Q1.
[Fig. 8.6, growing pattern of squares 3, 6, 12, 24] Can you predict the number of squares in Stages 5 and 6 of the pattern? In Stages 10, 11 and 12? In Stage 20? At any stage? How is this different from the growing pattern in Fig. 8.3?
Answer
Each stage of Fig. 8.6 is a rectangle 3 squares wide whose height doubles: 1, 2, 4, 8 rows. So the counts double rather than grow by a fixed amount.
3, 6, 12, 24, 48, 96, …
t1 = 3, t2 = 3 × 2, t3 = 3 × 22, t4 = 3 × 23
tn = 3 × 2n–1
t1 = 3, t2 = 3 × 2, t3 = 3 × 22, t4 = 3 × 23
tn = 3 × 2n–1
| Stage | 5 | 6 | 10 | 11 | 12 | 20 | n |
|---|---|---|---|---|---|---|---|
| Number of squares | 48 | 96 | 1536 | 3072 | 6144 | 15,72,864 | 3 × 2n–1 |
Stage 10: 3 × 29 = 3 × 512 = 1536
Stage 20: 3 × 219 = 3 × 524288 = 15,72,864
Stage 20: 3 × 219 = 3 × 524288 = 15,72,864
How it differs from Fig. 8.3:
| Fig. 8.3 | Fig. 8.6 | |
|---|---|---|
| Step from one stage to the next | add 4 | multiply by 2 |
| Sequence | 1, 5, 9, 13, 17, … | 3, 6, 12, 24, 48, … |
| Constant that describes it | common difference d = 4 | common ratio r = 2 |
| nth term | 4n – 3 | 3 × 2n–1 |
| Type | AP | GP |
| At Stage 20 | 77 | 15,72,864 |
Why it happens: repeated addition puts n in the formula as a plain multiplier, so the terms grow in proportion to n. Repeated multiplication puts n in the exponent, and an exponent compounds — each doubling acts on everything already accumulated. That is why two patterns which look similar at Stage 4 (13 against 24) are worlds apart at Stage 20 (77 against more than fifteen lakh).
Check it yourself: 3 × 2n–1 must give 3 at n = 1. And 20 = 1, so it does.