Sum of the Squares of First n Natural Numbers

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

11 and 12 Grade Math

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