The shape of a wall in a museum is shown to the right, where heights (in feet) are given at 6-foot intervals. The wall is to be paneled, and the cost to panel each square foot is $\$ 3.50$. Find the approximate cost of paneling the wall.
Which of the following sums would give the best approximation for the area of the wall?
$R_{n}=6\left(h_{2}+h_{3}+h_{4}+h_{5}+h_{6}+h_{7}+h_{8}\right)$
$R_{n}=6\left(\frac{h_{1}}{2}+2 h_{2}+2 h_{3}+2 h_{4}+2 h_{5}+2 h_{6}+2 h_{7}+\frac{h_{8}}{2}\right)$
$T_{n}=2\left(\frac{h_{1}}{2}+2 h_{2}+2 h_{3}+2 h_{4}+2 h_{5}+2 h_{6}+2 h_{7}+\frac{h_{8}}{2}\right)$
$T_{n}=6\left(\frac{h_{1}}{2}+h_{2}+h_{3}+h_{4}+h_{5}+h_{6}+h_{7}+\frac{h_{8}}{2}\right)$

Step 1 ：The question is asking for the best approximation for the area of the wall. The area of the wall can be approximated by the sum of the areas of the trapezoids formed by the heights at 6-foot intervals. The formula for the area of a trapezoid is \(\frac{1}{2}(b_1 + b_2)h\), where \(b_1\) and \(b_2\) are the lengths of the bases and \(h\) is the height. In this case, the height is the 6-foot interval and the bases are the heights of the wall at each interval. The sum of the areas of these trapezoids would give the best approximation for the area of the wall.

Step 2 ：Looking at the options, the formula that represents the sum of the areas of these trapezoids is \(T_{n}=6\left(\frac{h_{1}}{2}+h_{2}+h_{3}+h_{4}+h_{5}+h_{6}+h_{7}+\frac{h_{8}}{2}\right)\).

Step 3 ：Final Answer: The sum that would give the best approximation for the area of the wall is \(\boxed{T_{n}=6\left(\frac{h_{1}}{2}+h_{2}+h_{3}+h_{4}+h_{5}+h_{6}+h_{7}+\frac{h_{8}}{2}\right)}\).