The sum of all possible values of n N, so that the coefficients of x, x2 and x3 in the expansion of (1 + x2)2 (1 + x)n…

Year	2026	2025	2024	Total
Questions	9	17	11	37

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

Expand the expression:

(1 + x²)² (1 + x)ⁿ = (1 + 2x² + x⁴) \cdot Σr=0ⁿ ⁿCᵣ x^r

We need the coefficients of $x$ $x$, $x²$ $x^2$, and $x³$ $x^3$. For $x¹$ $x^1$: Multiplier is $1 \cdot ⁿC₁x = ⁿC₁$ $1 \cdot {^nC_1}x = {^nC_1}$. For $x²$ $x^2$: Multiplier is $1 \cdot ⁿC₂x² + 2x² \cdot ⁿC₀ = ⁿC₂ + 2$ . For $x³$ $x^3$: Multiplier is $1 \cdot ⁿC₃x³ + 2x² \cdot ⁿC₁x = ⁿC₃ + 2ⁿC₁$ .

These three coefficients are in Arithmetic Progression (A.P.).

Step 1: Set up the A.P. Condition

If $A, B, C$ $A, B, C$ are in A.P., then $2B = A + C$ $2B = A + C$.

2(ⁿC₂ + 2) = ⁿC₁ + (ⁿC₃ + 2ⁿC₁)

2((n(n-1))/(2) + 2) = 3n + (n(n-1)(n-2))/(6) (assuming n \geq 3)

Step 2: Solve the Polynomial Equation

n(n-1) + 4 = 3n + (n(n-1)(n-2))/(6)

Multiply by $6$ $6$:

6n² - 6n + 24 = 18n + n(n² - 3n + 2)

6n² - 6n + 24 = 18n + n³ - 3n² + 2n

n³ - 9n² + 26n - 24 = 0

Testing integer roots, for $n = 2$ $n = 2$: $8 - 36 + 52 - 24 = 0 \Rightarrow n=2$ $8 - 36 + 52 - 24 = 0 \Rightarrow n=2$ is a root (wait, A.P. works for $n=2$ $n=2$ also, let's check later). For $n = 3$ $n = 3$: $27 - 81 + 78 - 24 = 0 \Rightarrow n=3$ $27 - 81 + 78 - 24 = 0 \Rightarrow n=3$ is a root. For $n = 4$ $n = 4$: $64 - 144 + 104 - 24 = 0 \Rightarrow n=4$ is a root.

Step 3: Verification of Roots

Check for $n = 2$ $n = 2$ (since our formula assumed $n \geq 3$ $n \ge 3$, $²C₃=0$ ${^2C_3}=0$): Coeff of $x = 2$ $x = 2$, Coeff of $x² = 1+2 = 3$ $x^2 = 1+2 = 3$, Coeff of $x³ = 0 + 2(2) = 4$ $x^3 = 0 + 2(2) = 4$. Sequence is $2, 3, 4$ $2, 3, 4$ which is an A.P. So $n=2$ $n=2$ is valid. The values of $n$ $n$ are $2, 3, 4$ $2, 3, 4$. Required sum of values = $2 + 3 + 4 = 9$ $2 + 3 + 4 = 9$.

Pattern Recognition

Whenever binomial limits logically constrain formulas (like $ⁿC₃$ $^nC_3$ demanding $n\geq 3$ $n\geq 3$), explicitly check the lower bounds manually ( $n=2$ $n=2$) to avoid dropping edge-case roots from the algebraic polynomial factorization.

Chapter Mix

Class 11 Maths: Binomial Theorem

More Binomial Theorem Previous-Year Questions — Page 8

Q2 jee_main_2024_31_jan_morning Sum of Coefficients and Limits

Let

a

be the sum of all coefficients in the expansion of

(1 - 2x + 2x²)²⁰²³ (3 - 4x² + 2x³)²⁰²⁴

and

b = x \to 0 ( \int₀x (1 + t)t²⁰²⁴ + 1 dtx² )

. If the equations

cx² + dx + e = 0

and

2bx² + ax + 4 = 0

have a common root, where

c, d, e in R

, then

d : c : e

equals

A. $2:1:4$ $2:1:4$
B. $4:1:4$ $4:1:4$
C. $1:2:4$ $1:2:4$
D. $1:1:4$ $1:1:4$

Solution

Core Logic

To find the sum of all coefficients in a polynomial expansion, substitute $x = 1$ $x = 1$.

a = (1 - 2(1) + 2(1)²)²⁰²³ (3 - 4(1)² + 2(1)³)²⁰²⁴

a = (1)²⁰²³ (1)²⁰²⁴ = 1

Step 1: Evaluate Limit for b

Evaluate $b = x \to 0 \int₀x ln(1 + t)1 + t²⁰²⁴ dtx²$ Using L'Hôpital's Rule (differentiating numerator via Newton-Leibniz):

b = x \to 0 ln(1 + x)1 + x²⁰²⁴2x = x \to 0 (ln(1 + x))/(x) \times 12(1 + x²⁰²⁴)

b = 1 \times (1)/(2) = (1)/(2)

Step 2: Analyze Common Roots

The given second equation is $2bx² + ax + 4 = 0$ $2bx^2 + ax + 4 = 0$. Substitute $a = 1$ $a = 1$ and $b = (1)/(2)$ $b = \frac{1}{2}$:

2((1)/(2))x² + 1(x) + 4 = 0 x² + x + 4 = 0

The discriminant of $x² + x + 4 = 0$ $x^2 + x + 4 = 0$ is $D = 1 - 16 < 0$ $D = 1 - 16 < 0$. Roots are non-real complex conjugates.

Step 3: Final Ratio

Since $c, d, e in R$ $c, d, e \in \mathbb{R}$ and one root is common with a quadratic having non-real roots, both roots must be common. Thus, the coefficients must be proportional:

(c)/(1) = (d)/(1) = (e)/(4)

This implies $d : c : e = 1 : 1 : 4$ $d : c : e = 1 : 1 : 4$.

Pattern Recognition

If a quadratic equation with real coefficients shares a common root with another quadratic having complex roots ( $D < 0$ $D < 0$), both roots must be shared, meaning their coefficients are directly proportional.

Chapter Mix

Class 11 Maths: Binomial Theorem Class 12 Maths: Limits and Derivatives Class 11 Maths: Quadratic Equations

Q25 jee_main_2024_31_jan_morning Coefficients in Expansion

In the expansion of

(1 + x)(1 - x²)(1 + (3)/(x) + (3)/(x²) + (1)/(x³))⁵, x \neq 0

, the sum of the coefficient of

x³

and

x⁻¹³

is equal to

Numerical Answer. Answer: 118 to 118

Solution

Core Logic

(1+x)(1-x²) ( (1 + (1)/(x))³ )⁵

= (1+x)(1-x)(1+x) (x+1)¹⁵x¹⁵

= (1-x)(1+x)¹⁷x¹⁵

= (1+x)¹⁷ - x(1+x)¹⁷x¹⁵

Step 1: Find Coefficient of x^3

To find coeff of $x³$ $x^3$ in $(1+x)¹⁷ - x(1+x)¹⁷x¹⁵$ , we need the coeff of $x¹⁸$ $x^{18}$ in the numerator $(1+x)¹⁷ - x(1+x)¹⁷$ $(1+x)^{17} - x(1+x)^{17}$. The maximum power of $x$ $x$ in $(1+x)¹⁷$ $(1+x)^{17}$ is $17$ $17$, and in $x(1+x)¹⁷$ $x(1+x)^{17}$ is $18$ $18$. Coeff of $x¹⁸$ $x^{18}$ in $(1+x)¹⁷$ $(1+x)^{17}$ is 0. Coeff of $x¹⁸$ $x^{18}$ in $x(1+x)¹⁷$ $x(1+x)^{17}$ is the coeff of $x¹⁷$ $x^{17}$ in $(1+x)¹⁷$ $(1+x)^{17}$, which is $1717 = 1$ $\binom{17}{17} = 1$. Thus, coeff of $x¹⁸$ $x^{18}$ in the numerator is $0 - 1 = -1$ $0 - 1 = -1$.

Step 2: Find Coefficient of x^{-13}

To find coeff of $x⁻¹³$ $x^{-13}$, we need the coeff of $x²$ $x^2$ in the numerator $(1+x)¹⁷ - x(1+x)¹⁷$ $(1+x)^{17} - x(1+x)^{17}$. Coeff of $x²$ $x^2$ in $(1+x)¹⁷$ $(1+x)^{17}$ is $172$ $\binom{17}{2}$. Coeff of $x²$ $x^2$ in $x(1+x)¹⁷$ $x(1+x)^{17}$ is coeff of $x¹$ $x^1$ in $(1+x)¹⁷$ $(1+x)^{17}$, which is $171$ $\binom{17}{1}$. Value = $172 - 171 = (17 \times 16)/(2) - 17 = 136 - 17 = 119$ .

Step 3: Final Sum

Sum of coefficients = $-1 + 119 = 118$ $-1 + 119 = 118$.

Chapter Mix

Class 11 Maths: Binomial Theorem

The sum of all possible values of $n in N$ $n \in N$, so that the coefficients of $x, x²$ $x, x^2$ and $x³$ $x^3$ in the expansion of $(1 + x²)² (1 + x)ⁿ$ $(1 + x^2)^2 (1 + x)^n$, are in arithmetic progression is:

Solution & Explanation

Core Logic

Step 1: Set up the A.P. Condition

Step 2: Solve the Polynomial Equation

Step 3: Verification of Roots

Pattern Recognition

Chapter Mix

Reference Study Guides

More Binomial Theorem Previous-Year Questions — Page 8

Solution

Core Logic

Step 1: Evaluate Limit for b

Step 2: Analyze Common Roots

Step 3: Final Ratio

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Find Coefficient of x^3

Step 2: Find Coefficient of x^{-13}

Step 3: Final Sum

Chapter Mix

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The sum of all possible values of n in N$n \in N$, so that the coefficients of x, x²$x, x^2$ and x³$x^3$ in the expansion of (1 + x²)² (1 + x)ⁿ$(1 + x^2)^2 (1 + x)^n$, are in arithmetic progression is:

Solution & Explanation

Core Logic

Step 1: Set up the A.P. Condition

Step 2: Solve the Polynomial Equation

Step 3: Verification of Roots

Pattern Recognition

Chapter Mix

Reference Study Guides

More Binomial Theorem Previous-Year Questions — Page 8

Solution

Core Logic

Step 1: Evaluate Limit for b

Step 2: Analyze Common Roots

Step 3: Final Ratio

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Find Coefficient of x^3

Step 2: Find Coefficient of x^{-13}

Step 3: Final Sum

Chapter Mix

More Binomial Theorem Questions — jee_main_2026_23_january_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

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Formula Recognition

Exam Questions

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The sum of all possible values of $n in N$ $n \in N$, so that the coefficients of $x, x²$ $x, x^2$ and $x³$ $x^3$ in the expansion of $(1 + x²)² (1 + x)ⁿ$ $(1 + x^2)^2 (1 + x)^n$, are in arithmetic progression is: