Sum of Squares
Let n be a natural number. Squaring the number is denoted by n^{2}. The sum of squares means the sum of the squares of the given numbers. It could be finding the sum of squares of 2 numbers or 3 numbers or sum of squares of consecutive n numbers or n even numbers or n odd numbers. We evaluate the sum of the squares in statistics to find the variation in the data. We do these basic arithmetic operations which are required in statistics and algebra. There are different techniques to find the sum of squares of given numbers. Let us learn the formulae and the derivation to find them.
Sum of Squares of Two and Three Numbers
For small numbers, we can directly find the squares and add them, but for larger numbers, we need to know the identity to use to ease our calculations. Let a and b be the 2 numbers. Their squares are a^{2} and b^{2}. The sum of their squares is a^{2 }+ b^{2}. We could obtain a formula using the known algebraic identity (a+b)^{2 }= a^{2} + b^{2} + 2ab. From this we conclude that a^{2} + b^{2} = (a +b)^{2}  2ab.
Let a, b, c be the 3 numbers for which we are supposed to find the sum of squares. The sum of their squares is a^{2 }+ b^{2 }+ c^{2}. Using the known algebraic identity (a+b+c)^{2 }=a^{2 }+ b^{2 }+ c^{2 }+ 2ab + 2bc +2ca, we can evaluate that a^{2 }+ b^{2 }+ c^{2} = (a+b+c)^{2}  2ab 2bc 2ca
Sum of Squares of Natural Numbers
The natural numbers are the counting numbers from 1 to infinity. If we consider n consecutive natural numbers, then finding the sum of the squares of their numbers is represented as Σn^{2}, where n ranges from 1 to infinity. Here are the formulas for finding the sum of squares of n natural numbers, the sum of squares of first n even numbers, and the sum of squares of first n odd numbers.
Sum of squares of n natural numbers  \(\dfrac{n \times (n+1) \times (2n+1)}{6}\) 

Sum of squares of first n even numbers  \(\dfrac{2n \times (n+1) \times (2n+1)}{3}\) 
Sum of squares of first n odd numbers  \(\dfrac{n \times (2n+1) \times (2n1)}{3}\) 
Sum of Squares Formula
Let us learn to evaluate the sum of squares for larger sums. We can readily use the formula available to find the sum, however, it is essential to learn the derivation of the sum of squares formula.
Sum of Squares of n Natural Numbers Formula
Sum of n natural numbers can be defined as a form of arithmetic progression where the sum of n terms are arranged in a sequence with the first term being 1, n being the number of terms along with the n^{th} term. The sum of n natural numbers is represented as [n(n+1)]/2. Natural numbers are the numbers that start from 1 and end at infinity. Natural numbers include whole numbers in them except the number 0. If we need to calculate the sum of squares of n consecutive natural numbers, the formula is Σn^{2} = \(\dfrac{n \times (n+1) \times (2n+1)}{6}\). It is easy to apply the formula when the value of n is known. Let us prove this true using the known algebraic identity.
The sum of squares of n natural numbers is 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}+ .....n^{2} given by Σn^{2}, where n = 1 to ∞. Using the algebraic identity, a^{3 } b^{3 }= (ab) (a^{2 }+ ab + b^{2}), by replacing a with n and b with (n1), we have,
n^{3 } (n1)^{3 }= (n n+1)(n^{2} +n(n1)+ (n1)^{2})
n^{3 } (n1)^{3 }= 1(n^{2} +n^{2}n+ (n1)^{2})
= 1 (2n^{2 } n + n^{2 }+ 1  2n)
= 3n^{2 } 3n + 1
n^{3 } (n1)^{3 }= 3n^{2} 3n + 1 > (1)
(n1)^{3 } (n2)^{3 }= 3 (n1)^{2 } 3(n1) +1> (2)
(n2)^{3 } (n3)^{3 }= 3 (n2)^{2 } 3(n2) +1> (3)
...................
2^{3 } 1^{3 }= 3 (2)^{2 } 3(2) +1
1^{3} 0^{3 }= 3 (1)^{2 } 3(1) +1>(last step)
(1) + (2) + (3) +............+ (the last step)⇒ By adding all the above steps, we get, n^{3 } 0^{3 }= 3 Σ n^{2 } 3Σ n + n
n^{3 }= 3 Σ n^{2 } \(\dfrac{3n (n+1)}{2}\)+ n [ since Σ n = n(n+1)/2 (sum of n natural numbers)]
3 Σ n^{2 }= n^{3 }+ \(\dfrac{3n (n+1)}{2}\)  n
3 Σ n^{2 }= \(n[n^2 +\dfrac{3(n+1)}{2} 1]\)(Taking n as common from RHS)
Σ n^{2 }= \(\dfrac{n}{3}( n^2 + \dfrac{3n+3}{2} 1)\)
= \(\dfrac{n}{6}(2n^2+3n +1)\)
(Factorizing the quadratic equation)
Σ n^{2 }= \(\dfrac{n}{6} (2n+1)(n+1)\)
Sum of Squares of the First n Even Numbers Formula
The even numbers are denoted by 2n, where n is the natural number. The summation of the first n even numbers is given as 2^{2} + 4^{2}+ 6^{2}+ 8^{2} + 10^{2} + 12^{2 }+ ........(2n)^{2}. We are required to identify n and apply in the known formula \(\dfrac{2n \times (n+1) \times (2n+1)}{3}\). Let us derive the formula from the already learned formulas. When n takes the value from 1 to ∞, we evaluate Σ(2n)^{2} as, Σ(2^{2 }. n^{2}) as follows.
Σ(2n)^{2} = 2^{2}.1^{2} + 2^{2}.2^{2} + 2^{2}.3^{2} + 2^{2}.4^{2} +…+ 2^{2}.n^{2}
Σ(2n)^{2} = 2^{2}(1^{2} + 2^{2} + 3^{2} + 4^{2} … + n^{2})
Σ(2n)^{2} = \(\dfrac{4 n (n+1)(2n+1)}{6}\) (Formula for sum of squares of n natural numbers)
Thus Σ(2n)^{2} =\(\dfrac{2 n (n+1)(2n+1)}{3}\)
Sum of Squares of the First n Odd Numbers Formula
The odd numbers are denoted by (2n1), where n is the natural number. The sum of the squares of the first n odd natural numbers is given by 1^{2} + 3^{2} + 5^{2} + … + (2n – 1)^{2}. Identify n and apply in the known formula \(\dfrac{n \times (2n+1) \times (2n1)}{3}\). Let us get the proof as follows:
Σ(2n1)^{2 }= 1^{2} + 2^{2 }+ 3^{2} + … + (2n – 1)^{2} + (2n)^{2} – [2^{ 2} + 4^{2} + 6^{2} + … + (2n)^{2}]
Σ(2n1)^{2 }= (the sum of all the consecutive integers from 1 to 100)  (the sum of the squares of the even numbers)
<Σ(2n1)^{2 }= [1^{2} + 2^{2 }+ 3^{2} + … + (2n – 1)^{2} + (2n)^{2} ]– [2^{2}+ 4^{2} + 6^{2}+ … + (2n)^{2}]
On applying the formula for the addition of squares of 2n natural numbers and of n even natural numbers, we get;
Σ(2n1)^{2} = \(\dfrac{2n}{6} (2n +1) (4n +1)  \dfrac{2n}{3} (n+1)(2n+1)\)
= \(\dfrac{n}{3}[(2n+1) (4n+1)]  \dfrac{2n}{3}[(n+1)(2n+1)]\)
Taking out the common terms, we get;
Σ(2n1)^{2 }= \([\dfrac{n}{3}(2n+1)] (4n+1) 2(n+1)\)
= \([\dfrac{n}{3}(2n+1)] (4n+1 2n2)]\)
Σ(2n1)^{2} = \(\dfrac{n}{3}[(2n+1)(2n1)]\)
Related Topics on Sum of Squares
Check out these interesting articles to know more about the sum of squares and its related topics in math.
Important Notes
 Find the sum of squares of n numbers using the formulas.
 Use the known algebraic identities to prove the formulas of the sum of squares.
Important Topics
Solved Examples on Sum of Squares

Example 1: Find the sum of the squares of 19 and 22 directly and using the formula. Verify your answers.
Solution:
19^{2 }+ 22^{2 }= 361 + 484 = 845
Using the formula a^{2} + b^{2 }= (a +b)^{2} 2ab, we get 19^{2 }+ 22^{2 }= (19 + 22)^{2}  2 × 19 × 22
= 41^{2 } 836
= 1681 836 = 845
Thus verified. 
Example 2: What are the 3 consecutive numbers if the sum of their squares is 77?
Solution:
Let n be a number. We are required to find n, n+1 and n+2.
Given that n^{2} + (n+1)^{2} +(n+2)^{2} = 77
n^{2} + n^{2} + 2n + 1+ n^{2} +4n +4 = 77
3n^{2} + 6n +5 = 77
3n^{2} + 6n = 72
Dividing by 3 we get, n^{2} + 2n  24 = 0
Solving the quadratic equation, we get (n+6)(n4) = 0
since n cannot be negative, we have n = 4. Thus the required numbers are 4, 5 and 6. 
Example 3: Find the sum of the squares of the first 30 odd numbers.
Solution:
We know by formula that the sum of the squares of n odd numbers = [n(2n+1) (2n1)]/3
Here n = 30. Substituting we get,
30(61)(59)/ 3 = 35990
FAQs on Sum of Squares
How do you Calculate the Sum of Squares?
The sum of the squares of n positive integers is calculated using the formula [n(n+1)(2n+1)]/6.
What is the Sum of Square Numbers?
Sum of square numbers is given as a^{2}+ b^{2}+c^{2} +..... up to infinity. For the sum of squares of 2 numbers and 3 numbers we use the known algebraic identities and for the sum of n natural numbers or sum of n odd numbers or even numbers we use the respective formula.
What is the Sum of Squares from 1 to 100?
To calculate the sum of the squares of the numbers from 1 to 100, we apply the sum of squares formula for natural numbers, i.e., 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}+ ....n^{2 }=[n(n+1)(2n+1)]/6, so, 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}+ .....100^{2 }= [100(101)(201)]/6 = 338350.
What is the Sum of 1 to n?
The sum of 1 to n is the sum of n Natural numbers and is given by n(n+1)/2.
What is the Sum of First N Odd Numbers?
The sum of the first N odd numbers is given by n^{2}. The sum of first 5 odd numbers = 5^{2} = 25.
What is the Formula of Sum of n Natural Numbers?
The sum of natural numbers is derived with the help of arithmetic progression. Hence, the formula is
Sum of Natural Numbers Formula = [n(n+1)]/2
where n is the natural number.
What is the Formula for the Sum of the First n Even Natural Numbers?
The sum of the first n even natural number formula is n(n + 1). Here is how we derived at it:
n is even natural numbers that are 2,4,6,………., 2n which forms an arithmetic progression. Here, a = 2, d = 4  2 = 2
By using the arithmetic progression formula ,we get, Sn = n/2 [2a + (n – 1) d]
Sn = n/2 [2(2) + (n 1) 2]
Sn = n/2 [4 + 2n – 2]
Sn = n/2 × 2 [2 + n – 1]
Sn = n (n + 1)
Hence, Sum of Even Natural Numbers is n(n + 1).