Chapter 7.2, Problem 63PS

Solution

i.

To identify: . The rule for the number of cells in the nth ring.

HONEYCOMBS Domestic bees make their honeycomb by starting with a single hexagonal cell, then forming ring after ring of hexagonal cells around the initial cell, as shown. The numbers of cells in successive rings form an arithmetic sequence.

  

The required rule is $a_n=6n$ .

Given information:

The given image is:

  

Explanation:

Consider the given sequence.

The general formula for an arithmetic sequence is $a_n=a_1+\left(n-1\right)d$ .

The initial cell has 1 hexagonal cell

On the second image, can see that the initial cell and the first ring are combined to form one ring. It is made up of seven hexagonal cells. So, by subtracting one hexagonal cell or initial cell, we can conclude that the first ring is composed of six hexagonal cells.

Now, observe the third part of picture. There are initial cell, the first and the second ring. Thus, the second ring is made

  $\begin{array}{l}19-7=12\\ 12=2\cdot 6\end{array}$

hexagonal cells.

Therefore, the $n$ th ring will be made of

  $\begin{array}{l}{a}_n=nd\\ a_n=6n\end{array}$

ii.

To calculate: The total number of cells in the honeycomb after the 9th ring is formed?

  

  $270+1=271$ hexagonal cells.

Given information:

The given terms are $a_1=6,n=9$ .

Calculation:

Consider the given sequence.

It is given that $a_1=6,n=9$

  $\begin{array}{l}{a}_1=6\\ a_9=6\cdot 9\\ a_9=54\end{array}$

The general sum formula for a arithmetic sequence is $s_n=\frac{n\left(a_1+a_9\right)}{2}$ .

Put the value of $a_1=6\text{and}{a}_9=54$ into $s_n=\frac{n\left(a_1+a_9\right)}{2}$

  $\begin{array}{l}{s}_9=\frac{9\left(6+54\right)}{2}\\ s_9=\frac{9\cdot 60}{2}\\ s_9=270\end{array}$

Therefore after $9$ th ring there are $270+1=271$ hexagonal cells