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

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