Question

Find the sum of the squares of first five natural numbers.

Answer 1

⇒ n3 = 3S - 3 ∙ n(n+1)2 + n.

Therefore, S = n(n+1)(2n+1)6.

Thus, the sum of the squares of first n natural

numbers = n(n+1)(2n+1)6.

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Answer 2

The sum of the squares of first five natural numbers = 55

Given :

The first five natural numbers.

To find :

The sum of the squares of first five natural numbers

Solution :

Step 1 of 2 :

Write down first five natural numbers.

The first five natural numbers are 1 , 2 , 3 , 4 , 5

Step 2 of 2 :

Find the sum of the squares of first five natural numbers

We know that Sum of the squares of first n natural numbers

$\displaystyle \sf {1}^{2} + {2}^{2} + {3}^{2} + ... + {n}^{2} = \frac{n(n + 1)(2n + 1)}{6}$

Since we have to find the sum of the squares of first five natural numbers

So we take n = 5

Hence the sum of the squares of first five natural numbers

= 1² + 2² + 3² + 4² + 5²

$\displaystyle \sf{ = \frac{5 \times (5 + 1) \times (10 + 1)}{6} }$

$\displaystyle \sf{ = \frac{5 \times 6 \times 11}{6} }$

$\displaystyle \sf{ = 55 }$

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