For triangular numbers, verify that $P_3\left(n\right)=\frac{n\left(n+1\right)}{2}$.

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

Show that $P_3\left(n\right)+P_3\left(n+1\right)\equiv {\left(n+1\right)}^2$.

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

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

Show that $8P_3\left(n\right)+1$ is the square of an odd number for all $n\in {\mathrm{\mathbb{Z}}}^{+}$.

Hence show that $P_5\left(n\right)=\frac{n\left(3n-1\right)}{2}$ for $n\in {\mathrm{\mathbb{Z}}}^{+}$.

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.

A polygonal number, $P_r\left(n\right)$, can be represented by the series

$\underset{m=1}{\overset{n}{\text{Σ}}}\left(1+\left(m-1\right)\left(r-2\right)\right)$ where $r\in {\mathrm{\mathbb{Z}}}^{+}, r\ge 3$.

Use mathematical induction to prove that $P_r\left(n\right)=\frac{\left(r-2\right)n^2-\left(r-4\right)n}{2}$ where $n\in {\mathrm{\mathbb{Z}}}^{+}$.

$P_3\left(n\right)=\frac{\left(3-2\right)n^2-\left(3-4\right)n}{2}$  OR  $P_3\left(n\right)=\frac{n^2-\left(-n\right)}{2}$        A1

$P_3\left(n\right)=\frac{n^2+n}{2}$        A1

Note: Award A0A1 if $P_3\left(n\right)=\frac{n^2+n}{2}$ only is seen.

Do not award any marks for numerical verification.

so for triangular numbers, $P_3\left(n\right)=\frac{n\left(n+1\right)}{2}$        AG

[2 marks]

METHOD 1

uses a table of values to find a positive integer that satisfies $P_3\left(n\right)=351$        (M1)

for example, a list showing at least $3$ consecutive terms $\left(\dots 325, 351, 378\dots \right)$

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

$n=26$  ($26$th 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 $\frac{n\left(n+1\right)}{2}=351 \left(n^2+n-702=0\right)$ for $n$        (M1)

$n=\frac{-1\pm \sqrt{1^2-4\left(1\right)\left(-702\right)}}{2}$  OR  $\left(n-26\right)\left(n+27\right)=0$

$n=26$  ($26$th triangular number)        A1

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

[2 marks]

attempts to form an expression for $P_3\left(n\right)+P_3\left(n+1\right)$ in terms of $n$        M1

EITHER

$P_3\left(n\right)+P_3\left(n+1\right)\equiv \frac{n\left(n+1\right)}{2}+\frac{\left(n+1\right)\left(n+2\right)}{2}$

$\equiv \frac{\left(n+1\right)\left(2n+2\right)}{2} \left(\equiv \frac{2\left(n+1\right)\left(n+1\right)}{2}\right)$         A1

OR

$P_3\left(n\right)+P_3\left(n+1\right)\equiv \left(\frac{n^2}{2}+\frac{n}{2}\right)+\left(\frac{{\left(n+1\right)}^2}{2}+\frac{n+1}{2}\right)$

$\equiv \left(\frac{n^2+n}{2}\right)+\left(\frac{n^2+2n+1+n+1}{2}\right) \left(\equiv n^2+2n+1\right)$         A1

THEN

$\equiv {\left(n+1\right)}^2$         AG

[2 marks]

the sum of the $n$th and $\left(n+1\right)$th triangular numbers

is the $\left(n+1\right)$th square number         A1

[1 mark]

      A1

Note: Accept equivalent single diagrams, such as the one above, where the $4$th and $5$th triangular numbers and the $5$th square number are clearly shown.
Award A1 for a diagram that show $P_3\left(4\right)$ (a triangle with $10$ dots) and $P_3\left(5\right)$ (a triangle with $15$ dots) and $P_4\left(5\right)$ (a square with $25$dots).

[1 mark]

METHOD 1

$8P_3\left(n\right)+1=8\left(\frac{n\left(n+1\right)}{2}\right)+1 \left(=4n\left(n+1\right)+1\right)$          A1

attempts to expand their expression for $8P_3\left(n\right)+1$          (M1)

$=4n^2+4n+1$

$={\left(2n+1\right)}^2$          A1

and $2n+1$ is odd          AG

METHOD 2

$8P_3\left(n\right)+1=8\left({\left(n+1\right)}^2-P_3\left(n+1\right)\right)+1\left(=8\left({\left(n+1\right)}^2-\frac{\left(n+1\right)\left(n+2\right)}{2}\right)+1\right)$  A1

attempts to expand their expression for $8P_3\left(n\right)+1$          (M1)

$8\left(n^2+2n+1\right)-4\left(n^2+3n+2\right)+1 \left(=4n^2+4n+1\right)$

$={\left(2n+1\right)}^2$          A1

and $2n+1$ is odd          AG

Method 3

$8P_3\left(n\right)+1=8\left(\frac{n\left(n+1\right)}{2}\right)+1 \left(={\left(An+B\right)}^2\right)$ (where $A,B\in {\mathrm{\mathbb{Z}}}^{+}$)          A1

attempts to expand their expression for $8P_3\left(n\right)+1$          (M1)

$4n^2+4n+1 \left(=A^2{n}^2+2ABn+B^2\right)$

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

$={\left(2n+1\right)}^2$          A1

and $2n+1$ is odd          AG

[3 marks]

EITHER

$u_1=1$ and $d=3$          (A1)

substitutes their $u_1$ and their $d$ into $P_5\left(n\right)=\frac{n}{2}\left(2u_1+\left(n-1\right)d\right)$          M1

$P_5\left(n\right)=\frac{n}{2}\left(2+3\left(n-1\right)\right) \left(=\frac{n}{2}\left(2+3n-3\right)\right)$          A1

OR

$u_1=1$ and $u_n=3n-2$          (A1)

substitutes their $u_1$ and their $u_n$ into $P_5\left(n\right)=\frac{n}{2}\left(u_1+u_n\right)$          M1

$P_5\left(n\right)=\frac{n}{2}\left(1+3n-2\right)$          A1

OR

$P_5\left(n\right)=\left(3\left(1\right)-2\right)+\left(3\left(2\right)-2\right)+\left(3\left(3\right)-2\right)+\dots 3n-2$

$P_5\left(n\right)=\left(3\left(1\right)+3\left(2\right)+3\left(3\right)+\dots +3n\right)-2n \left(=3\left(1+2+3+\dots +n\right)-2n\right)$        (A1)

substitutes $\frac{n\left(n+1\right)}{2}$ into their expression for $P_5\left(n\right)$          M1

$P_5\left(n\right)=3\left(\frac{n\left(n+1\right)}{2}\right)-2n$

$P_5\left(n\right)=\frac{n}{2}\left(3\left(n+1\right)-4\right)$          A1

OR

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

$=\frac{1+\left(3n-2\right)}{2}$          A1

multiplies the above expression by the number of terms $n$

$P_5\left(n\right)=\frac{n}{2}\left(1+3n-2\right)$          A1

THEN

so $P_5\left(n\right)=\frac{n\left(3n-1\right)}{2}$          AG

[3 marks]

METHOD 1

forms a table of $P_3\left(n\right)$ values that includes some values for $n>5$         (M1)

forms a table of $P_5\left(m\right)$ values that includes some values for $m>5$         (M1)

Note: Award (M1) if at least one $P_3\left(n\right)$ value is correct. Award (M1) if at least one $P_5\left(m\right)$ value is correct. Accept as above for $\left(n^2+n\right)$ values and $\left(3m^2-m\right)$ 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 $P_3\left(n\right)=P_5\left(m\right)$ as a quadratic in $n$         (M1)

$n^2+n+\left(m-3m^2\right)\left(=0\right)$ (or equivalent)

attempts to solve their quadratic in $n$         (M1)

$n=\frac{-1\pm \sqrt{12m^2-4m+1}}{2}\left(=\frac{-1\pm \sqrt{1^2-4\left(m-3m^2\right)}}{2}\right)$

OR

attempts to express $P_3\left(n\right)=P_5\left(m\right)$ as a quadratic in $m$         (M1)

$3m^2-m-\left(n^2+n\right)\left(=0\right)$ (or equivalent)

attempts to solve their quadratic in $m$         (M1)

$m=\frac{1\pm \sqrt{12n^2-12n+1}}{6}\left(=\frac{1\pm \sqrt{{\left(-1\right)}^2+12\left(n^2+n\right)}}{6}\right)$

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

$\frac{n\left(n+1\right)}{2}=\frac{m\left(3m-1\right)}{2}$

let $n=m+k \left(n>m\right)$ and so $3m^2-m=\left(m+k\right)\left(m+k+1\right)$        M1

$2m^2-2\left(k+1\right)m-\left(k^2+k\right)=0$          A1

attempts to find the discriminant of their quadratic

and recognises that this must be a perfect square        M1

$\text{Δ}=4{\left(k+1\right)}^2+8\left(k^2+k\right)$

$N^2=4{\left(k+1\right)}^2+8\left(k^2+k\right) \left(=4\left(k+1\right)\left(3k+1\right)\right)$

determines that $k=8$ leading to $2m^2-18m-72=0\Rightarrow m=-3,12$ and so $m=12$          A1

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

METHOD 4

$\frac{n\left(n+1\right)}{2}=\frac{m\left(3m-1\right)}{2}$

let $m=n-k \left(m<n\right)$ and so $n^2+n=\left(n-k\right)\left(3\left(n-k\right)-1\right)$       M1

$2n^2-2\left(3k+1\right)n+\left(3k^2+k\right)=0$          A1

attempts to find the discriminant of their quadratic

and recognises that this must be a perfect square        M1

$\text{Δ}=4{\left(3k+1\right)}^2-8\left(3k^2+k\right)$

$N^2=4{\left(3k+1\right)}^2-8\left(3k^2+k\right) \left(=4\left(k+1\right)\left(3k+1\right)\right)$

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

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

[5 marks]

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

consider $n=1: P_r\left(1\right)=1+\left(1-1\right)\left(r-2\right)=1$ and $P_r\left(1\right)=\frac{\left(r-2\right)\left(1^2\right)-\left(r-4\right)\left(1\right)}{2}=1$

so true for $n=1$              R1 

Note: Accept $P_r\left(1\right)=1$ and $P_r\left(1\right)=\frac{\left(r-2\right)\left(1^2\right)-\left(r-4\right)\left(1\right)}{2}=1$.
Do not accept one-sided considerations such as '$P_r\left(1\right)=1$ 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. $P_r\left(k\right)=\frac{\left(r-2\right)k^2-\left(r-4\right)k}{2}$          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 $n=k+1:$

($P_r\left(k+1\right)$ can be represented by the sum

$\underset{m=1}{\overset{k+1}{\text{Σ}}}\left(1+\left(m-1\right)\left(r-2\right)\right)=\underset{m=1}{\overset{k}{\text{Σ}}}\left(1+\left(m-1\right)\left(r-2\right)\right)+\left(1+k\left(r-2\right)\right)$ and so

$P_r\left(k+1\right)=\frac{\left(r-2\right)k^2-\left(r-4\right)k}{2}+\left(1+k\left(r-2\right)\right) \left(P_r\left(k+1\right)=P_r\left(k\right)+\left(1+k\left(r-2\right)\right)\right)$         M1

$=\frac{\left(r-2\right)k^2-\left(r-4\right)k+2+2k\left(r-2\right)}{2}$         A1

 $=\frac{\left(r-2\right)\left(k^2+2k\right)-\left(r-4\right)k+2}{2}$

$=\frac{\left(r-2\right)\left(k^2+2k+1\right)-\left(r-2\right)-\left(r-4\right)k+2}{2}$         M1

$=\frac{\left(r-2\right){\left(k+1\right)}^2-\left(r-4\right)k-\left(r-4\right)}{2}$          (A1)

 $=\frac{\left(r-2\right){\left(k+1\right)}^2-\left(r-4\right)\left(k+1\right)}{2}$         A1

hence true for $n=1$ and $n=k$ true $\Rightarrow n=k+1$ true         R1

therefore true for all $n\in {\mathrm{\mathbb{Z}}}^{+}$

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]

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 n = −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 $8P_3\left(n\right)+1=4n^2+4n+1$. A good number of candidates were then able to express $4n^2+4n+1$ as ${(2n+1)}^2$ and conclude that $(2n+1)$ is odd. Rather than making the connection that $4n^2+4n+1$ is a perfect square, many candidates attempted instead to analyse the parity of either $4n(n+1)+1$ or $4n^2+4n+1$. 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 $P_5\left(n\right)=\frac{n(3n-1)}{2}$, a large number of
candidates used $P_5\left(n\right)=\frac{(5-2)n^2-(5-4)n}{2}$ . 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 $P_3\left(n\right)=P_5\left(n\right)$ 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.