# Sum of Cubes of Natural Number

In this post we will discuss the formula of sum of cube of natural number, odd number and even number. The post contain detailed derivation of each formula so that you can understand the basic aspects of equation.

I strongly urge to remember each of the formulas as they would help us solving questions from CBSE/NCERT and other competition exams

## Sum of cubes of Natural Number

1^{3} +2^{3} +3^{3} +4^{3} +5^{3} +6^{3} .\ .\ .\ .\ .n^{3}\\\ \\ Use\ the\ following\ equation \\ \\ ( k\ +\ 1)^{4} \ –\ k^{4} \ =\ 4k^{3} \ +\ 6k^{2} \ +\ 4k\ +\ 1\\\ \\ Putting\ k=1,2,3\ …n\\ \\ 2^{4} \ –\ 1^{4} \ =\ 4( 1)^{3} \ +\ 6( 1)^{2} \ +\ 4( 1) \ +\ 1\\ \\ 3^{4} \ –\ 2^{4} \ =\ 4( 2)^{3} \ +\ 6( 2)^{2} \ +\ 4( 2) \ +\ 1\\ \\ 4^{4} \ –\ 3^{4} \ =\ 4( 3)^{3} \ +\ 6( 3)^{2} \ +\ 4( 3) \ +\ 1 \\ \\ 5^{4} \ –\ 4^{4} \ =\ 4( 4)^{3} \ +\ 6( 4)^{2} \ +\ 4( 4) \ +\ 1\\ \\ .\ .\ .\ .\ .\ \ .\ .\ .\ \\ \\ .\ .\ .\ .\ .\ .\ .\ .\ .\ \\ \\ n^{4} \ –\ ( n-1)^{4} \ =\ 4( n-1)^{3} \ +\ 6( n-1)^{2} \ +\ 4( n-1) \ +\ 1\ \\\ \\ ( n+1)^{4} \ –\ ( n)^{4} \ =\ 4( n)^{3} \ +\ 6( n)^{2} \ +\ 4( n) \ +\ 1\ \\\ \\ \\\ \\ \\\ \\ Adding\ all\ the\ expressions\ we\ get\\ \\ ( n\ +\ 1)^{4} \ –\ 1^{4} \ =\ 4\left( 1^{3} \ +\ 2^{3} \ +\ 3^{3} \ +…+\ n^{3}\right) \ +\ 6\left( 1^{2} \ +\ 2^{2} \ +\ 3^{2} \ +\ …+\ n^{2}\right)\ + \\ \\ 4( 1\ +\ 2\ +\ 3\ +…+\ n) \ +\ n \\\ \\ ⟹\ 4\sum k^{3} \ +\ 6\sum k^{2} \ +\ 4\ \sum k\ +n\\\ \\ \\\ \\ \\\ \\ We\ already\ know\ that\ \\\ \\ \sum k \ =\ \frac{n( n+1)}{2}\\\ \\ \sum k^{2} \ =\ \frac{n\ ( n+1) \ ( 2n+1)}{6}\\\ \\ \\\ \\ Putting\ the\ values\\ \\ ( n\ +\ 1)^{4} \ –\ 1^{4} \ =\ 4\sum k^{3} \ +\ \ 6\ \times \ \frac{n\ ( n+1) \ ( 2n+1)}{6} \ \ +4\times \ \frac{n( n+1)}{2} + n \\\ \\ 4\sum k^{3}\ =\ ( n\ +\ 1)^{4} \ –\ 1^{4} \ -\ n\ ( n+1) \ ( 2n+1) \ -\ 2n( n+1)\ - n\\\ \\ 4\sum k^{3}\ =\ n^{4} +4n^{3} +6n^{2} +4n\ \ –\ 1^{4} \ -\ n\ ( n+1) \ ( 2n+1) \ -\ 2n( n+1)\ -\ n\\\ \\ \\\ \\ \\\ \\ After\ solving\ the\ expression\ we\ get\\ \\ \ 4\sum k^{3} \ =\ n^{2}( n\ +\ 1)^{2}\\\ \\ \sum k^{3} \ =\ \frac{n^{2}( n\ +\ 1)^{2}}{4}\\\ \\ \\\ \\

Hence the formula is:
1^{3} +2^{3} +3^{3} +4^{3} +5^{3} +6^{3} .\ .\ .\ .\ .n^{3} \Longrightarrow \ \frac{n^{2}( n\ +\ 1)^{2}}{4}\

## Formula for sum of Cube of even natural number

S\ =2^{3} +4^{3} +6^{3} +8^{3} +10^{3} +………( 2n)^{3}\\\ \\ S = 2^{3} \ \left[ 1^{3} +2^{3} +3^{3} +4^{3} +5^{3} +………( n)^{3}\right] \\ \\ \ .\ .\ .\ .\ .eq( 1)\\\ \\ we\ already\ know\ that\\ \\ 1^{3} +2^{3} +3^{3} +4^{3} +5^{3} …( n)^{3} =\frac{n^{2}( n+1)^{2}}{4}\\\ \\ Putting\ this\ value\ in\ eq( 01)\\\ \\ S\ =\ 2^{3} \times \frac{n^{2}( n+1)^{2}}{4}\\\ \\ S=\ 2\ \times n^{2}( n+1)^{2}\\\ \\

Hence the formula for sum of cube of even number is
2^{3} +4^{3} +6^{3} …( 2n)^{3} = 2\ \times n^{2}( n+1)^{2}\

## Formula for sum of Cube of Odd natural number

S\ =1^{3} +3^{3} +5^{3} +7^{3} \ ….( 2n-1)^{3}

Sum of cube of Odd number = Sum of cube of all natural number – Sum of cube of even natural number – – – -eq(01)

We know that
Sum of cube of natural number = \frac{n^{2}( n\ +\ 1)^{2}}{4}

Sum of cube of even number = 2\ \times n^{2}( n+1)^{2}\

Putting these values in equation 01, we get

Sum of cube of odd number = \frac{n^{2}( n\ +\ 1)^{2}}{4} - 2\ \times n^{2}( n+1)^{2}\

\Longrightarrow \ n^{2}\left[( 2n+1)^{2} -2( n+1)^{2}\right]\\\ \\ \Longrightarrow \ n^{2} \ \left[ 4n^{2} +1+4n\ -2n^{2} -2-4n\right]\\\ \\ \Longrightarrow \ n^{2} \ \left[ 2n^{2} -1\right]\

Hence, formula for sum of cube of odd number is
1^{3} +3^{3} \ ….( 2n-1)^{3} \Longrightarrow \ n^{2} \ \left[ 2n^{2} -1\right]\

## Questions for Sum of Cubes

(01) Find the sum of the cube of first 7 natural number
S=\ 1^{3} +2^{3} +3^{3} +4^{3} +5^{3} +6^{3} +7^{3}

The formula for sum of cube of natural number is \frac{n^{2}( n\ +\ 1)^{2}}{4}\\\ \\
Putting n= 7 in the formula, we get:
\Longrightarrow \ \frac{7^{2} \ ( 7+1)^{2}}{4}\\\ \\ \Longrightarrow \frac{7^{2} \ ( 8)^{2}}{4}\\\ \\ \Longrightarrow 784

784 is the right answer for this question

(02) Sum of cube of first six even number
S=\ 2^{3} +\ 4^{3} +6^{3} +8^{3} +10^{3} +12^{3}

The formula for cube of even number is 2\ \times n^{2}( n+1)^{2}

Putting n=6 in the above formula
\Longrightarrow 2\times 6^{2} \times ( 6+1)^{2}\\\ \\ \Longrightarrow 2\times 6^{2} \times ( 7)^{2}\\\ \\ \Longrightarrow \ 3528

3528 is the right answer for this question