Sum of Squares of n natural numbers is calculated using the formula [n(n+1)(2n+1)] / 6 where ‘n’ is a natural number.
In this article, we have covered Sum of squares of the ‘n’ natural number formula, its Sum of squares proof, related examples and others in detail.
Table of Content
 What is Sum of Squares of ‘n’ Natural Numbers?
 Sum of Squares of Natural Numbers Proof
 Sum of Squares of Even and Odd Natural Numbers
 Sum of Squares of Two and Three Natural Numbers
 Sum of Squares in Geometry
 Sum of Squares of n Natural Numbers Solved Questions
 Sum of Squares of n Natural Numbers – FAQs
What is Sum of Squares of ‘n’ Natural Numbers?
Sum of squares of ‘n’ natural numbers is the sum of squares of all natural numbers. Its formula is, Sum = 1^{2} + 2^{2} + 3^{2} + … + n^{2} = [n(n+1) (2n+1)]/6. Sum of Squares of First ‘n’ Natural Numbers is represented as S_{n} and the formula for the same is added in the image below:
Sum of Squares of n Natural Numbers Formula
Formula for Sum of Squares of ‘n’ Natural Numbers, Sum of Squares of First ‘n’Â Even Numbers, and Sum of Squares of First ‘n,Â Odd Numbers are added in the table below:
Sum of Squares Formula 


Sum of Squares of First ‘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 
Check: Python Program for Sum of squares of first n natural numbers
Sum of Squares of Natural Numbers Proof
Formula for Sum of Squares of Natural Numbers is,
Î£n^{2} = [n(n+1)(2n+1)]/6
We can prove the same using Principal of Mathematical Induction as:
Let, P(n): 1^{2} + 2^{2} + 3^{2} + … + n^{2} = [n(n+1)(2n+1)]/6
For P(1)
LHS = 1^{2} = 1
RHS = [1(1 + 1)(2(1) + 1)]/6 = (1Ã—2Ã—3)/6 = 6/6 = 1
So, LHS = RHS
Thus, P(1) is true
Let’s take P(k) to be true, i.e.,
1^{2} + 2^{2 }+ 3^{2} + … + k^{2} = [k(k+1)(2k+1)] / 6 is true…(i)
Now to prove that P(k+1) is true, i.e.
1^{2} + 2^{2} + 3^{2} + … + (k+1)^{2} = [(k+1)(k+2)(2k+3)]/6 is true
LHS = 1^{2} + 2^{2} + 3^{2} + … + (k+1)^{2}
â‡’ LHS = 1^{2} + 2^{2} + 3^{2 }+ … + k^{2} + (k+1)^{2}
â‡’ LHS = [k(k+1)(2k+1)] /6 + (k+1)^{2} …[from (i)]
â‡’ LHS = (k+1)/6 Ã— [k(2k+1) + 6(k+1)]
â‡’ LHS = (k+1)/6 Ã— [2k^{2} + k + 6k + 6]
â‡’ LHS = (k+1)/6 Ã— (2k^{2} + 7k + 6)
â‡’ LHS = (k+1)/6 Ã— (2k2 + 4k + 3k + 6)
â‡’ LHS = (k+1)/6 Ã— [2k(k + 2) + 3(k + 2)]
â‡’ LHS = (k+1)/6 Ã— (2k+3)(k + 2)
â‡’ LHS = [(k+1)(k+2)(2k+3)] / 6
â‡’ LHS = RHS
So, P(k+1) is true.
Thus, we can say that P(n) is true for all natural numbers ‘n’ by the Principle of Mathematical Induction
Sum of Squares of Even and Odd Natural Numbers
Formula for sum of square of even and odd number is covered below:
Sum of Squares of Even Natural Numbers
Formula for sum of square of even natural number is:
Sum = 2n(n+1)(2n+1)/3
This is calculated as:
 Sum = 2^{2}+ 4^{2}+ 6^{2} +… + (2n)^{2}
 Taking 2^{2} or 4 commons from each term
 Sum = 4 Ã— {1^{2} + 2^{2} + 3^{2} +…+ (n)^{2}}…(i)
 We know that sum of ‘n ‘Natural Number is given by:
 1^{2} + 2^{2} + 3^{2} +…+ (n)^{2} = n(n + 1)(2n + 1)/6…(ii)
 From eq. (i) and eq. (ii)
 Sum = 4n(n+1)(2n+1)/6
Sum = 2n(n+1)(2n+1)/3
Example: Find the sum of squares of the first 3 even numbers.
We know that,
Sum = 2n(n+1)(2n+1)/3
n = 3
Sum = 2Ã—3(3+1)(2Ã—3+1)/3
Sum = 56…(i)
Verification:
Sum = 2^{2} + 4^{2} + 6^{2}Â
â‡’ Sum = 4 + 16 + 36Â
â‡’ Sum = 56…(ii)
From eq(i) and eq(ii) Formula is Verified
Sum of Squares of Odd Natural Numbers
Formula for sum of square of odd natural number is:
Sum = n(2n+1)(2n1)/3
This is calculated as:
 Sum = 1^{2}+ 3^{2}+ 5^{2} +… + (2n – 1)^{2}
 â‡’ Sum = (Sum of Squares of 2n numbers) – (Sum of Squares of First n Even Numbers)
 â‡’ Sum = 2nÃ—(2n+1)Ã—(2Ã—2n + 1)/6 – 2n(n+1)(2n+1)/3
 â‡’ Sum = 2n(2n+1)/6 [4n+1 – 2(n+1)]
 â‡’ Sum = n(2n+1)/3 Ã— (2n1)
Sum = n(2n+1)(2n1)/3
Example: Find the sum of squares of the first 3 odd numbers.
We know that,
Sum = n(2n+1)(2n1)/3
n = 3
â‡’ Sum = 3(2Ã—3 + 1)(2Ã—3 – 1)/3
â‡’ Sum = 35…(i)
Verification:
Sum = 1^{2} + 3^{2} + 5^{2}Â
â‡’ Sum = 1 + 9 + 25
â‡’ Sum = 35…(ii)
From eq(i) and eq(ii) Formula is Verified
Sum of Squares of Two and Three Natural Numbers
Finding sum of square for smaller numbers is easy but finding sum od square for bigger numbers can be tricky some time and we can easily calculate their some by using some algebric identities. Let’s take two number ‘a’ and ‘b’ then sum of square of ‘a’ and ‘b’ is found as:
a^{2} + b^{2} = (a+b)^{2} – 2ab
For three numbers ‘a’, ‘b’, and ‘c’ then its square is found using algebric identities:
a^{2 }+ b^{2} + c^{2} = (a + b + c)^{2} – 2ab – 2bc – 2ca
Sum of Squares in Geometry
Sum of Square is geometry are generally seen in right angled triangle, take a right angle triangle PQR as shown in the image added below, and then using Pythagoras Theorem we can say that:
PQ^{2} + QR^{2} = PR^{2}
We can also represent the Pythagoras Thorem as:
(Base)^{2} + (Perpendicular)^{2} = (Hypotenuse)^{2}
Related Articles 


Conclusion of Sum of Squares of n Natural Numbers
Sum of the squares of the first n natural numbers is a crucial mathematical concept that finds applications in various areas such as geometry, physics, and statistics. It represents the cumulative total obtained by squaring each natural number from 1 to n and then summing these squared values. This sum is fundamental for calculating quantities like areas, moments of inertia, and statistical measures. Understanding the concept of summing the squares of natural numbers enhances mathematical reasoning and problemsolving skills across different fields of study.
Sum of Squares of n Natural Numbers Solved Questions
Question 1: Find the sum of the squares of 9 and 11.
Solution:
Using formula a^{2} + b^{2} = (a +b)^{2} – 2ab,
= 9^{2} + 11^{2 }
= (9 + 11)^{2} – 2 Ã— 9 Ã— 11
= 20^{2} – 198
= 400 198 = 202…(i)
Verification
9^{2} + 11^{2 }= 81 + 121 = 202…(ii)
From eq(i) and (ii) verified.
Question 2: Find 3 consecutive natural numbers if the sum of their squares is 50?
Solution:
Let the three number be, n, n+1 and n+2.
Given,
n^{2} + (n+1)^{2} +(n+2)^{2} = 50
â‡’ n^{2} + n^{2} + 2n + 1+ n^{2} + 4n + 4 = 50
â‡’ 3n^{2} + 6n + 5 = 50
â‡’ 3n^{2} + 6n = 45
Dividing by 3,
n^{2} + 2n = 15
â‡’ n^{2} + 2n – 15 = 0
â‡’ n^{2} + 5n – 3n – 15 = 0
â‡’ (n + 5)(n – 3) = 0
According to question, ‘n’ cannot be negative.
Thus, n – 3 = 0
â‡’ n = 3
Hence, required number are,
n, n+1 and n+2 = 3, 4, and 5
Question 3: Find the sum of the squares of the first 14 odd numbers.
Solution:
Formula for sum of the squares of n odd numbers = [n(2n+1)(2n1)]/3
Here n = 14
[n(2n+1)(2n1)]/3
= 14(29)(27)/3
= 3654
Practice Questions on Sum of Squares of n Natural Numbers
Q1: Square of sum of 2 numbers is 100. The product of the numbers is 48. What is the sum of their squares?
Q2: Simplify: 2^{2} + 3^{2} + 4^{2} + 5^{2}…….10^{2}
Q3: Simplify: 20^{2} + 30^{2} + 40^{2} + 50^{2}…….100^{2}
Q4: Simplify: 1^{2} + 3^{2} + 5^{2} + 7^{2}…….21^{2}
Sum of Squares of n Natural Numbers – FAQs
What is the formula for the sum of squares of n natural numbers?
Sum of squares of n natural numbersÂ is calculated using the formula, Î£n^{2Â }= 1^{2}Â + 2^{2}Â + 3^{2}Â + … + n^{2}Â = [n(n+1)(2n+1)] / 6.
What is the sum of square numbers in algebra?
Sum of square in algebra is an identity that is some times used to solve various problems these identities are,
 a^{2Â }+ b^{2}Â = (a + b)^{2}Â – 2ab
 a^{2Â }+ b^{2Â }+ c^{2}Â = (a + b + c)^{2}Â – 2ab 2bc 2ca
What is the sum of squares of 100 natural numbers from 1 to 100?
Sum of Squares of 100 Natural numbers from 1 to 100 is 338,350.
How to Calculate the sum of squares up to n?
Sum of squares = n(n + 1)(2n + 1) / 6
What is the formula of sum of squares?
Formula for sum of square is, a^{2} + b^{2} and is calculated as a^{2} + b^{2} = (a + b)^{2} – 2ab.
What is the sum of squares of even natural numbers?
Formula for sum of squares of even natural numbers is given as, Î£(2n)^{2}Â = [2n(n + 1)(2n + 1)] /3.
What is the sum of squares of odd natural numbers?
Formula for sum of squares of odd natural numbers is given as, Î£(2n – 1)^{2}Â = [n(2n + 1)(2n – 1)] /3.
What are the 2 natural numbers, the sum of whose squares is 52?
Let’s denote the two natural numbers as x and y. According to the given condition, we have: x^{2}+y^{2} = 52. We need to find two natural numbers whose sum of squares is 52. By trying different combinations, we find that 4^{2}+6^{2 }= 16+36 = 52. So, the two natural numbers are 4 and 6.
The sum of the squares of three consecutive natural numbers is 2030. what is the middle number?
Let’s denote the three consecutive natural numbers as xâˆ’1, x, and x+1. According to the given condition, we have: (xâˆ’1)^{2} + x^{2} + (x+1)^{2} =2030
Expanding and simplifying the equation, we get:
3x^{2}+2 = 2030
3x^{2 }= 2028
x^{2 }= 676
x = 26
So, the middle number is x = 26.
Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference?
Let’s denote the two natural numbers as x and y. According to the problem, we have the following equations:
x^{2}+y^{2 }= 25(x+y) ……………(1)
x^{2}+y^{2 }= 50(xâˆ’y) ……………(2)
Combining equations (1) and (2), we get:
25(x+y) = 50(xâˆ’y)
â‡’ x+y = 2(xâˆ’y) = 2xâˆ’2y
â‡’ y+2y = 2xâˆ’x
â‡’ 3y = x
Thus, x=3y. Substituting x in equation (1), we get:
x^{2}+y^{2 }= 25(x+y)
â‡’ (3y)^{2 }+ y^{2 }= 25(3y+y)
â‡’ 9y^{2}+y^{2 }= 25Ã—4y
â‡’ 10y^{2 }= 100y
â‡’ y^{2 }=10y
â‡’ y=0Â orÂ 10
However, y cannot be 0 as it’s a natural number. Therefore, y=10 and consequently x=3y=30.
Hence, the two numbers are 10 and 30.