# Sum of Squares of n Natural Numbers, Even Numbers and Odd Number

In this post we will derive and discuss the formula related to sum of squares of natural numbers, even numbers and odd numbers. The formulas we will derive below are important and you need to memorize them in order to solve questions fast.

## Sum of Squares of Natural Numbers

S\ =1^{2} +2^{2} +3^{2} +4^{2} +5^{2} +6^{2} \ .\ .\ .\ .\ .\ .\ .\ .\ .n^{2}\\\ \\ To\ find\ the\ sum\ to\ nth\ term\ we\ will\ use\\ \\ following\ equation \\ \\ k^{3} -( k-1)^{3} =3k^{2} -3k+1\\\ \\ Putting\ the\ value\ of\ k=1,2,3,4.\ .\ .\ .n,\ we\ get \\\ \\ 1-0\ =\ 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\\ \\ 5^{3} -4^{3} =\ 3\ ( 5)^{2} -3( 5) +1\\ \\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ \\ \\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ ..\ .\ .\ .\ \ \\ \\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ \\ \\ n^{3} -( n-1)^{3} =\ 3\ ( n)^{2} -3( n) +1\\ \\ -------------------------- \\\ \\ \\\ \\ Adding\ all\ the\ equation,\ we\ get\\\ \\ n^{3} -0^{3} =\ 3\ \left( 1^{2} +2^{2} +3^{2} +4^{2} +5^{2} \ .\ .\right) -\ 3( 1+2+3….) +n \\\ \\ n^{3} =3\ \sum k^{2} \ -\ 3\sum k\ +n\ \ \ \ \ \ \ \ \ \ \ \ \ ---eq( 1) \\\ \\ we\ already\ know\ that:\\ \\ ( 1+2+3+\ 4\ …) \ =\ \frac{n( n+1)}{2}\\\ \\ Putting\ the\ value\ in\ eq( 1) ,\ we\ get\\\ \\ n^{3} =3\ \sum ^{n}_{k=1} k^{2} \ -\ 3\frac{n( n+1)}{2} \ +n \\\ \\ \sum ^{n}_{k=1} k^{2} \ =\ \frac{1}{6}\left( 2n^{3} +3n^{2} +n\right)\\\ \\ \sum ^{n}_{k=1} k^{2} =\ \frac{n\ ( n+1) \ ( 2n+1)}{6}\\\ \\ \\\ \\

Formula for sum of square of natural number
1^{2} +2^{2} +3^{2} +\ .\ .\ .\ .\ .n^{2} =\frac{n\ ( n+1) \ ( 2n+1)}{6}\\\ \\

## Sum of square of even numbers

2^{2} +4^{2} +6^{2} +8^{2} +10^{2} +………( 2n)^{2}\\\ \\ 4\ \times \ \left[ 1^{2} +2^{2} +3^{2} +4^{2} +………( n)^{2}\right] \\\ \\ 4\ \times \frac{n( n+1)( 2n+1)}{6} \\\ \\ \frac{2n( n+1)( 2n+1)}{3} \\\ \\

Formula for Sum of square of even number
2^{2} +4^{2} +6^{2} +8^{2} +10^{2} +………( 2n)^{2} = \frac{2n( n+1)( 2n+1)}{3}

## Sum of square of odd numbers

1^{2} +3^{2} +5^{2} +……(2n-1)^{2}

Sum of Square of Odd Number = Sum of square of N natural Number – Sum of Square of Even number

\Longrightarrow \ \frac{n\ ( n+1)( 2n+1)}{6} -\frac{2n( n+1)( 2n+1)}{3}\\\ \\ \Longrightarrow \frac{2n( 2n+1)}{6}[ 4n+1-2( n+1)]\\\ \\ \Longrightarrow \frac{n( 2n+1)( 2n-1)}{3}

Formula for sum of square of odd numbers
1^{2} +3^{2} +5^{2} +……(2n-1)^{2} =\frac{n( 2n+1)( 2n-1)}{3}

## Questions on Sum of Squares

(01) Find the sum of squares of first 50 natural numbers

Solution
S\ =\ 1^{2} +2^{2} +3^{2} +4^{2} +\ .\ \ .\ .\ \ .50^{2}\\\ \\ Sum\ of\ Square\ of\ Natural\ Number\\ \\ \Longrightarrow \ \frac{n( n+1)( 2n+1)}{6} \\\ \\ Here\ n=50\\\ \\ Putting\ the\ values\\ \\ \Longrightarrow \ \frac{50\times 51\times 101}{6}\\ \\ \Longrightarrow 42925

(02) Find the sum of square of first 20 even number

Solution
S\ =\ 2^{2} +4^{2} +6^{2} +8^{2} +\ .\ \ .\ .\ \ .( 2n)^{2}\\\ \\ Sum\ of\ Square\ of\ Even\ Number\\ \\ \Longrightarrow \ \frac{2n( n+1)( 2n+1)}{3} \\\ \\ Here\ n=20\\ \\ Putting\ the\ values\\\ \\ \Longrightarrow \ \frac{40\times 21\times 41}{3}\\\ \\ \Longrightarrow 11480