Sum of the Infinite Series n³/3^(n−1) – Binomial Theorem JEE Main 2026 PYQ

Nishant Kumar Gupta

Former Faculty · Allen · Aakash · Narayana

Maths mentor behind PadhoLikhoJEE and Quantum Chemistry Classes (Arrah, Bihar). Teaches the “why” behind every JEE trap — not just the trick.

Quick Summary

Question Type:

Infinite series summation (generating-sum technique)

Chapter:

Binomial Theorem

Difficulty:

⭐⭐⭐ Medium

Time to Solve:

2 minutes

Key Concept:

Σ n³x^(n−1) = (1 + 4x + x²)/(1 − x)⁴; at x = 1/3 this is 99/8, and removing the n = 1 term leaves S = 91/8, so 8S = 91

Correct Answer:

Option (1) – 91

Why:

The full generating sum from n = 1 is 99/8; the n = 1 term is exactly 1, leaving S = 91/8

The Question

JEE Main 2026 (8 April – Evening Shift) – Binomial Theorem

If S = 2³/3 + 3³/3² + 4³/3³ + … to infinity, then the value of
8S is equal to:

Note: the series is Σ n³/3^(n−1) starting from n = 2.

1. (1) 91 ✓
2. (2) 89
3. (3) 93
4. (4) 87

Quick Answer

Correct Option: (1) – 91

Reasoning: one memorised generating sum evaluated at x = 1/3 finishes everything:

• Generating sum: Σ_n=1^∞ n³xⁿ⁻¹ = (1 + 4x + x²)/(1 − x)⁴ for |x| < 1
• At x = 1/3: numerator 1 + 4/3 + 1/9 = 22/9; denominator (2/3)⁴ = 16/81 → total (22/9)(81/16) = 99/8
• Remove the n = 1 term (which is 1³/3⁰ = 1): S = 99/8 − 1 = 91/8
• 8S = 91 ✓

Understanding the Concept

Polynomial Times Geometric — The Generating-Sum Family

A series like n³/3^(n−1) pairs a polynomial in the index with a geometric ratio — the classic
“power-weighted geometric” type. Its sums come from differentiating the geometric series.
Starting from Σ xⁿ = 1/(1 − x) and differentiating repeatedly (multiplying by x each time to
restore the exponent pattern) produces the family worth memorising verbatim:

All three hold for |x| < 1, with sums from n = 1. The cubic formula’s numerator 1 + 4x + x²
is the one to memorise cold — it appears in every exam from board papers to JEE Advanced.

The Starting-Index Audit

The given series begins at n = 2 (the terms 2³/3, 3³/3², … correspond to n³/3^(n−1) for
n = 2, 3, 4, …), while the generating sum starts at n = 1. The n = 1 term of the generating
sum is 1³/3⁰ = 1 — a value that must be subtracted, or the answer comes out 99/8 instead of
91/8 and 8S reads 99 instead of 91. Index audits are the whole game in series questions:
identify the general term, identify the summation start, and reconcile the two before
plugging numbers. Note the design of the distractors 89, 93, 87: they flank 91 at distance 2,
catching small arithmetic slips in the numerator 22/9 or the denominator 16/81, but the
dominant trap of this question type — forgetting the subtraction — lands on 99, which is
conspicuously absent. The options reward the index-aware solver.

Convergence Is Not Decoration

The formula requires |x| < 1, and x = 1/3 satisfies it comfortably. It is worth a sentence
in the rough work: the series converges because the ratio test gives limit (n+1)³/3^(n+1) ÷
n³/3ⁿ → 1/3 < 1. Convergence justifies term-by-term differentiation of the geometric series
behind the generating sums, so citing it is not pedantry — it is the licence that makes every
line of the method legal.

The Numeric Feel of the Answer

Direct addition of the first several terms — 2.667 + 3 + 2.370 + 1.543 + 0.889 + 0.471 +
0.234 + … — shows a total creeping towards about 11.375 = 91/8, and 8 × 11.375 = 91 exactly.
A 10-second partial-sum estimate (first four terms already give ≈ 9.6, and the tail adds
roughly 1.8) independently confirms the option before committing. Building this
estimate-then-verify rhythm makes series questions nearly self-checking.

Detailed Step-by-Step Solution

Final Answer

Essential Facts for This Topic

Generating-Sum Family (|x| < 1, sums from n = 1)

1. Geometric base: Σ x^(n−1) = 1/(1 − x)
2. Linear: Σ n x^(n−1) = 1/(1 − x)²
3. Quadratic: Σ n² x^(n−1) = (1 + x)/(1 − x)³
4. Cubic: Σ n³ x^(n−1) = (1 + 4x + x²)/(1 − x)⁴ — the numerator 1 + 4x + x² is the memorise-cold piece
5. Index audit: adjust between n = 1 and n = 2 starts by adding/subtracting the missing term explicitly
6. Ratio test: (n+1)³/3^(n+1) ÷ n³/3ⁿ → 1/3 < 1 — convergence licence for all the machinery

Common Mistakes to Avoid

Key Concept Summary

What You Must Remember

1. General term: n³/3^(n−1) = n³(1/3)^(n−1), starting at n = 2
2. Generating sum: Σ n³x^(n−1) = (1 + 4x + x²)/(1 − x)⁴ — the cubic member of the family
3. At x = 1/3: (22/9)/(16/81) = 99/8; subtract the n = 1 term (1) → S = 91/8
4. 8S = 91 — an integer by design; distractors 87/89/93 flank it for small slips
5. Two independent checks: partial-sum estimate ≈ 11.37 and the index audit both confirm

The Golden Rule for Power-Weighted Geometric Series

“Polynomial-times-geometric series die by the generating-sum family — memorise up to n³, then audit the starting index before plugging in x.”

Frequently Asked Questions

Q1: Where does the formula (1 + 4x + x²)/(1 − x)⁴ come from?

A: Differentiate the geometric series Σ xⁿ = 1/(1 − x) three times, multiplying by x after each differentiation to realign exponents. The first derivative gives Σ n x^(n−1) = 1/(1 − x)², the second gives Σ n² x^(n−1) = (1 + x)/(1 − x)³, and the third produces Σ n³ x^(n−1) = (1 + 4x + x²)/(1 − x)⁴, valid for |x| < 1.

Q2: Why must the n = 1 term be subtracted?

A: The memorised generating sum runs from n = 1, but the given series starts at n = 2. The n = 1 contribution is 1³/3⁰ = 1, so the tail sum is 99/8 − 1 = 91/8. Skipping this audit inflates the answer by exactly 8 in the 8S scale — from 91 to 99.

Q3: Does the series really converge, and why does that matter?

A: Yes: the ratio of successive terms is ((n+1)³/3^(n+1))/(n³/3ⁿ) → 1/3 < 1, so the series converges absolutely. Convergence is what licenses term-by-term differentiation of the geometric series — the step that builds the generating sums — so it is the legal foundation of the entire method.

Q4: What is the fastest partial-sum sanity check in the exam?

A: Add the first four terms: 8/3 + 3 + 64/27 + 125/81 ≈ 9.58, and estimate the remaining tail as roughly 1.8 (it is dominated by 216/243 + 343/729 + … ≈ 0.89 + 0.47 + …). The total ≈ 11.4 matches 91/8 ≈ 11.375, so 8S = 91 is confirmed without any algebra.

Q5: How would the question change if the series started from n = 1?

A: Then S would be the full generating sum 99/8 and 8S = 99. The starting index is the single lever that moves this question between answers — which is exactly why reading the first displayed term (2³/3, not 1³/1) is the most important ten seconds of the problem.

Prerequisites to Solve This Question

Before attempting this problem, you should be comfortable with:

1. Geometric series and its sum: Σ xⁿ = 1/(1 − x) for |x| < 1
2. Term-by-term differentiation: how the generating-sum family is generated
3. The cubic generating sum: Σ n³x^(n−1) = (1 + 4x + x²)/(1 − x)⁴ — memorised or rebuildable
4. Index notation fluency: matching displayed terms (2³/3, 3³/3², …) to n³/3^(n−1) from n = 2
5. Ratio test basics: confirming convergence of n³x^(n−1) at a given ratio

After Solving This, You Can:

• Sum Σ n x^(n−1), Σ n² x^(n−1), Σ n³ x^(n−1) at any |x| < 1 in under a minute each
• Handle any starting index by adding or subtracting the missing terms explicitly
• Tackle Σ n³/2ⁿ-style variants (convert to the x^(n−1) frame or use the xⁿ version directly)
• Build the whole family from scratch via differentiation if memory wobbles mid-exam
• Sanity-check any infinite series answer with a four-term partial sum plus tail estimate

Study Tips for This Topic

For JEE Main:

1. Memorise the numerator 1 + 4x + x²: it is the single most examinable piece of the family; rebuild-by-derivative is the backup
2. Index audit is a written step: literally write ‘start at n = 2, formula starts at n = 1, subtract 1’ — the 8-point swing deserves the ink
3. Budget 2 minutes: general term (20s) + plug x = 1/3 (40s) + index correction (20s) + 8S and partial-sum check (40s)

Common JEE Variants:

• Σ n³/2ⁿ from n = 1 — the famous value 26, the sibling of this question
• Σ n²/3ⁿ and Σ n/3ⁿ one-liners using the quadratic and linear family members
• Alternating versions Σ (−1)ⁿ n³/3ⁿ — evaluate at x = −1/3
• “Find the coefficient of xⁿ in (1 + x)(1 + x + x²)ⁿ”-style bridge questions between BT and series
• Finite versions: Σ_{n=1}^{k} n³/3^(n−1) left as an expression — generating sums minus the tail

Difficulty Rating & Exam Frequency

Difficulty Level: ⭐⭐⭐ (3/5) – Medium (one memorised formula does the work; the starting-index audit and the (2/3)⁴ arithmetic carry the risk)

JEE Main Frequency: High — infinite power-weighted series appear in most JEE Main papers, straddling Binomial Theorem and Sequences and Series

JEE Advanced Frequency: Medium — Advanced demands the same family inside generating-function arguments

Topic Importance: High — the generating-sum family is among the highest-ROI memorisations in the syllabus

Written by Nishant Kumar Gupta
Former Faculty · Allen · Aakash · Narayana — Quantum Chemistry Classes, Arrah (Bihar)
Last Updated: 2 Sep 2026 ·
Question Source: JEE Main 2026 (8 April – Evening Shift) Previous Year Question ·
Topic: Binomial Theorem / Sequences and Series – Generating Sums and Index Audits