Sum of Cubes of First n Natural Numbers with Examples
Sum of Cubes of First n Natural Numbers gives the total of 1³ + 2³ + 3³ + … + n³. The standard formula converts this series into the square of the sum of the first n natural numbers, making numerical questions faster to solve.
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What Is the Sum of Cubes of the First n Natural Numbers?
The expression n(n + 1)/2 is the sum of the first n natural numbers. Therefore, the sum of their cubes equals the square of this triangular-number sum. For n = 4, the sum is 1³ + 2³ + 3³ + 4³ = 100, and [4 × 5/2]² = 10² = 100.
Sum of Cubes of First n Natural Numbers Video Lessons
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Special Sums: n, n² and n³ Formulas
Learn the formulas for the sums of the first n natural numbers, their squares, and their cubes, with a clear understanding of how these special summations are represented.
Sum of Cubes of First n Natural Numbers Formula & Tricks
Important Formulas
Multiply n by n + 1, divide by 2, and square the result.
The sum of the first n cubes is the square of the sum of the first n natural numbers.
This is the same formula written by squaring both factors in n(n + 1)/2.
Quick Tricks
Calculate T = n(n + 1)/2 and then find T². This avoids adding each cube separately.
If n(n + 1)/2 is even or has convenient factors, simplify it before squaring.
Sum of Cubes of First n Natural Numbers Concepts
Identity Behind the Formula
If T = 1 + 2 + 3 + … + n = n(n + 1)/2, then 1³ + 2³ + 3³ + … + n³ = T². This identity applies for every positive integer n.
Direct Calculation Method
Step 1: Find n(n + 1). Step 2: Divide by 2. Step 3: Square the quotient. For n = 8, T = 8 × 9/2 = 36, so the required sum is 36² = 1296.
Difference Between Consecutive Cube Sums
If Sₙ = 1³ + 2³ + … + n³, then Sₙ₊₁ − Sₙ = (n + 1)³ because the next cube is the only added term. Hence, Sₙ₊₁ = Sₙ + (n + 1)³.
Finding n from a Given Cube Sum
For a given sum S, √S = n(n + 1)/2. The square root must be a triangular number for S to be the sum of cubes of the first n natural numbers.
Sum of Cubes of First n Natural Numbers Data Table
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Sample Sums of Cubes
Values of n, the corresponding triangular sum, and the sum of cubes.
| n | n(n + 1)/2 | Sum of cubes |
|---|---|---|
| 1 | 1 | 1 |
| 2 | 3 | 9 |
| 3 | 6 | 36 |
| 4 | 10 | 100 |
| 5 | 15 | 225 |
| 10 | 55 | 3025 |
Note: The final column is the square of the middle column.
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Sum of Cubes Formula: Revision Points
Use these direct rules for quick calculation and verification.
- For n ≥ 1, 1³ + 2³ + … + n³ = [n(n + 1)/2]².
- The sum of cubes is the square of the sum of the first n natural numbers.
- Calculate n(n + 1)/2 before squaring.
- The next cube sum is obtained by adding (n + 1)³.
- If a number is a cube sum, its square root must be a triangular number.
Sum of Cubes of First n Natural Numbers FAQs
What is the sum of cubes of the first n natural numbers formula?
The formula is 1³ + 2³ + … + n³ = [n(n + 1)/2]².
What is the sum of the cubes of the first 5 natural numbers?
Using [5 × 6/2]² = 15², the sum is 225.
What is the sum of cubes from 1 to 10?
The sum is [10 × 11/2]² = 55² = 3025.
How is the sum of cubes related to the sum of natural numbers?
The sum of the first n cubes equals the square of the sum of the first n natural numbers: Σk³ = (Σk)².
What is the sum of cubes of the first 20 natural numbers?
The triangular sum is 20 × 21/2 = 210. Therefore, the cube sum is 210² = 44,100.
How can you check whether a number is a sum of cubes of the first n natural numbers?
Take its square root. If the root equals n(n + 1)/2 for a positive integer n, the number is such a cube sum.