Answer:
[tex]\begin{array}{|r|c|c|c|c|c|c|c|}\cline{1-8}\text{Steps}&1&2&3&4&5&6&7\\\cline{1-8}\text{Rods}&4&10&18&28&40&54&70\\\cline{1-8}\end{array}[/tex]
Step-by-step explanation:
Given a diagram of staircase frames with 1, 2, and 3 steps, along with a table showing the number of rods in each frame, you are asked to fill in the table for up to 7 steps.
Counting rods
Looking at the attached diagram, we see the number of blue rods is equal to the number of red rods, and each of those numbers is 1 more than the number of steps in the frame. These are the added rods from the previous frame.
Starting with 0 steps, we have 0 rods. (This is not shown in the table.)
Step 1 adds 2×(1+1) = 4 rods for a total of 4 in a 1-step frame.
Step 2 adds 2×(2+1) = 6 rods, for a total of 4+6 = 10 in a 2-step frame.
Step 3 adds 2×(3+1) = 8 rods, for a total of 10+8 = 18 in a 3-step frame.
Step 4 adds 2×(4+1) = 10 rods, for a total of 18+10 = 28 in a 4-step frame.
Each addition is 2 more than the previous addition, so the next few table entries are ...
28 +12 = 40 (for 5 steps)
40 +14 = 54 (for 6 steps)
54 +16 = 70 (for 7 steps)
Then the table is ...
[tex]\begin{array}{|r|c|c|c|c|c|c|c|}\cline{1-8}\text{Steps}&1&2&3&4&5&6&7\\\cline{1-8}\text{Rods}&4&10&18&28&40&54&70\\\cline{1-8}\end{array}[/tex]
Graph and rule
A graph of the table values is shown in the second attachment. In the above, we have described a recursive relation between the steps (n) and rods (r).
- r(1) = 4
- r(n) = r(n-1) +2(n+1)
In the second attachment, we show that relation can also be described by the explicit rule ...
- r(n) = n(n+3) . . . . . a quadratic pattern
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Additional comment
The first differences of the numbers of rods are ...
10 -4 = 6
18 -10 = 8
Then the second difference is ...
8 -6 = 2
As we noted above, the second difference is a constant, meaning succeeding first differences are 10, 12, 14, 16, ... as we have shown.
The constant 2nd difference means the relation can be described by a 2nd degree polynomial. The leading coefficient of that polynomial is half the value of the 2nd difference: 2/2 = 1.
Knowing that the sequence is described by ...
r(n) = 1·n² +bn +c
we can find values of 'b' and 'c' to fill out the equation. Using the first two table entries, we have ...
r(1) = 4 = 1² +1b +c
r(2) = 10 = 2² +2b +c
Subtracting the first equation from the second gives ...
(10) -(4) = (4 +2b +c) -(1 +b +c)
6 = 3 +b ⇒ b = 3
Using the first equation to find c, we have ...
4 = 1 + b + c = 1 + 3 + c ⇒ c = 0
This means our explicit equation is ...
r(n) = n² +3n . . . . as we have seen in the graph
