Can you predict the number of squares in Stages 5 and 6 of the sequence (1, 5, 9, 13)? In Stages 10, 11 and 12? In Stage 20? At any stage?                                                                                                                                                              [Page No. 180]

The counts 1, 5, 9, 13 form an arithmetic progression with first term $a = 1$ and common difference $d = 4$.

So, the nth term is $t_n = 1 + (n - 1) \times 4 = 4n - 3$.

Using this rule:

The number of squares increases by 4 at each stage: 1, 5, 9, 13, 17, 21,...

This is an arithmetic sequence with first term 1 and common difference 4. Therefore, the number of squares at Stage n is

$$t_n = 1 + (n - 1) \cdot 4 = 4n - 3.$$

Using this formula:
Stage 5: $t_{5} = 4(5) - 3 = 17$
Stage 6: $t_{6} = 4(6) - 3 = 21$
Stage 10: $t_{10} = 4(10) - 3 = 37$
Stage 11: $t_{11} = 4(11) - 3 = 41$
Stage 12: $t_{12} = 4(12) - 3 = 45$
Stage 20: $t_{20} = 4(20) - 3 = 77$

Hence, the number of squares at Stage n is given by $t_n = 4n - 3$.