 8P3(n)+1=4n2+4n+1. A good number of candidates were then able to express 4n2+4n+1 as (2n+1)2 and conclude that (2n+1) is odd. Rather than making the connection that 4n2+4n+1 is a perfect square, many candidates attempted instead to analyse the parity of either 4n(n+1)+1 or 4n2+4n+1. As with part (b) (i), many candidates set their LHS to equal the RHS throughout the solution. A number of candidates unfortunately adopted numerical verification.

Part (d) was not answered as well as anticipated with many candidates not understanding what was
required. Instead of using the given arithmetic series to show that P5(n)=n(3n-1)2, a large number of
candidates used P5(n)=(5-2)n2-(5-4)n2 . Unfortunately, a number of candidates adopted numerical verification.

In part (e), the overwhelming majority of candidates who successfully determined that 210 is the smallest positive integer greater than 1 that is both triangular and pentagonal used a table of values. Unfortunately, a large proportion of these candidates seemingly spent quite a few minutes listing the first 20 triangular numbers and the first 12 pentagonal numbers. And it can be surmised that a number of these candidates constructed their table of values either without the use of a GDC or with the arithmetic functionality of a GDC rather than with a GDC's table of values facility. Candidates should be aware that a relevant excerpt from a table of values is sufficient evidence of correct working. A number of candidates started constructing a table of values but stopped before identifying 210. Disappointingly, a significant number of candidates attempted to solve P3(n)=P5(n) for n.

Part (f) proved beyond the reach of most with only a small number of candidates successfully proving the given result. A significant number of candidates were unable to show that the result is true for n=1. A number of candidates established the validity of the base case for the RHS only while a number of other candidates attempted to prove the base case for r = 3. A large number of candidates did not state the inductive step correctly with the assumption of truth not clear. A number of candidates then either attempted to work backwards from the given result or misinterpreted the question and attempted to prove the result stated in the question stem rather than the result stated in the question. Some candidates who were awarded the first answer mark when considering the n=k+1 case were unable to complete the square or equivalent simplification correctly. Disappointingly, a significant number listed the steps involved in an induction proof without engaging in the actual proof.

a.i.
[N/A]
a.ii.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
b.iii.
[N/A]
c.
[N/A]
d.
[N/A]
e.
[N/A]
f.

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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