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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