What Is the Sum of Cubes of n Natural Numbers?
The sum of cubes of n natural numbers is the total you get by cubing each of the first $n$ counting numbers and adding the results: $1^3 + 2^3 + 3^3 + \cdots + n^3$. It has a clean closed form, meaning you can compute it directly without adding every term:
$$\sum_{k=1}^{n} k^3 = 1^3 + 2^3 + \cdots + n^3 = \left[\frac{n(n+1)}{2}\right]^2.$$
The expression inside the brackets, $\dfrac{n(n+1)}{2}$, is the sum of the first $n$ natural numbers — the $n$th triangular number. So the formula says something striking: the sum of the first $n$ cubes is the square of the sum of the first $n$ numbers. This identity is part of the family of summation formulas that also covers the sum of the numbers themselves and the sum of their squares.
What Is the Formula for the Sum of Cubes of n Natural Numbers?
The formula is written two equivalent ways. Both give the same number; pick whichever is cleaner for your problem.
$$\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2 = \frac{n^2(n+1)^2}{4}.$$
Symbol | Meaning |
|---|---|
$n$ | how many natural numbers you're cubing and adding |
$\dfrac{n(n+1)}{2}$ | the sum $1 + 2 + \cdots + n$ (the $n$th triangular number) |
$\left[;\right]^2$ | square the triangular number to get the answer |
The bracket form makes the "square of the sum" idea visible; the $\dfrac{n^2(n+1)^2}{4}$ form is sometimes faster to plug numbers into. For $n = 4$: the triangular number is $\dfrac{4 \cdot 5}{2} = 10$, and $10^2 = 100$ — which matches $1 + 8 + 27 + 64$ exactly.
How Do You Prove the Sum of Cubes Formula?
There are two standard proofs. The telescoping proof derives the formula from scratch; the induction proof verifies a formula you already suspect. Both are worth seeing.
Proof 1 — Telescoping with an algebraic identity
Start from the identity $n^4 - (n-1)^4 = 4n^3 - 6n^2 + 4n - 1$, which you can check by expanding $(n-1)^4$. Now write it out for $n = 1, 2, 3, \ldots, n$ and add every row. The left side telescopes — almost everything cancels — leaving just $n^4 - 0^4 = n^4$:
$$n^4 = 4\sum k^3 - 6\sum k^2 + 4\sum k - \sum 1.$$
Substitute the known sums $\sum k = \dfrac{n(n+1)}{2}$, $\sum k^2 = \dfrac{n(n+1)(2n+1)}{6}$, and $\sum 1 = n$, then solve for $\sum k^3$. The algebra simplifies — after collecting terms — to:
$$\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4} = \left[\frac{n(n+1)}{2}\right]^2.$$
Proof 2 — Mathematical induction
Induction has two steps. Base case: for $n = 1$, the left side is $1^3 = 1$ and the right side is $\left[\dfrac{1 \cdot 2}{2}\right]^2 = 1$. They match.
Inductive step: assume the formula holds for some $n = m$: $$1^3 + \cdots + m^3 = \left[\frac{m(m+1)}{2}\right]^2.$$ Add the next cube, $(m+1)^3$, to both sides and show the result equals $\left[\dfrac{(m+1)(m+2)}{2}\right]^2$:
$$\left[\frac{m(m+1)}{2}\right]^2 + (m+1)^3 = \frac{(m+1)^2}{4}\big[m^2 + 4(m+1)\big] = \frac{(m+1)^2(m+2)^2}{4}.$$
That's exactly the formula with $n = m+1$, so by induction it holds for every natural number. Induction is the standard tool for any sequence-and-summation identity, and this is one of the cleanest examples to learn it on.
What Are the Even and Odd Cube Sum Formulas?
The same idea adapts to cubing only the even numbers or only the odd ones.
Sum of cubes of the first $n$ even natural numbers — factor a $2^3 = 8$ out of each term: $$2^3 + 4^3 + \cdots + (2n)^3 = 8\sum_{k=1}^{n} k^3 = 2\big[n(n+1)\big]^2.$$
Sum of cubes of the first $n$ odd natural numbers: $$1^3 + 3^3 + \cdots + (2n-1)^3 = n^2(2n^2 - 1).$$
The even formula follows directly from the main one (pull out the 8). The odd formula is the main sum up to $2n$ minus the even part — a tidy subtraction that Example 6 works through.
Examples of Sum of Cubes of n Natural Numbers
The examples progress from a direct plug-in, through a sum starting partway up, the most common squaring slip, an even-cube case, and finish by solving backward for $n$.
Example 1
Find the sum of cubes of the first 5 natural numbers.
Use $\left[\dfrac{n(n+1)}{2}\right]^2$ with $n = 5$. The triangular number is $\dfrac{5 \cdot 6}{2} = 15$:
$$\sum_{k=1}^{5} k^3 = 15^2 = 225.$$
Final answer: $225$. Check by hand: $1 + 8 + 27 + 64 + 125 = 225$.
Example 2
Find the sum of cubes of the first 10 natural numbers.
Triangular number: $\dfrac{10 \cdot 11}{2} = 55$. Square it:
$$\sum_{k=1}^{10} k^3 = 55^2 = 3025.$$
Final answer: $3025$. Adding ten cubes by hand would take a minute; the formula takes seconds.
Example 3
A common slip — find the sum of cubes of the first 6 natural numbers.
Wrong attempt. A student remembers "square of the sum" but squares the wrong thing — they compute the sum of the numbers, $1+2+3+4+5+6 = 21$, then cube it instead of squaring, getting $21^3 = 9261$. Sanity check: the largest single cube in the list is $6^3 = 216$, and there are only six terms, so the total can't reach nine thousand. Something is off.
Correct. The formula squares the triangular number, it does not cube it:
$$\sum_{k=1}^{6} k^3 = \left[\frac{6 \cdot 7}{2}\right]^2 = 21^2 = 441.$$
Final answer: $441$. Verify: $1 + 8 + 27 + 64 + 125 + 216 = 441$. The operation is square, because the cubes sum to the square of the triangular number, never its cube.
Example 4
Find the sum of cubes from 5 to 10, i.e. $5^3 + 6^3 + \cdots + 10^3$.
This is a partial sum, so subtract: (sum to 10) minus (sum to 4).
$$\sum_{k=1}^{10} k^3 = 55^2 = 3025, \qquad \sum_{k=1}^{4} k^3 = 10^2 = 100.$$
$$5^3 + \cdots + 10^3 = 3025 - 100 = 2925.$$
Final answer: $2925$. To sum from $a$ to $n$, take the sum to $n$ and remove the sum to $a-1$.
Example 5
Find the sum of cubes of the first 4 even natural numbers, $2^3 + 4^3 + 6^3 + 8^3$.
Use the even formula $2[n(n+1)]^2$ with $n = 4$:
$$2[4 \cdot 5]^2 = 2 \cdot 20^2 = 2 \cdot 400 = 800.$$
Final answer: $800$. Check: $8 + 64 + 216 + 512 = 800$.
Example 6
The sum of cubes of the first $n$ natural numbers is 1296. Find $n$.
Work backward. Set the formula equal to $1296$ and take the square root, since the sum is a perfect square:
$$\left[\frac{n(n+1)}{2}\right]^2 = 1296 ;\Longrightarrow; \frac{n(n+1)}{2} = \sqrt{1296} = 36.$$
So $n(n+1) = 72$. The two consecutive numbers multiplying to $72$ are $8$ and $9$, giving $n = 8$.
Final answer: $n = 8$. Check: $\left[\dfrac{8 \cdot 9}{2}\right]^2 = 36^2 = 1296$. Because the sum is always a perfect square, taking the square root is a legal first move.
The Mathematicians Behind the Sum of Cubes
Nicomachus of Gerasa (c. 60–120 CE, Roman-era Greek) recorded the sum-of-cubes pattern in his Introduction to Arithmetic, which is why the identity $\sum k^3 = (\sum k)^2$ is often called Nicomachus's theorem. Aryabhata (476–550 CE, India) independently stated the same closed form in the Aryabhatiya, giving rules for the sums of squares and cubes used in Indian astronomy.
Why the Sum of Cubes Formula Matters
"Adding forever, in one line, without adding at all."
The deeper point isn't the single formula — it's that an endless-looking sum collapses to a tidy expression. That move, replacing a long addition with a closed form, is the engine of how mathematics handles "add up all of these" at scale.
Where it earns its keep:
Computer science — algorithm cost. Counting the operations in certain nested loops produces $\sum k^3$. The closed form tells you the cost grows like $n^4$, which is how engineers predict whether a program will finish in seconds or centuries.
Numerical methods. Summation formulas for powers underpin numerical integration — approximating areas under curves by adding many thin pieces, then collapsing the sum.
Number theory and patterns. Nicomachus's "sum of cubes is square of sum" is a gateway result that gets students asking why such patterns exist — the same curiosity that drives the study of perfect cubes and figurate numbers.
Exam technique. In CBSE Class 11 and competitive papers, recognising a series as $\sum k^3$ turns a page of arithmetic into one substitution.
Where Students Trip Up on the Sum of Cubes
Mistake 1: Squaring versus cubing the triangular number
Where it slips in: Remembering "square of the sum" loosely and cubing the triangular number, or forgetting to square it at all.
Don't do this: Compute $1 + 2 + \cdots + n$ and then cube it, or leave it un-squared.
The correct way: The sum of cubes is the square of the triangular number: $\left[\dfrac{n(n+1)}{2}\right]^2$. Square it once — no more, no less.
Mistake 2: Forgetting to subtract for a partial sum
Where it slips in: Asked for $5^3 + \cdots + 10^3$, a student plugs $n = 10$ and stops, including the unwanted $1^3$ through $4^3$.
Don't do this: Apply the formula to the top number alone when the sum doesn't start at 1.
The correct way: Take the sum up to the top, then subtract the sum up to one below the start: (sum to 10) minus (sum to 4) $= 3025 - 100 = 2925$. The second-guesser often suspects something is missing here but can't name it — the missing piece is the subtraction.
Mistake 3: Treating the formula as a sum of natural numbers
Where it slips in: Confusing $\sum k^3$ with $\sum k$ and using $\dfrac{n(n+1)}{2}$ as the final answer.
Don't do this: Stop at the triangular number. That's the sum of the numbers, not the sum of their cubes.
The correct way: The triangular number is only the inside of the formula. Square it to get the sum of cubes.
Key Takeaways
The sum of cubes of n natural numbers is $\left[\dfrac{n(n+1)}{2}\right]^2 = \dfrac{n^2(n+1)^2}{4}$.
It equals the square of the sum of the first $n$ numbers — Nicomachus's theorem.
The telescoping identity $n^4 - (n-1)^4$ proves it directly; induction verifies it cleanly.
For a partial sum starting above 1, subtract the sum up to one below the start.
The even-cube sum is $2[n(n+1)]^2$; the odd-cube sum is $n^2(2n^2 - 1)$.
Because the result is always a perfect square, you can square-root it to solve for $n$.
Practice These Before Moving On
Find the sum of cubes of the first 7 natural numbers.
Find $3^3 + 4^3 + \cdots + 8^3$.
The sum of cubes of the first $n$ natural numbers is $441$. Find $n$.
Answer to Question 1: $\left[\tfrac{7 \cdot 8}{2}\right]^2 = 28^2 = 784$. Answer to Question 2: (sum to 8) − (sum to 2) $= 36^2 - 3^2 = 1296 - 9 = 1287$. Answer to Question 3: $\sqrt{441} = 21$, so $\tfrac{n(n+1)}{2} = 21$, giving $n = 6$. If Question 1 came out as $28^3$, return to Mistake 1.
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