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# Sum of Squares

We can find squares of natural numbers. Whether they are even, odd, prime or composite numbers, we can find their squares. This means we can multiply the number by itself to find another value known as a square of a number. We can also find the squares of added and subtracted numbers using algebraic formulas. Let us study these formulas.

## Sum of squares formula for two and three numbers

It is easier to find squares of smaller numbers than larger numbers. We can simply add the numbers and multiply them by themselves to get the final result. Let a, b and c be the three numbers that we wish to add.

The formula for sum of squares of two digits = a² + b² = (a +b)² - 2ab

The formula for sum of squares of three digits = a² + b² + c² = (a + b + c)² - 2ab - 2bc - 2ca

## Sum of squares of natural numbers

Let n be the number of natural numbers whose sum of squares we need to find out. The sum of squares is represented by Σn². Different formulas for natural numbers are:

## Derivation of sum of squares of n natural numbers formula

We know the identity, a³ - b³ = (a-b) (a² + ab + b²). Replacing a with n, and b with (n-1), then we have,

n³ - (n - 1)³ = (n – n + 1) (n² + n (n - 1) + (n - 1)²)

n³ - (n - 1)³ = (n² + n² – n + (n - 1)²)

= (2n² - n + n² + 1 - 2n)

= 3n² - 3n + 1

n³ - (n - 1)³ = 3n² - 3n + 1   (1)

(n - 1)³ - (n - 2)³ = 3 (n - 1)² – 3 (n - 1) +1   (2)

(n - 2)³ - (n - 3)³ = 3 (n - 2)² – 3 (n - 2) +1   (3)

Repeating the steps to n terms, then the last term will be,

2³ - 1³ = 3 (2)² - 3(2) +1

1³- 0³ = 3 (1)² - 3(1) +1   (last step)

(1)+ (2) + (3) +............+ (the last step)

By adding all the above steps, we get, n³ - 0³ = 3 Σ n2 – 3 Σn + n

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

After factoring the equation, we get,

Σn2 = n (2n+1) (n+1) / 6

Example: Find the sum of squares of odd numbers 1, 3, and 5.

Solution:

From the sum of squares of n odd number formula, we have,

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