How to find the Sum and Difference of Squares?
The arithmetic value which is used for representing the quantity and used in making calculations are defined as NUMBERS. A symbol like β4, 5, 6β 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, β¦ }
- 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, 5β¦ }
- 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 = { β¦., -3, -2, -1, 0, 1, 2, β¦.}
- 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 : β3, β5, β7, 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 β¦
- Odd numbers: Odd numbers are numbers that can not be fully divided by two. Example : 3, 5, 7, 9, β¦
- Prime Numbers: A whole number greater than one that can only be divided by itself or by one. Example : 2, 3, 5, 7, 11, β¦
- 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, β¦
- 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 Γ n, where, n is any number.
For example : 22 = 2 Γ 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 β’ Sum of squares of two numbers say, a and b
a2 + b2 + c2 β’ Sum of squares of three numbers say, a, b and c
(a1)2 + (a2)2 + (a3)2 + β¦β¦.. +(an)2 β’ Sum of squares of βnβ natural numbers
Formula 1: To Find The Sum of Squares of Two Numbers is a2 +b2 = (a + b)2 β 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 β 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 β 2ab
Formula 2: To Find the Sum of Squares of Three Numbers is a2 + b2 + c2 = (a + b + c)2 -2ab β 2bc β 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 β 2bc β 2ca
Hence, after knowing the values of (a + b + c)2 β 2ab -2bc -2ca we can find out the value of a2 + b2 + c2
Hence,
a2 + b2 + c2 = (a + b + c)2 -2ab β 2bc β 2ca
Formula 3: To Find the Sum of Squares of first βnβ Natural Numbers is β n2 = [n(n+1)(2n+1)]/6
Proof:
We will show it by Principal of Mathematical Induction,
Let P(n): 12 + 22 + 32 + β¦ +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)(2Γ1 + 1)]/6
=[2Γ3]/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 + β¦ + k2 = [k(k+1)(2k+1)]/6 β¦β¦β¦β¦.(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 + β¦. + k2 + (k+1)2 = [(k+1)(k+2)(2k+3)]/6 is true
Taking LHS, 12 + 22 + 32 + β¦. + k2 + (k+1)2
Since, from equation (i) we have, 12 + 22 + 32 + β¦ + k2 = [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 βnβ natural number is :
β n2 = [n(n+1)(2n+1)]/6
Formula 5: To Find the Sum of First n Even Numbers is β (2n)2 = 2[n(n+1)(2n+1)]/3
Proof :
The addition of squares of first n even numbers is given by :
β (2n)2 = 22 + 42 + 62 + β¦ + (2n)2
Proof :
β (2n)2 = 22.12 + 22.22 + 22.32 + β¦ + 22.n2
β (2n)2 = 22.(12 + 22 + 32 + β¦ + n2)
β (2n)2 = 22.β n2
β (2n)2 = 22.[n(n+1)(2n+1)]/6
β (2n)2 = 4[n(n+1)(2n+1)]/6
β (2n)2 = 2[n(n+1)(2n+1)]/3
Formula 6: To Find the Sum of First n odd n Numbers is β (2n-1)2 = [n(2n + 1)(2n β 1)]/3
Proof:
The addition of squares of first n odd numbers is given by
β (2n-1)2 = 12 + 32 + 52 + β¦ + (2n-1)2
β (2n-1)2 = 12 + 22 + 32 + β¦ + (2n-1)2 β [22 + 42 + 62 + β¦ + (2n-2)2]
Now applying the formula for the addition of squares of 2n natural numbers and of n natural numbers, we get :
β (2n-1)2 = 2n/6 (2n β 1)(4n β 1) β (2n/3)(n β 1)(2n + 1)
β (2n-1)2 = n/3 [(2n β 1)(4n β 1)] β 2n/3 [(n β 1)(2n + 1)]
After taking out the common terms, we get :
β (2n-1)2 = n/3 (2n + 1)[4n β 1 β 2n + 2]
β (2n-1)2 = [n(2n + 1)(2n β 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 β b2 = (a + b)(a β b)
Proof:
Consider RHS = (a+b)(a-b)
= a(a-b) + b(a-b)
= a2 β ab + ba β b2
= a2 -ab + ab β b2
= a2 + 0 β b2
= a2 β b2
= LHS
Sample Questions
Question 1: What is the value of 152 β 62?
Solution:
The formula for difference of squares is,
a2 β b2 = (a + b)(a β b)
From the given expression,
a = 15 and b = 6
a2 β b2 = 152 β 62
= (15 + 6)(15 β 6)
= 21 Γ 9
= 189
Therefore, the value of 152 β 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 :
βn2 = [n(n+1)(2n+1)]/6
Here, n = 25
β252 = [25(25+1)(2Γ25 + 1)]/6
β252 = [25Γ26Γ51]/6
β252 = 33150/6
β252 = 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 :
β (2n)2 = 2[n(2n+1)]/3
Here, n = 30
β(2Γ30)2 = 2[30(2Γ30 + 1)]/3
β602 = 2Γ[30Γ61]/3
β602 = (2 Γ 1830)/3
β602 = 3660/3
β602 = 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 β 2ab
From the given expression we have,
a = 5
b = 6
52 + 62 = (5 + 6)2 β 2Γ5Γ6
52 + 62 = 112 β 60
52 + 62 = 121 β 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 :
β (2n-1)2 = [n(2n + 1)(2n β 1)]/3
Here, n=50
β(2Γ50 β 1)2 =[50(2Γ50 + 1)(2Γ50 β 1)]/3
β992 = [50Γ101Γ99]/3
β992 = 499950/3
β992 = 166650
Sum of squares of first 50 odd numbers is 166650.
Question 6: Consider a = 11 and b = 5, then verify that a2 β b2 = (a+b)(a-b)
Solution:
Given,
a = 11 , b = 5
To Prove : 112 β 52 = (11 + 5)(11 β 5)
Take LHS = 112 β 52
= 121-25
= 96
Take RHS = (11 + 5)(11 β 5)
=(16)Γ(6)
=96
Since, LHS = RHS
Hence, the formula is verified.
Question 7: Evaluate the value of x , where (x2 β 22) = 0
Solution:
Given : x2 β 22 = 0
we know that, a2 β b2 = (a+b)(a-b)
Here, a = x and b = 2
Hence,
x2 β 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 β 2bc β 2ca
Hence,
32 + 52 + (-1)2 = (3+5-1)2 β 2(3)(5) β 2(5)(-1) β 2(-1)(3)
= (7)2 β 30 + 10 +6
= 49 β 30 +10 + 6
= 35
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