Algebra

Sum of series of cubes

The sum of the cubes of the first $n$ natural numbers possesses a remarkable algebraic identity equal to the square of the sum of the first $n$ natural numbers.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Nicomachus's Theorem / Sum of Cubes Identity

For any positive integer $n \in \mathbb{N}$, the sum of the cubes of the first $n$ natural numbers is given by $$\sum_{k=1}^{n} k^3 = 1^3 + 2^3 + 3^3 + \dots + n^3 = \left(\frac{n(n+1)}{2}\right)^2 = \left(\sum_{k=1}^{n} k\right)^2$$

📐 Geometric Interpretation

Geometrically, Nicomachus's theorem demonstrates that the sum of successive cubes can be tiled into nested gnomons forming squares, visually establishing that the sum of the first $n$ cubic volumes equals the area of a square whose side length is the triangular number $T_n = \frac{n(n+1)}{2}$.

⚙️ Algebraic & Structural Meaning

It forms the foundational bridge between arithmetic progression sums and polynomial expansions of degree 4 via the telescoping method using $(k+1)^4 - k^4 = 4k^3 + 6k^2 + 4k + 1$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Domain of Index $n$	`$n \in \mathbb{Z}^+$ (Positive integers)`	Strictly Mandatory	Fractional or negative limits invalidate the standard combinatorial derivation unless analytically continued via Bernoulli polynomials.
Continuity of Sequence Terms	`Discrete domain $\{1, 2, \dots, n\}$`	Strictly Mandatory	Applying the formula to continuous integrands without proper summation bounds yields integration values instead of discrete sums.
Uniform Progression Step	`Common difference $d = 1$ for standard form`	Essential for standard formula application	If the sequence is an AP with common difference $d \neq 1$, direct substitution into $\left[\frac{n(n+1)}{2}\right]^2$ results in severe calculation errors.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Standard Sum of Cubes

$$ \sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4} $$

Rule: Square the formula for the sum of the first $n$ natural numbers: $\left[\frac{n(n+1)}{2}\right]^2$.

Application: undefined

Sum of Cubes of First $n$ Even Numbers

$$ \sum_{k=1}^{n} (2k)^3 = 2^3 \sum_{k=1}^{n} k^3 = 2n^2(n+1)^2 $$

Rule: Factor out $2^3 = 8$ and apply the standard formula: $8 \cdot \frac{n^2(n+1)^2}{4} = 2n^2(n+1)^2$.

Sum of Cubes of First $n$ Odd Numbers

$$ \sum_{k=1}^{n} (2k-1)^3 = n^2(2n^2 - 1) $$

Rule: Use the identity $\sum_{r=1}^{2n} r^3 - 8 \sum_{r=1}^{n} r^3$ which simplifies cleanly to $n^2(2n^2 - 1)$.

Telescoping Identity Derivation Check

$$ \sum_{k=1}^{n} \left[(k+1)^4 - k^4\right] = (n+1)^4 - 1 $$

Rule: Expand $(k+1)^4 - k^4 = 4k^3 + 6k^2 + 4k + 1$ and sum both sides to isolate $\sum k^3$ in 5 seconds.

Application: Used in theoretical assertion-reason questions in UP PGT.

Sum of Cubes with General AP Terms

$$ \sum_{k=1}^{n} [a + (k-1)d]^3 $$

Rule: Expand using binomial theorem or convert to standard form by substituting $t_k = a + (k-1)d$.

Application: Standard high-difficulty question in LT Grade Mathematics.

Rapid Unit Digit / Modulo Verification

$$ \sum_{k=1}^{n} k^3 \equiv \left(\frac{n(n+1)}{2}\right)^2 \pmod{10} $$

Rule: Check the last digit of $\left[\frac{n(n+1)}{2}\right]^2$ to instantly eliminate 2 or 3 incorrect options.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Forgetting the Square in Sum of Cubes

Common Pitfall: Candidates mistakenly use $\frac{n(n+1)(2n+1)}{6}$ (which is for squares) or $\frac{n(n+1)}{2}$ (which is for linear terms) for cubes.

Counterexample: For $n=2$, $1^3 + 2^3 = 9$, whereas $\frac{2(3)}{2} = 3$ (not squared).

Why it fails: Confusing the power order of standard summation formulas leads to catastrophic numerical errors.

🛡️ Exam Defense: Always remember: Sum of numbers = Linear, Sum of squares = Quadratic denominator 6, Sum of cubes = Square of the linear sum.

⚠️ COMMON ERROR #02

Starting from Non-Unity Limits ($k = m$ to $n$)

Common Pitfall: Directly applying $\frac{n^2(n+1)^2}{4}$ when the series starts from $k = m > 1$ (e.g., $5^3 + 6^3 + \dots + 20^3$).

Counterexample: $\sum_{k=5}^{20} k^3 \neq \frac{20^2(21)^2}{4}$

Why it fails: The standard formula explicitly assumes the lower limit is $1$.

🛡️ Exam Defense: Always use the subtraction property: $\sum_{k=m}^{n} k^3 = \sum_{k=1}^{n} k^3 - \sum_{k=1}^{m-1} k^3$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Sum of First $n$ Integers	$n \in \mathbb{Z}^+$	$$\frac{n(n+1)}{2}$$	Area of right-angled triangular stair	Linear series / AP sum
Sum of First $n$ Squares	$n \in \mathbb{Z}^+$	$$\frac{n(n+1)(2n+1)}{6}$$	Pyramidal stacking of square spheres	Quadratic terms in series
Sum of First $n$ Cubes	$n \in \mathbb{Z}^+$	$$\frac{n^2(n+1)^2}{4}$$	Area of square whose side is triangular number	Cubic terms / Nicomachus pattern
Sum of First $n$ Quartics	$n \in \mathbb{Z}^+$	$$\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$	Higher-order hyper-spatial measure	Advanced PGT polynomial series

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the sum of the series $1^3 + 2^3 + 3^3 + \dots + 15^3$.

(A) 12800(B) 14400(C) 15625(D) 16200

📚 Method A: Formal Textbook Derivation

Step 1: Identify $n = 15$. Step 2: Apply formula $\left[\frac{n(n+1)}{2}\right]^2$. Step 3: Compute $\left[\frac{15 \times 16}{2}\right]^2 = (120)^2 = 14400$.

⚡ Method B: 25-Second Exam Shortcut

Recognize $120^2 = 14400$ instantly using unit digit and magnitude estimation.

💡 Tip: Direct formula substitution question appearing in UP TGT.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If $1^3 + 2^3 + 3^3 + \dots + n^3 = 44100$, then the value of $n$ is:

(A) 15(B) 20(C) 21(D) 25

📚 Method A: Formal Textbook Derivation

Step 1: Set $\left[\frac{n(n+1)}{2}\right]^2 = 44100$. Step 2: Take square root: $\frac{n(n+1)}{2} = \sqrt{44100} = 210$. Step 3: Solve $n(n+1) = 420 \implies 20 \times 21 = 420 \implies n = 20$ wait, check: $20 \times 21 = 420$, so $n=20$. Let us correct option check: If $n=20$, $\frac{20\times 21}{2} = 210$, $210^2 = 44100$. Thus $n=20$ (Option B).

⚡ Method B: 25-Second Exam Shortcut

Take square root of sum immediately: $\sqrt{44100} = 210$, then find consecutive integers multiplying to $420$ ($20 \times 21$).

💡 Tip: Watch out for arithmetic calculation slips on square roots.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: What is the sum of the cubes of the first 10 natural numbers, i.e., $\sum_{k=1}^{10} k^3$?

A 2025

B 3025

C 385

D 55

Derivation & Analysis:

Using the formula $\left[\frac{10(11)}{2}\right]^2 = (55)^2 = 3025$. Option (B) is correct.

MCQ 2: The sum $2^3 + 4^3 + 6^3 + \dots + 20^3$ is equal to:

A 6050

B 12100

C 24200

D 48400

Derivation & Analysis:

Factor out $2^3 = 8$ to get $8 \sum_{k=1}^{10} k^3 = 8 \times 3025 = 24200$. Option (C) is correct.

MCQ 3: If $\sum_{k=1}^{n} k^3 = 1296$, then the value of $n$ is:

A 6

B 8

C 9

D 12

Derivation & Analysis:

We have $\left[\frac{n(n+1)}{2}\right]^2 = 1296 \implies \frac{n(n+1)}{2} = \sqrt{1296} = 36 \implies n(n+1) = 72 = 8 \times 9 \implies n = 8$. Option (B) is correct.

MCQ 4: The sum of the cubes of the first $n$ odd natural numbers $1^3 + 3^3 + 5^3 + \dots + (2n-1)^3$ is given by:

A $n^2(2n^2 - 1)$

B $n^2(n^2 - 1)$

C $\frac{n^2(2n^2+1)}{2}$

D $2n^4 - n^2$

Derivation & Analysis:

Standard identity for the sum of cubes of odd integers is $n^2(2n^2 - 1)$. Option (A) is correct.

MCQ 5: Evaluate the finite sum $\sum_{k=4}^{8} k^3$.

A 1260

B 1274

C 1296

D 1320

Derivation & Analysis:

$\sum_{k=1}^{8} k^3 - \sum_{k=1}^{3} k^3 = \left(\frac{8 \times 9}{2}\right)^2 - \left(\frac{3 \times 4}{2}\right)^2 = (36)^2 - (6)^2 = 1296 - 36 = 1260$. Wait, let us recalculate: $36^2 = 1296$, $6^2 = 36$, $1296 - 36 = 1260$. Option (A) is correct.

MCQ 6: Which of the following polynomial identities is used to derive the sum of cubes formula?

A $(k+1)^2 - k^2 = 2k + 1$

B $(k+1)^3 - k^3 = 3k^2 + 3k + 1$

C $(k+1)^4 - k^4 = 4k^3 + 6k^2 + 4k + 1$

D $(k+1)^5 - k^5 = 5k^4 + \dots$

Derivation & Analysis:

The telescoping series expansion of $(k+1)^4 - k^4$ isolates $\sum k^3$. Option (C) is correct.

MCQ 7: If the sum of the cubes of two numbers is 72 and the sum of the numbers is 6, then the sum of their squares is:

A 18

B 20

C 24

D 30

Derivation & Analysis:

Let numbers be $a, b$. $a+b=6$, $a^3+b^3 = (a+b)(a^2 - ab + b^2) = 72 \implies 6(a^2 - ab + b^2) = 72 \implies a^2 - ab + b^2 = 12$. Also $(a+b)^2 = a^2 + 2ab + b^2 = 36$. Subtracting gives $3ab = 24 \implies ab = 8$. Then $a^2 + b^2 = 36 - 2(8) = 20$. Option (B) is correct.

MCQ 8: Assertion (A): $\sum_{k=1}^{n} k^3 = \left(\sum_{k=1}^{n} k\right)^2$. Reason (R): Nicomachus's theorem establishes equality between sum of consecutive cubes and the square of triangular numbers.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Both statement and reason are mathematically rigorous and directly linked. Option (A) is correct.

MCQ 9: The value of $\lim_{n \to \infty} \frac{1^3 + 2^3 + 3^3 + \dots + n^3}{n^4}$ is:

A $\frac{1}{2}$

B $\frac{1}{4}$

C $\frac{1}{3}$

D 1

Derivation & Analysis:

$\lim_{n \to \infty} \frac{n^2(n+1)^2}{4n^4} = \lim_{n \to \infty} \frac{n^2(n^2+2n+1)}{4n^4} = \frac{1}{4}$. Option (B) is correct.

Q10

MCQ 10: What is the last digit of the sum $1^3 + 2^3 + 3^3 + \dots + 100^3$?

A 0

B 5

C 4

D 6

Derivation & Analysis:

For $n=100$, sum is $\left[\frac{100 \times 101}{2}\right]^2 = (5050)^2$, which ends in $0$. Option (A) is correct.

Common questions

This algebraic identity stems from Nicomachus's theorem and can be proven rigorously using mathematical induction or telescoping sums with fourth powers.

Aspirants frequently confuse the formula with the sum of squares $\frac{n(n+1)(2n+1)}{6}$ or fail to adjust the limits when the series starts from $k > 1$.

Memorize standard values for $n=5, 10, 20$, utilize unit digit modulo tests, and instantly factor out common multipliers in even/odd cube series.

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