We will discuss here how to find the sum of the squares of first n natural numbers.

Let us assume the required sum = S

Therefore, S = 1\(^{2}\) + 2\(^{2}\) + 3\(^{2}\) + 4\(^{2}\) + 5\(^{2}\) + ................... + n\(^{2}\)

Now, we will use the below identity to find the value of S:

n\(^{3}\) - (n - 1)\(^{3}\) = 3n\(^{2}\) - 3n + 1

Substituting, n = 1, 2, 3, 4, 5, ............., n in the above identity, we get

1\(^{3}\) - 0\(^{3}\) = 3 . 1\(^{2}\) - 3 ∙ 1 + 1

2\(^{3}\) - 1\(^{3}\) = 3 . 2\(^{2}\) - 3 ∙ 2 + 1

3\(^{3}\) - 2\(^{3}\) = 3 . 3\(^{2}\) - 3 ∙ 3 + 1

4\(^{3}\) - 3\(^{3}\) = 3 . 4\(^{2}\) - 3 ∙ 4 + 1

......................................

n\(^{3}\) - (n - 1)\(^{3}\) = 3 ∙ n\(^{2}\) - 3 ∙ n + 1

______ _______

Adding we get, n\(^{3}\) - 0\(^{3}\) = 3(1\(^{2}\) + 2\(^{2}\) + 3\(^{2}\) + 4\(^{2}\) + ........... + n\(^{2}\)) - 3(1 + 2 + 3 + 4 + ........ + n) + (1 + 1 + 1 + 1 + ......... n times)

⇒ n\(^{3}\) = 3S - 3 ∙ \(\frac{n(n + 1)}{2}\) + n

⇒ 3S = n\(^{3}\) + \(\frac{3}{2}\)n(n + 1) – n = n(n\(^{2}\) - 1) + \(\frac{3}{2}\)n(n + 1)

⇒ 3S = n(n + 1)(n - 1 + \(\frac{3}{2}\))

⇒ 3S = n(n + 1)(\(\frac{2n - 2 + 3}{2}\))

⇒ 3S = \(\frac{n(n + 1)(2n + 1)}{2}\)

Therefore, S = \(\frac{n(n + 1)(2n + 1)}{6}\)

i.e., 1\(^{2}\) + 2\(^{2}\) + 3\(^{2}\) + 4\(^{2}\) + 5\(^{2}\) + ................... + n\(^{2}\) = \(\frac{n(n + 1)(2n + 1)}{6}\)

Thus, the sum of the squares of first n natural numbers = \(\frac{n(n + 1)(2n + 1)}{6}\)

Solved examples to find the sum of the squares of first n natural numbers:

**1.** Find the sum of the squares of first 50 natural numbers.

**Solution:**

We know the sum of the squares of first n natural numbers (S) = \(\frac{n(n + 1)(2n + 1)}{6}\)

Here n = 50

Therefore, the sum of the squares of first 50 natural numbers = \(\frac{50(50 + 1)(2 × 50 + 1)}{6}\)

= \(\frac{50 × 51 × 101}{6}\)

= \(\frac{257550}{6}\)

= 42925

**2.** Find the sum of the squares of first 100 natural numbers.

**Solution:**

We know the sum of the squares of first n natural numbers (S) = \(\frac{n(n + 1)(2n + 1)}{6}\)

Here n = 100

Therefore, the sum of the squares of first 50 natural numbers = \(\frac{100(100 + 1)(2 × 100 + 1)}{6}\)

= \(\frac{100 × 101 × 201}{6}\)

= \(\frac{2030100}{6}\)

= 338350

**●** **Arithmetic Progression**

**Definition of Arithmetic Progression****General Form of an Arithmetic Progress****Arithmetic Mean****Sum of the First n Terms of an Arithmetic Progression****Sum of the Cubes of First n Natural Numbers****Sum of First n Natural Numbers****Sum of the Squares of First n Natural Numbers****Properties of Arithmetic Progression****Selection of Terms in an Arithmetic Progression****Arithmetic Progression Formulae****Problems on Arithmetic Progression****Problems on Sum of 'n' Terms of Arithmetic Progression**

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