Sum of Squares of n Natural Numbers
The sum of squares of n natural numbers can be calculated using the formula [n(n+1)(2n+1)] / 6. Let n be a natural number. Squaring the number is denoted by n^{2}. The sum of squares of n natural numbers means the sum of the squares of the given series of natural 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.
In this article, we will discuss the formula to calculate the sum of squares of n natural numbers and derive using the principle of mathematical induction. We will also discuss the formula to find the sum of squares of even and odd natural numbers, and the sum of squares in geometry. We will also solve a few examples for a better understanding of the concept.
What is Sum of Squares of n Natural Numbers?
Let us first recall the meaning 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 these numbers is represented as Σn^{2}, where n ranges from 1 to infinity. We can find the sum of squares of the first n natural numbers using the formula, SUM = 1^{2} + 2^{2} + 3^{2} + ... + n^{2} = [n(n+1)(2n+1)] / 6. We can prove this formula using the principle of mathematical induction. Let's go through the formulas of finding the sum of squares of even and odd natural numbers in the next section.
Sum of Squares of n Natural Numbers Formula
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  [n(n+1)(2n+1)] / 6 

Sum of squares of first n even numbers  [2n(n + 1)(2n + 1)] / 3 
Sum of squares of first n odd numbers  [n(2n+1)(2n1)] / 3 
Sum of Squares of Two and Three Natural 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 Proof
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 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. If we need to calculate the sum of squares of n consecutive natural numbers, the formula is Σn^{2} = [n(n+1)(2n+1)] / 6. It is easy to apply the formula when the value of n is known. Let us prove this formula using the principle of mathematical induction.
Let P(n): 1^{2} + 2^{2} + 3^{2} + ... + n^{2} = [n(n+1)(2n+1)] / 6
Consider P(1). LHS = 1^{2} = 1, RHS = [1(1+1)(2(1)+1)] / 6 = (1 × 2 × 3) / 6 = 6/6 = 1. So, LHS = RHS. Therefore, P(1) is true.
Assume P(k) is true, i.e., 1^{2} + 2^{2} + 3^{2} + ... + k^{2} = [k(k+1)(2k+1)] / 6 holds true.  (1)
Now we will prove that P(k+1) is true, that is, we need to prove that 1^{2} + 2^{2} + 3^{2} + ... + (k+1)^{2} = [(k+1)(k+2)(2k+3)] / 6 is true.
Consider LHS = 1^{2} + 2^{2} + 3^{2} + ... + (k+1)^{2}
= 1^{2} + 2^{2} + 3^{2} + ... + k^{2} + (k+1)^{2}
= [k(k+1)(2k+1)] / 6 + (k+1)^{2}  [Using (1)]
= (k+1)/6 × [k(2k+1) + 6(k+1)]
= (k+1)/6 × [2k^{2} + k + 6k + 6]
= (k+1)/6 × (2k^{2} + 7k + 6)
= (k+1)/6 × (2k^{2} + 4k + 3k + 6)
= (k+1)/6 × [2k(k + 2) + 3(k + 2)]
= = (k+1)/6 × (2k+3)(k + 2)
= [(k+1)(k+2)(2k+3)] / 6
= RHS
So, P(k+1) is true.
Thus, we can say that P(n) is true for all natural numbers n. So, we have 1^{2} + 2^{2} + 3^{2} + ... + n^{2} = [n(n+1)(2n+1)] / 6. Hence, we have derived the formula for the sum of squares of n natural numbers.
Alternate Proof:
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}  [3n(n+1)/2]+ n [ since Σ n = n(n+1)/2 (sum of n natural numbers)]
3 Σ n^{2 }= n^{3 }+ [3n(n+1)/2]  n
3 Σ n^{2 }= n[n^{2} + 3(n+1)/2  1]  (Taking n as common from RHS)
Σ n^{2 }= (n/3)( n^{2} + (3n+3)/2 1)
= (n/6) (2n^{2}+ 3n + 1)
(Factorizing the quadratic equation)
Σ n^{2 }= 1^{2} + 2^{2} + 3^{2} + ... + n^{2} = [n(n+1)(2n+1)] / 6
Sum of Squares of Even Natural Numbers
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 it in the known formula [2n(n + 1)(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} = 4 [n(n+1)(2n+1)] / 6 (Formula for sum of squares of n natural numbers)
Thus Σ(2n)^{2} = [2n(n + 1)(2n + 1)] / 3
Sum of Squares of Odd Natural Numbers
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 [n(2n+1)(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 2n)  (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} = [2n(2n+1)(4n+1)] / 6  [2n(n + 1)(2n + 1)] / 3
= [n(2n+1)(4n+1)] / 3  [2n(n + 1)(2n + 1)] / 3
Taking out the common terms, we get;
Σ(2n1)^{2 }= (n/3) (2n+1) [4n+1  2(n+1)]
= (n/3) (2n+1) (4n+12n2)
= [n(2n+1)(2n1)] / 3
Σ(2n1)^{2} = [n(2n+1)(2n1)] / 3
Sum of Squares in Geometry
As we know, in a rightangled triangle, the sum of squares of the perpendicular and the base is equal to the square of the hypotenuse. This result is known as the Pythagoras theorem. So, we have the sum of squares in geometry given by,
Base^{2} + Perpendicular^{2} = Hypotenuse^{2}
Important Notes on Sum of Squares of n Natural Numbers
 The sum of squares of n natural numbers is, Σ n^{2 }= 1^{2} + 2^{2} + 3^{2} + ... + n^{2} = [n(n+1)(2n+1)] / 6
 The sum of squares of even and odd natural numbers is given by,
 Σ(2n1)^{2} = [n(2n+1)(2n1)] / 3
 Σ(2n)^{2} = [2n(n + 1)(2n + 1)] / 3
 We can derive the formula for the sum of squares of n natural numbers using the principle of mathematical induction.
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Sum of Squares of n Natural Numbers Examples

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
Answer: Thus verified. 
Example 2: What are the 3 consecutive natural 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.
Answer: 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
Answer: The sum of the squares of the first 30 odd numbers is 35990. 
Example 4: Find the sum of squares of the first 20 natural numbers.
Solution: The sum of squares of first n natural numbers formula is Σ n^{2 }= 1^{2} + 2^{2} + 3^{2} + ... + n^{2} = [n(n+1)(2n+1)] / 6.
Here n = 20. So, we have
Σ 20^{2} = [n(n+1)(2n+1)] / 6 = [20(20+1)(40+1)] / 6 = [20(21)(41)] / 6 = 2870
Answer: The sum of squares of first 20 natural numbers is 2870.
FAQs on Sum of Squares of n Natural Numbers
What is the Sum of Squares of n Natural Numbers?
We can calculate the sum of squares of n natural numbers using the formula, Σ n^{2 }= 1^{2} + 2^{2} + 3^{2} + ... + n^{2} = [n(n+1)(2n+1)] / 6.
How do you Calculate the Sum of Squares of Natural Numbers?
The sum of the squares of n natural numbers 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. Some of the formulas are:
 a^{2 }+ b^{2} = (a + b)^{2}  2ab
 a^{2 }+ b^{2 }+ c^{2} = (a + b + c)^{2}  2ab 2bc 2ca
What is the Sum of Squares of 100 Natural numbers 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 Squares of Even Natural Numbers?
The sum of squares of even natural numbers is given by Σ(2n)^{2} = [2n(n + 1)(2n + 1)] / 3.
What is the Sum of Squares of Odd Natural Numbers?
The sum of squares of odd natural numbers is given by Σ(2n1)^{2} = [n(2n+1)(2n1)] / 3.
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, S_{n} = n/2 [2a + (n – 1) d]
S_{n} = n/2 [2(2) + (n 1) 2]
S_{n} = 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).
What is the Sum of Squares in Geometry?
We have the sum of squares in a right triangle given by Base^{2} + Perpendicular^{2} = Hypotenuse^{2}
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