NCERT Solutions Ganita Prakash (Part 2) Chapter 4 .1 Fractals — In-text Questions

Book page 714 Updated on2026-09-05

Q1.
Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Sierpinski Carpet.
Answer

Start with one square. At every step, cut each square that is still present into 9 equal sub-squares and rub out the middle one.

Step 0 Step 1 Step 2 . . .
Step 0 is a full square. Step 1 has one hole. Step 2 has that hole plus one new hole in each of the 8 remaining squares.
Tip: Draw on squared paper with a 27 × 27 square. Then a Step-1 sub-square is 9 × 9 and a Step-2 sub-square is 3 × 3, so every line falls on a grid line.
Q2.
Do you see any pattern in the number of holes and squares that remain at each step?
Answer

Yes. The number of squares that remain is multiplied by 8 at every step, and the number of holes goes up by exactly the number of squares that were present before the step.

Step n0123
Squares remaining Rn1864512
Holes Hn01973
Why it happens: One square is cut into 9 sub-squares; 1 is removed and 8 survive. So every single square of Step n turns into 8 squares of Step (n + 1) — hence Rn+1 = 8Rn. At the same time that one square produces exactly one new hole, so the number of new holes added equals the number of squares present, Rn. Old holes are never filled in, so Hn+1 = Hn + Rn.
Q3.
Can this be used to get a formula for Rₙ?
Answer

Yes — Rn = 8n.

R0 = 1
R1 = 8 × R0 = 8
R2 = 8 × R1 = 8 × 8 = 8²
R3 = 8 × R2 = 8³
In general, Rn = 8n
Why the rule Rn+1 = 8Rn gives a power: the rule says “multiply by 8 once for every step”. Going from Step 0 to Step n takes n steps, so 8 has been used as a factor n times — and n factors of 8 is what 8n means. A rule of the form “multiply by a fixed number each time” always produces a power like this.
Q4.
Similarly, how do we find the number of holes at a given step?
Answer

Add up all the squares that ever existed before that step:

Hn = Hn−1 + Rn−1
so Hn = R0 + R1 + … + Rn−1
= 1 + 8 + 8² + … + 8n−1

This sum can be closed up. Multiply it by 7 and watch the middle terms cancel:

7Hn = (8 − 1)(1 + 8 + … + 8n−1)
= (8 + 8² + … + 8n) − (1 + 8 + … + 8n−1)
= 8n − 1
So Hn = (8n − 1)⁄7
Check: H1 = (8 − 1)⁄7 = 1
H2 = (64 − 1)⁄7 = 9 = 1 + 8 ✓
H3 = (512 − 1)⁄7 = 73 = 1 + 8 + 64 ✓
Did you know? Hn = (8n − 1)⁄7 is exactly the number written as n eights in base 9 — the same trick that makes 1 + 10 + 100 = 111.
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