Review question

Ref: R9507

## Solution

1. Write down the next three members of the sequence of numbers$1,4,9,16,\dotsc$
2. Calculate the 300th member of the sequence.

The value of each member is simply the square of its position. So the 300th member of the sequence is given by $300^2=90\ 000.$

1. Calculate the first five members of the sequence of differences$4-1\ ,9-4\ ,16-9,\dotsc$
2. If $a$,$b$,$c$, are consecutive members of the sequence in (i) write down a formula connecting $c-b$ and $b-a$. Rewrite this formula so $b$ is the subject.

The first five members of the sequence are $4-1,9-4,16-9,25-16,36-25,$ or $3,5,7,9,11.$

If $a$,$b$ and $c$ are consecutive members of the initial sequence, $b-a$ and $c-b$ are consecutive members of the sequence of differences.

This means $c-b$ is $2$ larger than $b-a$, or $c-b=b-a+2,$ which gives $b=\frac{1}{2} \left(a+c \right) -1$.

1. Use the result of (iv) and the fact that $8677^2 = 75290329 \qquad \text{ and} \qquad 8679^2 = 75325041$ to calculate $8678^2$ making your method clear.

The numbers $8677$, $8678$ and $8679$ are three consecutive integers, and we know the squares of the first and last of these.

This gives us our $a$ and $c$. Now we can substitute those numbers into the equation and obtain

\begin{align*} b &= \frac{1}{2}(a + c) -1 \\ 8678^2 &= \frac{1}{2} \left(75290329+75325041 \right) -1 \\ &=75307584. \end{align*}