Sum of Squares of n Natural Numbers

The sum of the squares of n natural numbers means the total of the squares of first n natural numbers, where n is the natural numbers starting from 1. 

To find the sum of squares of n natural numbers, we use the formula:-
S = 12 + 22 + 32 + 42 + …. + n2 = n (n+1) (2n +1)6 

In this, n represents the natural numbers. To find the sum of squares, we square the numbers one by one, adding them together using the ‘+’ sign.

For example, we square the numbers 1, 2, and 3 as 12, 22, and 32 as 1, 4, and 9 to find the sum as 1 + 4 + 9 = 14.

Now find the sum using the formula.

Since there are a total of 3 numbers in the series, ‘n’ = 3.

Apply the value of n in the formula:- S = n (n+1) (2n +1)6

S = 3(3+1) ((2 x 3) +1)6 = 3 x 4 x 76 = 12 x 76 = 846 = 14

From this, we can say that by counting the total number of numbers given in the series, we can easily find the sum of squares using the formula. 

Here, in this section, we will learn how to derive the formula to find the sum of squares like 1² + 2² + 3² + … 

The formula used is 1² + 2² + 3² + … + n² =  n (n+1) (2n +1) / 6 

To see how the formula is derived, we use of the process called mathematical induction. It makes sure that the formula works for one number, proving that it applies to other numbers as well. Follow the steps given below:-

Step 1: Check if the formula works for number 1. Therefore, we can take the value of ‘n’ as 1. 

LHS is 1² = 1

RHS = n (n+1) (2n +1) / 6 

Now substituting the value of ‘n’ as 1 in the formula, we get 1 (1+1) ((2 x 1) +1) / 6 =
1 x 2 x 36 =  6/6 = 1

Since both the LHS and RHS are the same, we can say that the formula works for n = 1

Step 2: Now let’s check if the formula works for the number ‘k’.
Assume that,

1² + 2² + 3³ + … + k² = k(k+1) (2k + 1) / 6

Since we are assuming for number ‘k’ to be true, this process is called the inductive hypothesis.
Now we have to check if it works for the number ‘k + 1’

Step 3: Let’s prove that the formula works for ‘k + 1’
1² + 2² + 3³ + … + (k + 1)² =  (k+1) (k+2) (2k+3) / 6

The sum of squares up to k is 1² + 2² + 3³ + … + k² = k(k+1) (2k + 1) / 6

Now we can add (k + 1)² to both sides:-

1² + 2² + 3² + … + k² + (k + 1)² = k(k+1) (2k + 1) / 6 + (k + 1)²

Step 4: Take out the common factor (k + 1) from both side
k+1/ 6 [k (2k + 1) + 6 (k +1)]
By simplifying the brackets, we get:-

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

Step 5: Now factor (k+2) from (2k² + 7k + 6) by splitting them
By doing so, we get:-
k+1 6 [2k (k + 2)  + 3 (k + 2)]
Factoring out (k + 2), we get:-

k+1 6 (k + 2) (2k + 3)

Step 6: We get the final answer as:- (k+1) (k + 2) (2k + 3) / 6

Since the formula is true for n=1, and we've shown that if it works for any number k, it also works for k+1, we can conclude that the formula is true for all natural numbers. Therefore, the sum of squares formula is:

S = 1² + 2² + 3² + 4² + …. + n² = n (n+1) (2n +1) / 6 

The sum of squares of n natural numbers starting from other 1 can be calculated easily by using the formula. If we want to find the sum ranging from a to b, a ≠ 1, and the largest number will be b
For example, let’s find the sum of squares of numbers 5, 6, 7, and 8

To find the sum of squares other than 1, we use the formula for the full range, that is we find the sum of squares ranging from 1 to 8 in this case. Then we subtract the sum of squares for the numbers before the starting point, which is 1 to 4 in this case.

Using the formula, we find the sum of squares from 1 to 8 as:-
 S = n (n+1) (2n +1)6 = 8 (8+1) ((2 x 8) +1)6  = 8 x 9 x 176 = 204
The sum of squares from 1 to 4 is:-
S = n (n+1) (2n +1)6 = 4 (4+1) ((2 x 4) +1)6 = 4 x 5 x 96 = 30

Now, Sum of squares from 5 to 8 = Sum of squares from 1 to 8 − Sum of squares from 1 to 4
Sum of squares from 5 to 8 = 204 - 30 = 174

Natural numbers are positive integers starting from 1. It goes on like 1, 2, 3, 4, 5, 6, … and so on. To find the sum of n Natural numbers, we use the formula

Sum = n(n+1)2 
 For example, find the sum of the first 10 natural numbers
The first ten natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10
So the value of n is 10
Applying the value of n in the formula, we get:-
Sum = 10(10+1)2 = 10 x 112 = 55

Any number which is divisible by 2 is even. The even number goes like 2, 4, 6, 8, …
To find the sum of even numbers, use the formula:- 
S = n (n+1)
Here, n is the number of even numbers
For example, find the sum of even numbers 2, 4, 6, and 8
The value of n is 4
Applying the value of n in the formula, we get:-
S = n(n+1) = 4 (4+1) = 4 × 5 = 20

Any number which is not divisible by 2 is odd. The odd number goes like 1, 3, 5, 7, 9, …

Sum of odd numbers = n2, where n is the total number of odd numbers given.
For example, the sum of odd numbers 1, 3, 5, 7, 9 will be:-
The value of n is 5
Sum = n2 = 52 = 5 × 5  = 25

If the GP is finite, use the formula:
Sum = 1-rn1-r , where r ≠ 1
Here ‘r’ is the common ratio and ‘n’ is the number of terms

For example, the terms of the sequence are 3, 6, 12, 24. The common ratio ‘r’ is 2
To find the common ratio, we divide any term by the previous term
6/3 = 2
12/6 = 2
24/12 = 2
Formula for finite GP:- Sum = 1-rn1-r = 1-241-2 = 1-16-1 = -15-1 = 15

If the GP is not finite, use the formula:-
Sum = a1-r , where r < 1. In infinite GP, the terms get reducing because ‘r’ is less than 1.
Here, ‘a’ is the first term, and ‘r’ is the common ratio

For example, let’s take ‘a’ as 5 and ‘r’ as 12
Since, ‘r’ is  12, the GP will be 5,  52 , 54, 58, …
The next term is obtained by multiplying the current term with 12
Therefore, the sum = a1-r = 51-12 = 5 × 2 = 10

In arithmetic progression (AP), each number in the sequence increases by the same amount called the common difference.

To find the sum of n terms of AP, use the formula:-
Sum = n2  × (2a + (n - 1) × d)
In the formula,
‘a’ is the first term
‘n’ is the number of terms
‘d’ is the common difference
For example, find the sum if the sequence is 11, 13, 15
From the sequence we find the common difference as 2
Since there are three terms in the sequence, the value of ‘n’ will be 3
The value of ‘a’ is 11
Now substituting the values of a, n and d in the formula, we get:-
Sum =  n2  × (2a + (n - 1) × d) =  32  × ((2 × 11) + (3 - 1) × 2) =  32  × (22 + 4) =  32  × 26 = 3 × 13 = 39

Even though it is easy to find the sum using the formula, there will be situations where you will feel that it’s tricky to solve the problems. To tackle this situation, few tips and tricks are given below:-

Always remember that ‘n’ means the number of terms in the sequence. If ‘n’ found correctly, then apply the value of ‘n’ in the formula:-
S = n (n+1) (2n +1)6

Always start with small ‘n’ for practice and then move on to bigger ‘n’. By doing so, you will gain confidence to solve the problem.

Keep in mind that natural numbers are always positive and starts from 1.

• Natural Numbers: Positive integers that range from 1 to infinity. For example, natural numbers goes on like 1, 2, 3, 4, 5, and so on.

• Sum of Even Numbers: Total of numbers that are divisible by 2

• Sum of Odd Numbers: Total of numbers that are not divisible by 2

• Geometric Progression: Process in which each number gets multiplied by the same number to get the next number.