# Sum of squares of n natural numbers

The squares of natural numbers are $1^{2},2^{2},3^{2},4^{2},...,n^{2}$.
sum of squares of n natural numbers are
$S_{n} = 1^{2} + 2^{2} + 3^{2} + 4^{2} + ...+ n^{2}$
Consider the identity
$k^{3} - (k - 1)^{3} =$k^{3} - (k^{3} - 3k^{2}$+ 3k -1) =$k^{3} - k^{3} + 3k^{2}$- 3k + 1 ∴$k^{3} - (k - 1)^{3} = 3k^{2}$- 3k + 1 Put k = 1, 2, 3,...,n k = 1 ⇒$1^{3} - 0^{3} = 3(1)^{2} - 3(1) + 1
k = 2 ⇒ $2^{3} - 1^{3} = 3(2)^{2} - 3(2) + 1 k = 3 ⇒$3^{3} - 2^{3} = 3(3)^{2} - 3(3) + 1
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k = n ⇒ $n^{3} - (n-1)^{3} = 3(n)^{2}$ - 3(n) + 1
Adding all the the equations we get
$n^{3} - 0^{3} = 3(1^{2} + 2^{2} + 3^{2} + 4^{2} + ...+ n^{2}$) + 3 ( 1 + 2 + 3+...+ n)

$n^{3} = 3\sum_{k=1}^{n}k^{2} - 3 \sum_k$ + n

$\sum_{k=1}^{n}k^{2} = \frac{1}{3} [n^{3} + 3 \sum_{k=1}^{n}k$ -n ] --------(i)

As we know that sum of 1st 'n' natural numbers is $\frac{n(n+1)}{2}$
∴ $\sum_k = \frac{n(n+1)}{2}$

(i) ⇒ $\sum_k^{2} = S_{n} = \frac{1}{3} [ n^{3} + 3\frac{n(n+1)}{2}$ - n]

$S_{n} = \frac{1}{3}[\frac{2n^{3} + 3n^{2} + 3n -2n}{3}$]

$S_{n} = \frac{1}{3}[\frac{2n^{3} + 3n^{2} + n }{3}$]

after finding the factors of above quadratic equation we get ,
$S_{n} = \frac{1}{6}$n(n+1)(2n+1)

## Examples on Sum of squares of n natural numbers

1) Find the sum of squares of first 20 numbers.

Solution : As we know that sum of squares of n numbers is given by the formula
$S_{n} = \frac{1}{6}$n(n+1)(2n+1)

Here n = 20 so,
$S_{20} = \frac{1}{6}\times20(20 + 1)(2\times 20$+1)

$S_{20} = \frac{1}{6}\times 20 \times 21$(40 + 1)

$S_{20} = \frac{1}{6}\times 20\times 21 \times 41$

$S_{20}$ = 2870
So the sum of squares of first 20 terms is 2870.

2) Find the sum of squares of first 15 numbers.

Solution : As we know that sum of squares of n numbers is given by the formula
$S_{n} = \frac{1}{6}$n(n+1)(2n+1)

Here n = 15 so,
$S_{15} = \frac{1}{6}\times15(15 + 1)(2\times 15$+1)

$S_{15} = \frac{1}{6}\times 15 \times 16$(30 + 1)

$S_{15} = \frac{1}{6}\times 15\times 16 \times 31$

$S_{15}$ = 1240
So the sum of squares of first 15 terms is 1240.