5.3-Some More Interesting Patterns

5.3-Some More Interesting Patterns Important Formulae

You are currently studying
Grade 8 → Math → Squares and Square Roots → 5.3-Some More Interesting Patterns

5.3 - Some More Interesting Patterns

In this section, we explore some interesting patterns related to squares and square roots. These patterns help us understand the relationship between numbers and can be used in problem-solving and mental calculations. Let's examine a few such patterns in detail:

Pattern 1: Sum of the first n odd numbers

One of the most fascinating patterns is that the sum of the first n odd numbers is always a perfect square. This can be expressed as:

If we add the first n odd numbers:

  • 1 + 3 = 4 = $2^2$
  • 1 + 3 + 5 = 9 = $3^2$
  • 1 + 3 + 5 + 7 = 16 = $4^2$
  • 1 + 3 + 5 + 7 + 9 = 25 = $5^2$
  • And so on...
The sum of the first n odd numbers is always equal to the square of n, i.e., the sum of the first n odd numbers is equal to $n^2$.

Pattern 2: Difference of squares

The difference of squares is another well-known pattern. The difference between the squares of two consecutive numbers can be represented as:

$a^2 - (a - 1)^2 = (a + 1)(a - 1) = 2a - 1$

This pattern shows that the difference between the squares of two consecutive numbers is always an odd number, specifically $2a - 1$ where $a$ is the larger number.

Pattern 3: Square of a binomial

The square of a binomial can be expanded using the identity:

$(a + b)^2 = a^2 + 2ab + b^2$

Similarly, for the negative binomial:

$(a - b)^2 = a^2 - 2ab + b^2$

These identities provide a quick way to expand and simplify expressions involving squares of binomials. For example:

$(x + 3)^2 = x^2 + 6x + 9$ and $(x - 4)^2 = x^2 - 8x + 16$.

Pattern 4: Consecutive squares

The difference between two consecutive square numbers follows a predictable pattern. If $n^2$ and $(n+1)^2$ are two consecutive squares, the difference between them is:

$(n + 1)^2 - n^2 = (n + 1 + n)(n + 1 - n) = (2n + 1)$

For example:

$(5^2 - 4^2) = (25 - 16) = 9$ and $(6^2 - 5^2) = (36 - 25) = 11$

This shows that the difference between the squares of two consecutive numbers is always an odd number and increases by 2 as we move to the next consecutive numbers.

Pattern 5: Square roots of perfect squares

Another interesting pattern is the relationship between perfect squares and their square roots. The square root of a perfect square always gives an integer. For example:

  • The square root of $16$ is $4$.
  • The square root of $25$ is $5$.
  • The square root of $36$ is $6$.

This pattern helps us recognize perfect squares and also compute square roots quickly.

Pattern 6: Relationship between squares and area

The square of a number can be thought of geometrically as the area of a square with side length equal to that number. For example, if the side of a square is 4 units, its area is $4^2 = 16$ square units. This concept is useful in solving problems related to area and geometry.

These patterns highlight some interesting and useful relationships in the world of squares and square roots, which can be applied in various mathematical problems.

What will be the unit digit of the squares of the following numbers?

(i) 81
(ii) 272
(iii) 799
(iv) 3853
(v) 1234
(vi) 26387
(vii) 52698
(viii) 99880
(ix) 12796
(x) 55555

Solution:

What will be the unit digit of the squares of the following numbers?
(i) 81: The unit digit of 81 is 1. The unit digit of $1^2$ is 1.
(ii) 272: The unit digit of 272 is 2. The unit digit of $2^2$ is 4.
(iii) 799: The unit digit of 799 is 9. The unit digit of $9^2$ is 1.
(iv) 3853: The unit digit of 3853 is 3. The unit digit of $3^2$ is 9.
(v) 1234: The unit digit of 1234 is 4. The unit digit of $4^2$ is 6.
(vi) 26387: The unit digit of 26387 is 7. The unit digit of $7^2$ is 9.
(vii) 52698: The unit digit of 52698 is 8. The unit digit of $8^2$ is 4.
(viii) 99880: The unit digit of 99880 is 0. The unit digit of $0^2$ is 0.
(ix) 12796: The unit digit of 12796 is 6. The unit digit of $6^2$ is 6.
(x) 55555: The unit digit of 55555 is 5. The unit digit of $5^2$ is 5.

The following numbers are obviously not perfect squares. Give reason.

(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050

Solution:

Reasons for Numbers Not Being Perfect Squares Based on Unit Place
(i) 1057
The unit digit of 1057 is 7. A perfect square cannot have a unit digit of 7.

(ii) 23453
The unit digit of 23453 is 3. A perfect square cannot have a unit digit of 3.

(iii) 7928
The unit digit of 7928 is 8. A perfect square cannot have a unit digit of 8.

(iv) 222222
The unit digit of 222222 is 2. A perfect square cannot have a unit digit of 2.

(v) 64000
The unit digit of 64000 is 0. While perfect squares can end in 0, the number 64000 is not a perfect square as it is not a square of a whole number.

(vi) 89722
The unit digit of 89722 is 2. A perfect square cannot have a unit digit of 2.

(vii) 222000
The unit digit of 222000 is 0. While perfect squares can end in 0, the number 222000 is not a perfect square as it is not a square of a whole number.

(viii) 505050
The unit digit of 505050 is 0. While perfect squares can end in 0, the number 505050 is not a perfect square as it is not a square of a whole number.

The squares of which of the following would be odd numbers?

(i) 431
(ii) 2826
(iii) 7779
(iv) 82004

Solution:

The squares of which of the following would be odd numbers?
(i) 431: $431^2$ = 185761 (Odd)
(ii) 2826: $2826^2$ = 7991076 (Even)
(iii) 7779: $7779^2$ = 60408841 (Odd)
(iv) 82004: $82004^2$ = 6720656016 (Even)

Observe the following pattern and find the missing digits.

11$^2$ = 121
101$^2$ = 10201
1001$^2$ = 1002001
100001$^2$ = 1 ......... 2 ......... 1
10000001^2 = ...

Solution:

Observe the following pattern and find the missing digits:
11$^2$ = 121
101$^2$ = 10201
1001$^2$ = 1002001
100001$^2$ = 10000200001
10000001$^2$ = 100000020000001

Observe the following pattern and supply the missing numbers.

11$^2$ = 1 2 1
101$^2$ = 1 0 2 0 1
10101$^2$ = 102030201
1010101$^2$ = ...........................
............$^2$ = 10203040504030201

Solution:

Observe the following pattern and supply the missing numbers.
11$^2$ = 1 2 1
101$^2$ = 1 0 2 0 1
10101$^2$ = 102030201
1010101$^2$ = 1020304030201
101010101$^2$ = 10203040504030201

Using the given pattern, find the missing numbers.

1^2+2^2+2^2 =3^2
2^2+3^2+6^2 =7^2
3^2 +4^2 +12^2=13^2
4^2 +5^2 +(..)^2 =21^2
5^2 +(..)^2 +30^2=31^2
6^2+7^2 +(..)^2 =(..)^2

Solution:

Using the given pattern, find the missing numbers:
12 + 22 + 22 = 32
22 + 32 + 62 = 72
32 + 42 + 122 = 132
42 + 52 + 20$^2$ = 212
52 + 6$^2$ + 302 = 312
62 + 72 + 42$^2$ =43$^2$

Without adding, find the sum.

(i) 1+3+5+7+9
(ii) 1+3+5+7+9+11+13+15+17+19
(iii) 1+3+5+7+9+11+13+15+17+19+21+23

Solution:
Without adding, find the sum.

(i) 1+3+5+7+9 = 5$^2$ = 25
(ii) 1+3+5+7+9+11+13+15+17+19 = 10$^2$ = 100
(iii) 1+3+5+7+9+11+13+15+17+19+21+23 = 12$^2$ = 144

(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.

Solution:

(i) Express 49 as the sum of 7 odd numbers.
$49 = 1 + 3 + 5 + 7 + 9 + 11 + 13$
(ii) Express 121 as the sum of 11 odd numbers.
$121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21$

How many numbers lie between squares of the following numbers?

(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100

Solution:

How many numbers lie between squares of the following numbers?

(i) Between the squares of 12 and 13:

Square of 12 = $12^2 = 144$

Square of 13 = $13^2 = 169$

Numbers between 144 and 169 = $169 - 144 - 1 = 24$

(ii) Between the squares of 25 and 26:

Square of 25 = $25^2 = 625$

Square of 26 = $26^2 = 676$

Numbers between 625 and 676 = $676 - 625 - 1 = 50$

(iii) Between the squares of 99 and 100:

Square of 99 = $99^2 = 9801$

Square of 100 = $100^2 = 10000$

Numbers between 9801 and 10000 = $10000 - 9801 - 1 = 198$