If the Coefficient of x30 in the expansion of (1+(1/x))6(1+x2)7(1-x3)8, x0 is alpha, then |alpha| equals

Year	2026	2025	2024	Total
Questions	9	17	11	37

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

General term in a binomial expansion $(1+t)ⁿ$ $(1+t)^n$ is given by:

Tᵣ₊₁ = nr t^r

Core Logic

Let's simplify the algebraic structure of the product expression first:

(1+(1)/(x))⁶(1+x²)⁷(1-x³)⁸ = ((x+1)⁶ (1+x²)⁷ (1-x³)⁸)/(x⁶)

Finding the coefficient of $x³⁰$ $x^{30}$ in this full product is equivalent to finding the coefficient of $x³⁶$ $x^{36}$ in the numerator expansion:

Target = Coefficient of x³⁶ in (1+x)⁶ (1+x²)⁷ (1-x³)⁸

Step 1: Setting up General Term Constraints

The product of the three general terms is:

6r₁ xr₁ \cdot 7r₂ (x²)r₂ \cdot 8r₃ (-x³)r₃ = 6r₁ 7r₂ 8r₃ (-1)r₃ xr₁ + 2r₂ + 3r₃

We require the total exponent to equal $36$ $36$:

r₁ + 2r₂ + 3r₃ = 36

with boundaries $0 \leq r₁ \leq 6$ $0 \le r_1 \le 6$, $0 \leq r₂ \leq 7$ $0 \le r_2 \le 7$, $0 \leq r₃ \leq 8$ $0 \le r_3 \le 8$.

Step 2: Case Analysis by r3

Let's evaluate non-vanishing integer combinations case-by-case:

Case I: $r₃ = 8$ $r_3 = 8$

r₁ + 2r₂ = 36 - 24 = 12

$r₂ = 6, r₁ = 0 60 76 88(-1)⁸ = 1 \times 7 \times 1 = 7$
$r₂ = 5, r₁ = 2 62 75 88(-1)⁸ = 15 \times 21 \times 1 = 315$
$r₂ = 4, r₁ = 4 64 74 88(-1)⁸ = 15 \times 35 \times 1 = 525$
$r₂ = 3, r₁ = 6 66 73 88(-1)⁸ = 1 \times 35 \times 1 = 35$
Case II: $r₃ = 7$ $r_3 = 7$

r₁ + 2r₂ = 36 - 21 = 15

$r₂ = 7, r₁ = 1 61 77 87(-1)⁷ = 6 \times 1 \times 8 \times (-1) = -48$
$r₂ = 6, r₁ = 3 63 76 87(-1)⁷ = 20 \times 7 \times 8 \times (-1) = -1120$
$r₂ = 5, r₁ = 5 65 75 87(-1)⁷ = 6 \times 21 \times 8 \times (-1) = -1008$
Case III: $r₃ = 6$ $r_3 = 6$

r₁ + 2r₂ = 36 - 18 = 18

$r₂ = 7, r₁ = 4 64 77 86(-1)⁶ = 15 \times 1 \times 28 = 420$
$r₂ = 6, r₁ = 6 66 76 86(-1)⁶ = 1 \times 7 \times 28 = 196$

Step 3: Summation for Alpha

Summing all calculated values:

α = (7 + 315 + 525 + 35) + (-48 - 1120 - 1008) + (420 + 196)

α = 882 - 2176 + 616 = -678

Thus, the absolute value is:

$|α| = 678$ $|\alpha| = 678$

Pattern Recognition

Sees: Multi-product polynomial coefficient extraction problem. Trap: Remember that the negative sign inside $(1-x³)⁸$ $(1-x^3)^8$ alters the polarity of terms based on whether $r₃$ $r_3$ is odd or even. Always track the $(-1)r₃$ $(-1)^{r_3}$ factor carefully.

Chapter Mix

Class 11 Mathematics: 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

If the Coefficient of $x³⁰$ $x^{30}$ in the expansion of $(1+(1)/(x))⁶(1+x²)⁷(1-x³)⁸, x\neq0$ is $α$ $\alpha$, then $|α|$ $|\alpha|$ equals

Solution & Explanation

Related Formula

Core Logic

Step 1: Setting up General Term Constraints

Step 2: Case Analysis by r3

Step 3: Summation for Alpha

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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If the Coefficient of x³⁰$x^{30}$ in the expansion of (1+(1)/(x))⁶(1+x²)⁷(1-x³)⁸, x≠0$\left(1+\frac{1}{x}\right)^{6}(1+x^{2})^{7}(1-x^{3})^{8}, x\ne0$ is α$\alpha$, then |α|$|\alpha|$ equals

Solution & Explanation

Related Formula

Core Logic

Step 1: Setting up General Term Constraints

Step 2: Case Analysis by r3

Step 3: Summation for Alpha

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_2024_01_february_morning

Select Your Target Exam

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If the Coefficient of $x³⁰$ $x^{30}$ in the expansion of $(1+(1)/(x))⁶(1+x²)⁷(1-x³)⁸, x\neq0$ is $α$ $\alpha$, then $|α|$ $|\alpha|$ equals