Sum of Squares of n Natural numbers

The sum of the squares of n natural numbers is the result obtained by adding the squares of all numbers from 1 to n. It shows the total when each number is multiplied by itself and then added together.

The formula for the sum of squares of the first n natural numbers is:

Examples:
n = 1, Sum = (1 x (1 + 1) x (1 x 2 + 1))/6 = 1
n = 2, Sum = (2 x (2 + 1) x (2 x 2 + 1))/6 = 5 (Which is sum of 1¹ + 2²)
n = 3, Sum = (3 x (3 + 1) x (3 x 2 + 1))/6 = 14 (Which is sum of 1¹ + 2² + 2²)

Other Formulas

Proof

We need to prove that 1² + 2² + 3² + … + n² = n(n + 1)(2n + 1) / 6

We can prove the same using the Principle of Mathematical Induction as follows:

Step 1: Base Case (n = 1)
LHS = 1² = 1
RHS = 1(1 + 1)(2×1 + 1) / 6 = (1 × 2 × 3) / 6 = 1
Thus, LHS = RHS, so the statement is true for n = 1.

Step 2: Inductive Hypothesis
Assume the statement is true for n = k:
1² + 2² + 3² + … + k² = k(k + 1)(2k + 1) / 6 ...(1)

Step 3: Inductive Step (n = k + 1)
We need to prove:
1² + 2² + … + k² + (k + 1)² = (k + 1)(k + 2)(2k + 3) / 6

Starting from LHS:
= [k(k + 1)(2k + 1)] / 6 + (k + 1)² (using (1))
= (k + 1)/6 [k(2k + 1) + 6(k + 1)]
= (k + 1)/6 (2k² + k + 6k + 6)
= (k + 1)/6 (2k² + 7k + 6)
= (k + 1)/6 (2k + 3)(k + 2)
= (k + 1)(k + 2)(2k + 3) / 6 = RHS

Since the statement is true for n = 1 and true for n = k implies true for n = k + 1,
therefore, by mathematical induction,
1² + 2² + 3² + … + n² = n(n + 1)(2n + 1) / 6 is true for all natural numbers n.

Alternative Proof (Telescoping Method)

n³ − (n − 1)³ = (n − (n − 1))[n² + n(n − 1) + (n − 1)²]
= 1[n² + n² − n + n² − 2n + 1] = 3n² − 3n + 1 ...(1)

Similarly,
(n − 1)³ − (n − 2)³ = 3(n − 1)² − 3(n − 1) + 1
(n − 2)³ − (n − 3)³ = 3(n − 2)² − 3(n − 2) + 1
…
2³ − 1³ = 3(2)² − 3(2) + 1
1³ − 0³ = 3(1)² − 3(1) + 1

Adding all terms:
n³ − 0³ = 3∑n² − 3∑n + n

Since ∑n = n(n + 1)/2,

n³ = 3∑n² − 3[n(n + 1)/2] + n

⇒ 3∑n² = n³ + 3n(n + 1)/2 − n

= n[(2n² + 3n + 1)/2]

⇒ ∑n² = n(2n² + 3n + 1)/6
= n(n + 1)(2n + 1)/6

Solved Questions

Question 1: Find the sum of the squares of 9 and 11.

Solution:

Using formula a² + b² = (a +b)² - 2ab,
= 9² + 11²
= (9 + 11)² - 2 × 9 × 11
= 20² - 198
= 400- 198 = 202...(i)

Verification
9² + 11² = 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,
n² + (n+1)² +(n+2)² = 50
⇒ n² + n² + 2n + 1+ n² + 4n + 4 = 50
⇒ 3n² + 6n + 5 = 50
⇒ 3n² + 6n = 45
Dividing by 3,
n² + 2n = 15
⇒ n² + 2n - 15 = 0
⇒ n² + 5n - 3n - 15 = 0
⇒ (n + 5)(n - 3) = 0

According to question, 'n' cannot be negative.
Thus, n - 3 = 0
⇒ 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)(2n-1)]/3
Here n = 14
[n(2n+1)(2n-1)]/3
= 14(29)(27)/3
= 3654

Question 4: Find the sum of the squares of 6 and 12.

Solution:

Using the formula a² + b² = (a +b)² - 2ab, where a = 6 and b = 12:
6² + 12² = (6 + 12)² - 2 × 6 × 12
= 18² - 144
= 324 - 144
= 180

Verification:

6² + 12² = 36 + 144
=180
The formula is verified.

Question 5: Find 3 consecutive natural numbers if the sum of their squares is 50.

Solution:

Let the three numbers be n, n+1, and n+2. Given:

n² + (n+1)² + (n+2)² = 50
n² + n² + 2n + 1 + n² + 4n + 4 = 50
3n² + 6n + 5 = 50
3n² + 6n = 45
n² + 2n = 15
n² + 2n - 15 = 0
(n + 5)(n - 3) = 0

Since n cannot be negative:
n = 3

Hence, the required numbers are 3, 4, and 5.

Practice Questions

Question 1: The square of the sum of two numbers is 100, and their product is 48. Find the sum of their squares.

Question 2: Simplify: 2² + 3² + 4² + 5² + ... + 10².

Question 3: Simplify: 1² + 3² + 5² + 7² + ... + 21².

Question 4: Find the sum of the squares of 4 consecutive natural numbers if the sum is 354.

Question 5: The sum of the squares of three consecutive natural numbers is 365. Find the numbers.