A sequence of figures is made up of squares. The number of squares in the first four figures are 2, 5, 10, 17. Find the rule for the number of squares in the nth figure and calculate the number of squ

Here's the solution to Question 4:

4.1.1:
First, count the number of squares in each given figure:

• Fig 1: 2 squares
• Fig 2: 5 squares
• Fig 3: 10 squares
• Fig 4: 17 squares

Next, find the pattern by looking at the differences between consecutive terms:
$T_2 - T_1 = 5 - 2 = 3$
$T_3 - T_2 = 10 - 5 = 5$
$T_4 - T_3 = 17 - 10 = 7$
The first differences are 3, 5, 7. The second differences are $5-3=2$ and $7-5=2$. Since the second difference is constant (2), the sequence is quadratic, of the form $T_n = an^2 + bn + c$.
The coefficient $a$ is half of the second difference, so $a = \frac{2}{2} = 1$.
Thus, $T_n = n^2 + bn + c$.
Using $T_1 = 2$: $1^2 + b(1) + c = 2 \implies 1 + b + c = 2 \implies b + c = 1$.
Using $T_2 = 5$: $2^2 + b(2) + c = 5 \implies 4 + 2b + c = 5 \implies 2b + c = 1$.
Subtracting the first equation from the second: $(2b + c) - (b + c) = 1 - 1 \implies b = 0$.
Substitute $b=0$ into $b+c=1 \implies 0+c=1 \implies c=1$.
So, the explicit rule is $T_n = n^2 + 1$.

Now, calculate the number of squares for Figure 5 and Figure 6:
$T_5 = 5^2 + 1 = 25 + 1 = 26$
$T_6 = 6^2 + 1 = 36 + 1 = 37$

The completed table is:

$$\begin{array}{|c|c|c|c|c|c|c|} \hline \text{FIGURE NUMBER} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{NUMBER OF SQUARES} & 2 & 5 & 10 & 17 & 26 & 37 \\ \hline \end{array}$$

4.1.2:
To find the number of squares in Figure 40, use the rule $T_n = n^2 + 1$.

$$T_{40} = 40^2 + 1$$ $$T_{40} = 1600 + 1$$ $$T_{40} = \boxed{\text{1601 squares}}$$

4.1.3:
The recursive rule describes how to get the next term from the previous one.
The number of squares in Figure $n$ is found by adding an odd number to the number of squares in Figure $n-1$. The odd number added increases by 2 for each subsequent figure. Specifically, to get the number of squares for Figure $n$, you add $2n-1$ squares to the number of squares in Figure $n-1$. The first figure has 2 squares.

4.1.4:
The algebraic form of the recursive rule described in 4.1.3 is:

$$T_n = T_{n-1} + (2n-1) \quad \text{for } n > 1, \text{ with } T_1 = 2$$