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How to find the Sum and Difference of Squares?

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The arithmetic value which is used for representing the quantity and used in making calculations are defined as NUMBERS. A symbol like “4, 5, 6” which represents a number is known as numerals. Without numbers, counting things is not possible, date, time, money, etc. these numbers are also used for measurement and used for labeling. The properties of numbers make them helpful in performing arithmetic operations on them. 

Types of numbers

There are different types of numbers which are defined below:

  • Natural Numbers: Natural Numbers are the family of numbers from 1 to infinity, they are also called counting numbers. A set of natural numbers is denoted by N. N = {1, 2, 3, 4, 5, … }
  • Whole Numbers: The set of natural numbers including zero is called a set of Whole Numbers and denoted by W. W = {0, 1, 2, 3, 4, 5… }
  • Integers: An integer is a set of whole numbers that can be positive, negative, or zero. The set of integers is denoted by Z. Z = { …., -3, -2, -1, 0, 1, 2, ….}
  • Rational Numbers: A rational number is a number that can be expressed as the ratio of two integers or a number that can be written as p/q form where q is not equal to zero. The set of rational numbers is denoted by Q. Example: 4/7, 6/5, 2.3 = 23/10, etc.
  • Irrational Numbers: Irrational numbers is the set of numbers that can not be expressed as a simple fraction. Example : √3, √5, √7, etc.
  • Even Number: An even number is a number that can be fully divided by 2 and leave zero as a remainder. Example : 2, 4, 6, 8, 10 …  
  • Odd numbers: Odd numbers are numbers that can not be fully divided by two. Example : 3, 5, 7, 9, … 
  • Prime Numbers: A whole number greater than one that can only be divided by itself or by one. Example : 2, 3, 5, 7, 11, …
  • Composite Number: A  number that has at least one factor other than 1 and itself is known as a composite number. Example : 4, 6, 8, 9, …
  • Real Numbers: The power set of natural numbers, whole numbers, integers, rational numbers, irrational numbers, even numbers, odd numbers, prime numbers, and composite numbers is known as the set of Real Numbers. It is denoted by R

Square of a Number 

When you multiply a number by itself, the resulting number is called the square of the number. The square of the number is represented as n2,

n2 =n × n,  where, n is any number.

For example : 22 = 2 × 2 = 4, and 4 is called the square of 2.

The Square of a real number is always positive because if you multiply a negative number by itself then you will always get a positive number. Hence, the square of any real number is always a positive number.

Sum of Square Numbers

The sum of the square of numbers is defined as the addition of squared numbers. 

a2 + b2 ⇢ Sum of squares of two numbers say, a and b

a2 + b2 + c2 ⇢ Sum of squares of three numbers say, a, b and c

(a1)2 + (a2)2 + (a3)2 + …….. +(an)2 ⇢ Sum of squares of ‘n’ natural numbers

Formula 1: To Find The Sum of Squares of Two Numbers is  a2 +b2 = (a + b)2 – 2ab

Proof:

Let us suppose we have to find the sum of squares of two real numbers a and b

As we know the mathematical formula, (a + b)2 = a2 + b2 + 2ab

We can reduce the above formula as, a2 +b2 = (a + b)2 – 2ab

Hence, knowing the value of (a + b)2 and 2ab we can find out the value of a2 + b2.

Therefore,

 a2 +b2 = (a + b)2 – 2ab

Formula 2: To Find the Sum of Squares of Three Numbers is a2 + b2 + c2 = (a + b + c)2 -2ab – 2bc – 2ca

Proof:

Let us suppose we have to find the sum of squares of three real numbers a, b and c

As we know the mathematical formula, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca

we can reduce the above formula as, a2 + b2 + c2 = (a + b + c)2 -2ab – 2bc – 2ca

Hence, after knowing the values of (a + b + c)2 – 2ab -2bc -2ca we can find out the value of a2 + b2 + c2

Hence,

a2 + b2 + c2 = (a + b + c)2 -2ab – 2bc – 2ca

Formula 3: To Find the Sum of Squares of first ‘n’ Natural Numbers is  ∑ n2 = [n(n+1)(2n+1)]/6

Proof:

We will show it by Principal of  Mathematical Induction,

Let P(n): 12 + 22 + 32 + … +n2 = [n(n+1)(2n+1)]/6

let n = 1

Then LHS, n2 = 12 = 1

Taking RHS, [n(n+1)(2n+1)]/6 = [1(1+1)(2×1 + 1)]/6

                                                =[2×3]/6

                                                =6/6

                                                =1

Since, LHS = RHS

This implies, the result is true for n = 1.

Now, let us suppose that this result is also true for n=k.

This implies, 12 + 22 + 32 + … + k = [k(k+1)(2k+1)]/6           ………….(i)

Now, we will show that the result is also true for n = k+1

This implies, we have to prove that 12 + 22 + 32 + …. + k2 +  (k+1)2 = [(k+1)(k+2)(2k+3)]/6 is true

Taking LHS, 12 + 22 + 32 + …. + k2 + (k+1)2

Since, from equation (i) we have, 12 + 22 + 32 + … + k = [k(k+1)(2k+1)]/6

Therefore, LHS = [k(k+1)(2k+1)]/6 + (k+1)2

take (k+1) common from both terms we get, (k+1)[k(2k+1)/6 + (k+1)]

= (k+1)[(2k2 + k)/6 + (k+1)]

=(k+1)[2k2 + k + 6k + 6]/6

= (k+1)[2k2 + 7k + 6]/6

= (k+1)[2k2 + 4k + 3k + 6]/6

= [(k+1)(k+2)(2k+3)]/6 

= RHS

Hence, the result is also true for n = k+1

And therefore, we can say that the result is true for all natural numbers n.

Therefore, the formula for calculating the sum of squares of ‘n’ natural number is :

∑ n2 = [n(n+1)(2n+1)]/6

Formula 5: To Find the Sum of First n Even Numbers is ∑ (2n)2 = 2[n(n+1)(2n+1)]/3

Proof :

The addition of squares of  first n even numbers is given by :

∑ (2n)2 = 22 + 42 + 62 + … + (2n)2

Proof : 

∑ (2n)2 = 22.12 + 22.22 + 22.32 + … + 22.n2

∑ (2n)2 = 22.(12 + 22 + 32 + … + n2)

∑ (2n)2 = 22.∑ n2

∑ (2n)2 = 22.[n(n+1)(2n+1)]/6

∑ (2n)2 = 4[n(n+1)(2n+1)]/6

∑ (2n)2 = 2[n(n+1)(2n+1)]/3

Formula 6: To Find the Sum of First n odd n Numbers is ∑ (2n-1)2 = [n(2n + 1)(2n – 1)]/3

Proof:

The addition of squares of first n odd numbers is given by

∑ (2n-1)2 = 12 + 32 + 52 + … + (2n-1)2

∑ (2n-1)2 = 12 + 22 + 32 + … + (2n-1)2 – [22 + 42 + 62 + … + (2n-2)2]

Now applying the formula for the addition of squares of 2n natural numbers and of n natural numbers, we get :

∑ (2n-1)2 = 2n/6 (2n – 1)(4n – 1) – (2n/3)(n – 1)(2n + 1)

∑ (2n-1)2 = n/3 [(2n – 1)(4n – 1)] – 2n/3 [(n – 1)(2n + 1)]

After taking  out the common terms, we get :

∑ (2n-1)2 = n/3 (2n + 1)[4n – 1 – 2n + 2]

∑ (2n-1)2 = [n(2n + 1)(2n – 1)]/3

Difference in Squares of Numbers

The difference of two squares formula is one of the main algebraic expressions used to expand terms of the form a2-b2. In other words, this is an algebraic form of the equation used to equalize the difference between two squared values. This equation is useful for converting complex equations into simple equations.

Formula To Calculate the Difference of Squares

The formula for calculating the difference of squares is given by :

a2 – b2 = (a + b)(a – b)

Proof:

Consider RHS = (a+b)(a-b)

= a(a-b) + b(a-b)

= a2 – ab + ba – b2

= a2 -ab + ab – b2

= a2 + 0 – b2

= a2 – b2

= LHS

Sample Questions 

Question 1: What is the value of 152 – 62?

Solution:

The formula for difference of squares is,

a2 – b2 = (a + b)(a – b)

From the given expression,

a = 15  and b = 6

a2 – b2 = 152 – 62

= (15 + 6)(15 – 6)

= 21 × 9

= 189

Therefore, the value of 152 – 62 is 189.

Question 2: Find the sum of the squares of the first 25 natural numbers.

Solution:

The formula for the sum of squares of first n natural numbers is :

∑n2 = [n(n+1)(2n+1)]/6

Here, n = 25

∑252 = [25(25+1)(2×25 + 1)]/6

∑252 = [25×26×51]/6

∑252 = 33150/6

∑252 = 5525

Hence, the sum of squares of first 25 natural numbers is 5525.

Question3: Find the sum of the squares of first 30 even numbers.

Solution:

The formula for the sum of squares of first n even numbers is :

∑ (2n)2 = 2[n(2n+1)]/3

Here, n = 30

∑(2×30)2 = 2[30(2×30 + 1)]/3

∑602 = 2×[30×61]/3

∑602 = (2 × 1830)/3

∑602 = 3660/3

∑602 = 1220

Hence, the sum of squares of first 30 even numbers is 1220.

Question 4: Evaluate 52 + 62 with the help of formula and directly as well. Also, verify the answers.

Solution:

The formula for the sum of squares of two numbers is :

a2 +b2 = (a + b)2 – 2ab

From the given expression we have,

a = 5

b = 6

52 + 62 = (5 + 6)2 – 2×5×6

52 + 62 = 112 – 60

52 + 62 = 121 – 60

52 + 62 = 61

Now, solving the given equation directly, we get :

52 + 62 = 25 + 36 = 61

Since, both the answers are same. Hence, verified.

Question 5: Find the sum of squares of the first 50 odd numbers.

The formula for the sum of squares of first n even numbers is :

∑ (2n-1)2 = [n(2n + 1)(2n – 1)]/3

Here, n=50

∑(2×50 – 1)2 =[50(2×50 + 1)(2×50 – 1)]/3

∑992 = [50×101×99]/3

∑992 = 499950/3

∑992 = 166650

Sum of squares of first 50 odd numbers is 166650.

Question 6: Consider a = 11 and b = 5, then verify that a2 – b2 = (a+b)(a-b)

Solution: 

Given,

a = 11 , b = 5

To Prove : 112 – 52 = (11 + 5)(11 – 5)

Take LHS = 112 – 52

= 121-25

= 96

Take RHS = (11 + 5)(11 – 5)

=(16)×(6)

=96

Since, LHS = RHS

Hence, the formula is verified.

Question 7: Evaluate the value of x , where (x2 – 22) = 0

Solution:

Given : x2 – 22 = 0

we know that, a2 – b2 = (a+b)(a-b)

Here, a = x and b = 2

Hence, 

x2 – 22 = (x+2)(x-2) = 0

(x+2)(x-2) = 0

Comparing both terms to zero

(x+2) = 0  and (x-2) = 0

This implies, x = -2  and  x = 2

Hence, the value of x can be 2 or -2.

Question 8: Evaluate a2 + b2 + c2 for a=3, b=5, c=-1.

Solution:

we know that, 

a2 + b2 + c2 = (a + b + c)2 -2ab – 2bc – 2ca

Hence,

32 + 52 + (-1)2 = (3+5-1)2 – 2(3)(5) – 2(5)(-1) – 2(-1)(3)

= (7)2 – 30 + 10 +6

= 49 – 30 +10 + 6

= 35



Last Updated : 30 Jan, 2024
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