Consider any two numbers ‘R’ and ‘S’. The sum of the numbers is written as ‘R + S’ and their difference is written as ‘R - S’. The sum of squares of these two numbers can be calculated under two different cases.

Case 1: Squaring the Sum of the Numbers:

Sum of the two numbers ‘R’ and ‘S’ is written as ‘R + S’. Its square is calculated as:

(R + S)2 = (R + S) (R + S)

= R. R + R. S + S. R + S. S

= R2 + RS + RS + S2

(R + S)2 = R2 + 2 RS + S2

Transfering 2 RS to the other side of the equation, we get

i.e. The sum of squares of two numbers is equal to the difference obtained when twice the product of the numbers is subtracted from the square of their sum.

Case 2: Squaring the Difference Between the Numbers:

The difference between the two numbers ‘R’ and ‘S’ is written as ‘R - S’. Its square is calculated as:

(R - S)2 = (R - S) (R - S)

= R. R - R. S - S. R + (-S). (-S)

= R2 - RS - RS + S2

(R - S)2 = R2 - 2 RS + S2

Transfering - 2 RS to the other side of the equation, we get

i.e. Sum of squares of two numbers can also be calculated as the sum of the square of their difference and twice their product.

Sum of three numbers ‘L’, ‘M’ and ‘N’ is written as L + M + N. Its square is calculated as follows:

(L + M + N)2 = (L + M + N) (L + M + N)

= L2 + LM + LN + ML + M2 + MN + NL + NM + N2

= L2 + M2 + N2 + 2 LM + 2 MN + 2 NL

(L + M + N)2 = L2 + M2 + N2 + 2 (LM + MN + NL)

The above formula gives the sum of squares of three numbers.

The sum of squares of natural numbers formula is given as

The sum of the squares of first n natural numbers is calculated as the quotient obtained when the product of the number of natural numbers whose sum of the squares is to be determined, one more than the number of natural numbers and one more than twice the number of natural numbers is divided by the positive integer 6.

An even number is generally represented as a multiple of 2. Let us consider an even number ‘2p’. The sum of squares of even numbers is calculated by substituting 2p in the place of ‘p’ in the formula for finding the sum of squares of first n natural numbers. In this case n = p.

Sum of squares of natural numbers formula is

\[\sum p^{2}\] = \[\frac{p(p+1)(2p+1)}{6}\]

Substituting p = 2p in the above equation, we get

\[\sum(2p)^{2}\] = \[\sum 2^{2}.p^{2}\] = \[4\sum(p)^{2}\]

\[\sum (2p)^{2}\] = \[4(\frac{n(n+1)(2n+1)}{6})\]

\[\sum (2p)^{2}\] = \[\frac{2n(n+1)(2n+1)}{3}\]

The general representation of an odd number is given as ‘(2p - 1)’. The sum of squares of odd numbers can be calculated as:

\[\sum (2p-1)^{2}\] = \[\frac{p(2p+1)(2p-1)}{3}\]

A square number always ends with 0, 1, 4, 5, 6, and 9. Any number with digits 2, 3, 7, and 8 are not square numbers.

All counting numbers are called natural numbers.

The numbers that are completely divisible by two are called even numbers. The numbers that have ‘1’ as a remainder when divided by two are called odd numbers.

FAQ (Frequently Asked Questions)

1. How do we Calculate the Sum of Squares of the First 100 Natural Numbers?

The sum of the squares of first n natural numbers is calculated by a standard formula in Mathematics. The sum of squares of natural numbers formula is given as:

∑p² = [ {p(p+1)(2p+1)}/{6} ]

In the above formula, ‘p’ is the number of natural numbers whose sum is to be calculated and the symbol Σ indicates the sum of the squares of ‘p’ natural numbers.

If we have to calculate the sum of squares of the first 100 natural numbers, the value of ‘p’ should be taken as 100. So, the sum of squares of the first 100 natural numbers is calculated as:

∑p² = [ {100(100 + 1)(2(100 + 1))}/{6} ]

∑p² = [ {100(101)(201)}/{6} ]

∑p² = [ 2030100/6 ] = 338350

So, the sum of squares of first n natural numbers where n = 100 is 338350.

2. How is the Sum of Squares of two Numbers Calculated when the Sum and Difference Between the Numbers are Given?

Consider two numbers ‘m’ and ‘n’. Sum of the squares of these two numbers can be calculated in two different ways.

**Case 1: Considering the square of sum of the numbers**

According to algebraic identities, it is evident that the square of the sum of any two numbers is equal to the sum of the squares of the two numbers added to twice the product of the numbers. So, the sum of the squares of two numbers can, therefore, be calculated as the difference obtained when twice the product of the numbers is subtracted from the square if their sum.

m^{2} + n^{2} = (m + n)^{2} - 2 mn

**Case 2: Considering the square of difference between the numbers**

The algebraic identity to find the square of difference between two numbers is as follows: The square of the difference between two numbers is equal to the difference obtained when twice the product of the two numbers is subtracted from the sum of their squares. By slight alterations in this formula, the sum of squares of two numbers can be calculated as the sum of the square of the difference between the two numbers and twice their product.

m^{2} + n^{2} = (m - n)^{2} + 2 mn