Investigations (Cambridge (CIE) IGCSE International Maths: Extended): Exam Questions

The hexagonal numbers are an increasing sequence of numbers, with each number greater than the one before.

The first six hexagonal numbers are

$H_1=1,H_2=6,H_3=15,H_4=28,H_5=45,H_6=66$

Show that the first three hexagonal numbers are triangular numbers.

Is it true that all triangular numbers are hexagonal numbers?

Explain your answer.

The hexagonal numbers are an increasing sequence of numbers, with each number greater than the one before.

The first six hexagonal numbers are

$H_1=1,H_2=6,H_3=15,H_4=28,H_5=45,H_6=66$

To test whether a number, $x$, is a hexagonal number or not, substitute it into the formula:

$\frac{1+\sqrt{8x+1}}{4}$

If the output is a positive integer, then it is a hexagonal number.

For example, to test if 6 is a hexagonal number, substitute in $x=6$:

$\begin{array}{rcl}\frac{1+\sqrt{8\times 6+1}}{4}& =& \frac{1+\sqrt{48+1}}{4}\\ & =& \frac{1+\sqrt{49}}{4}\\ & =& \frac{1+7}{4}\\ & =& \frac{8}{4}\\ & =& 2\end{array}$

The output is 2, which is a positive integer, so 6 is a hexagonal number.

Use the test above to determine whether 120 is a hexagonal number.

You must show your working clearly.

Is it enough for the square root part of the formula, $\sqrt{8x+1}$, to be an integer in order for the output of the formula to be an integer?

Explain your answer.

A student believes that the number 1 is the only square number that is also a hexagonal number.

Use the number 1225 to show that the student is not correct.

You must show your working clearly.

The hexagonal numbers are an increasing sequence of numbers, with each number greater than the one before.

To test whether a number, $x$, is a hexagonal number or not, you can substitute it into the formula:

$\frac{1+\sqrt{8x+1}}{4}$

If the output is a positive integer, then it is a hexagonal number.

A hexagonal number, $H_n$, and its position value, $n$, satisfy this formula in the following way:

$\frac{1+\sqrt{8H_n+1}}{4}=n$

By making $H_n$ the subject, show that the formula for the $n$th term of a hexagonal number is given by

$H_n=2n^2-n$

The sequence of hexagonal numbers, $H_n$, is shown below.

A new sequence, $K_n$, is formed using the $n$th term formula:

$K_n=n+H_n$

The first three terms, $K_1$, $K_2$ and $K_3$, are shown in the table below.

Find $K_4$, $K_5$ and $K_6$.

A fraction is said to be a perfect square if it can be written in the form ${\left(\frac{a}{b}\right)}^2$ where $a$ and $b$ are non-zero integers.

Terms in the sequence of $K_n$ can be made into fractions by dividing the previous term, $K_{n-1}$, by the current term:

$\frac{K_{n-1}}{K_n}=\frac{\mathrm{previous}\mathrm{term}}{\mathrm{current}\mathrm{term}}$

Use the table below to show that fractions formed in this way are perfect squares.

The first example has been done for you, as follows:

$\frac{K_1}{K_2}=\frac{2}{8}=\frac{1}{4}={\left(\frac{1}{2}\right)}^2$

By substituting the $n$th term formula

$H_n=2n^2-n$

into the formula

$K_n=n+H_n$

and simplifying, use algebra to prove that

$\frac{K_{n-1}}{K_n}=\frac{{\left(n-1\right)}^2}{n^2}$

The hexagonal numbers are an increasing sequence of numbers, with each number greater than the one before.

The $n$th term formula for the hexagonal numbers is

$H_n=2n^2-n$

A new sequence is given by

$R_n=\frac{2}{2n-1}-\frac{1}{n}$

Use algebra to prove that $R_n$ is the sequence of the reciprocals of hexagonal numbers.