Sum of Cubes Calculator

Calculate the sum of cubes of consecutive integers between \(A\) and \(B\). See every cube, the total, and the mathematical formula used to verify the result.

Calculate the Sum of Cubes

Enter the starting value \(A\) and ending value \(B\). The calculator adds \(A^3+(A+1)^3+\cdots+B^3\).

Enter integers from 0 to 10,000, with \(A \leq B\).

What Is the Sum of Cubes?

The sum of cubes is the result of adding the third powers of consecutive integers. For the first \(n\) positive integers, the expression is:

\[ 1^3+2^3+3^3+\cdots+n^3 \]

A famous identity gives this sum directly:

\[ \boxed{ 1^3+2^3+\cdots+n^3 = \left(\frac{n(n+1)}{2}\right)^2 } \]

This is particularly useful because it converts a potentially long summation into a simple formula involving the triangular number \(\frac{n(n+1)}{2}\).

Sum of Cubes Formula

The standard formula for the sum of the first \(n\) cubes is:

\[ \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \]

Because \[ \frac{n(n+1)}{2} \] is the \(n\)-th triangular number, the sum of the first \(n\) cubes is the square of the \(n\)-th triangular number.

\[ \sum_{k=1}^{n} k^3 = T_n^2 \]

where:

\[ T_n=\frac{n(n+1)}{2} \]

Sum of Cubes From A to B

When the calculation starts at a number other than \(1\), use the difference between two cumulative cube sums.

\[ \sum_{k=A}^{B}k^3 = \sum_{k=1}^{B}k^3 - \sum_{k=1}^{A-1}k^3 \]

Applying the cube-sum formula gives:

\[ \boxed{ \sum_{k=A}^{B}k^3 = \left(\frac{B(B+1)}{2}\right)^2 - \left(\frac{(A-1)A}{2}\right)^2 } \]

This identity is useful for quickly calculating a range without individually cubing every integer.

Examples of Sum of Cubes

Example 1: First Five Cubes

The first five positive cubes are:

\[ 1^3+2^3+3^3+4^3+5^3 \] \[ = 1+8+27+64+125 \] \[ =225 \]

Using the formula:

\[ \left( \frac{5(5+1)}{2} \right)^2 = 15^2 = 225 \]

Example 2: Cubes From 3 to 6

\[ 3^3+4^3+5^3+6^3 \] \[ = 27+64+125+216 \] \[ =432 \]

Example 3: Sum of the First 10 Cubes

\[ \sum_{k=1}^{10}k^3 = \left( \frac{10(11)}{2} \right)^2 \] \[ = 55^2 = 3025 \]

Why Does the Cube-Sum Formula Work?

The identity can be expressed using triangular numbers. Since \[ T_n=\frac{n(n+1)}{2}, \] the cube sum becomes:

\[ 1^3+2^3+\cdots+n^3=T_n^2 \]

For example, the triangular number for \(5\) is:

\[ T_5=1+2+3+4+5=15 \]

Therefore:

\[ 1^3+2^3+3^3+4^3+5^3 = 15^2 = 225 \]

This connection between triangular numbers and cube sums is one of the most useful patterns in elementary summation formulas.

Sum of Cubes and Sigma Notation

Sigma notation provides a compact way to represent a sum of cubes:

\[ \sum_{k=1}^{n}k^3 \]

For a general range from \(A\) to \(B\), we can write:

\[ \sum_{k=A}^{B}k^3 \]

The corresponding closed-form expression is:

\[ \sum_{k=A}^{B}k^3 = \left(\frac{B(B+1)}{2}\right)^2 - \left(\frac{(A-1)A}{2}\right)^2 \]

Important Properties of Cube Sums

Closed Form The first \(n\) cubes have a simple closed-form formula.
Triangular Number Connection The sum of the first \(n\) cubes equals the square of the \(n\)-th triangular number.
Polynomial Degree The cube-sum formula is a polynomial of degree four in \(n\).
Range Sums Any consecutive range can be found by subtracting two cumulative cube sums.

How to Use the Sum of Cubes Calculator

  1. Enter the starting integer \(A\).
  2. Enter the ending integer \(B\).
  3. Make sure that \(A\leq B\).
  4. Click Calculate Sum.
  5. Review the total, individual cubes, and formula verification.

For example, entering \(A=1\) and \(B=5\) calculates:

\[ 1^3+2^3+3^3+4^3+5^3=225 \]

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Frequently Asked Questions

What is the formula for the sum of cubes?

The sum of the first \(n\) positive cubes is:

\[ 1^3+2^3+\cdots+n^3 = \left(\frac{n(n+1)}{2}\right)^2 \]

What is the sum of cubes from A to B?

For non-negative integers \(A\) and \(B\), where \(A\leq B\):

\[ \sum_{k=A}^{B}k^3 = \left(\frac{B(B+1)}{2}\right)^2 - \left(\frac{(A-1)A}{2}\right)^2 \]

What is the sum of the first 10 cubes?

The sum is:

\[ 1^3+2^3+\cdots+10^3 = 3025 \]

Why is the sum of cubes related to triangular numbers?

Because the sum of the first \(n\) cubes is exactly the square of the \(n\)-th triangular number:

\[ \sum_{k=1}^{n}k^3=T_n^2 \]

Can I calculate a range of cubes?

Yes. Enter the starting and ending integers into the calculator. It calculates every cube in the range and gives the total.

What is \(0^3\)?

Since \(0\) multiplied by itself three times is zero:

\[ 0^3=0 \]

Therefore, including zero at the beginning of a cube-sum range does not change the total.