图中显示了一个由相同大小的正方形组成的图形序列的前三个图形。 下表记录了垂直边数 V 和总边数 S。

1. Continue the pattern for the number of vertical sides to find $p$.

For $V$: $2,\ 5,\ 9,\ldots$

The increases are
$+3,\ +4,\ldots$
Continue the increases:
$9+5=14$
$14+6=20$
$20+7=27$
So, for figure 6,
$p=27.$

2. Continue the pattern for the total number of sides to find $q$.

For $S$: $4,\ 10,\ 18,\ldots$

The increases are
$+6,\ +8,\ldots$
Continue the increases:
$18+10=28$
$28+12=40$
$40+14=54$
So, for figure 6,
$q=54.$

1. Substitute the given total number of sides into the relationship.

Given $S=n^2+3n$ and $S=270$,
$n^2+3n=270$
$n^2+3n-270=0$

2. Solve the quadratic equation and choose the valid figure number.

$n^2+3n-270=0$
$(n+18)(n-15)=0$
So $n=-18$ or $n=15$.

Since $n$ is a figure number, $n$ must be positive.
$n=15$

1. Use the pattern in the number of vertical sides to form the nth-term expression.

From the table, $V=2,5,9,\ldots$

The increases are $3,4,5,\ldots$, so
$V=1+1+2+3+\cdots+n+n.$
Using $1+2+\cdots+n=\dfrac{n(n+1)}{2}$,
$V=\dfrac{n(n+1)}{2}+n=\dfrac{n^2+3n}{2}.$

1. Write the difference $S-V$ in terms of $n$.

From earlier parts,
$S=n^2+3n,\qquad V=\frac{n^2+3n}{2}.$
Hence
$S-V=(n^2+3n)-\frac{n^2+3n}{2}=\frac{n^2+3n}{2}.$

2. Test whether the difference can be 100 by forming an equation.

If the difference were 100, then
$\frac{n^2+3n}{2}=100.$
So
$n^2+3n=200.$

3. Show that the equation has no valid figure number.

Solving
$n^2+3n-200=0,$
n=\frac{-3\pm\sqrt{3^2-4(1)(-200)}}{2}=rac{-3\pm\sqrt{809}}{2}.
Since $\sqrt{809}$ is not an integer, $n$ is not a whole number. But $n$ must be a positive integer because it is a figure number.