Let alpha = =0n((nCk)2k+1) and beta = =0n-1(nCknCk+1k+2) . If 5alpha = 6beta , then n equals

Year	2026	2025	2024	Total
Questions	9	17	11	37

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

(ⁿCᵣ)/(r+1) = ⁿ⁺¹Cᵣ₊₁n+1

Σk=0ⁿ ⁿ⁺¹Ck+1 \cdot ⁿCn-k = ²ⁿ⁺¹Cₙ₊₁

Core Logic

Simplify $α$ $\alpha$ by absorbing the denominator:

α = Σk=0ⁿ ⁿCk \cdot ⁿCkk + 1

Multiply and divide by $(n+1)$ $(n+1)$:

α = (1)/(n+1) Σk=0ⁿ (n+1)/(k+1) ⁿCk \cdot ⁿCn-k

α = (1)/(n+1) Σk=0ⁿ ⁿ⁺¹Ck+1 \cdot ⁿCn-k

This corresponds to choosing $k+1$ $k+1$ items from a set of $n+1$ $n+1$, and $n-k$ $n-k$ items from a set of $n$ $n$. Total items chosen = $(k+1) + (n-k) = n+1$ $(k+1) + (n-k) = n+1$ from a total pool of $(n+1) + n = 2n+1$ $(n+1) + n = 2n+1$.

α = (1)/(n+1) \cdot ²ⁿ⁺¹Cₙ₊₁

Step 1: Evaluating Beta

Now for $β$ $\beta$:

β = Σk=0ⁿ⁻¹ ⁿCk \cdot ⁿCk+1k + 2

Convert $ⁿCk$ ${}^nC_k$ to $ⁿCn-k$ ${}^nC_{n-k}$ and absorb the denominator into $ⁿCk+1$ ${}^nC_{k+1}$:

β = (1)/(n+1) Σk=0ⁿ⁻¹ ⁿCn-k \cdot (n+1)/(k+2) ⁿCk+1

β = (1)/(n+1) Σk=0ⁿ⁻¹ ⁿCn-k \cdot ⁿ⁺¹Ck+2

Here, total items chosen is $(n-k) + (k+2) = n+2$ $(n-k) + (k+2) = n+2$ from a total pool of $n + (n+1) = 2n+1$ $n + (n+1) = 2n+1$.

β = (1)/(n+1) \cdot ²ⁿ⁺¹Cₙ₊₂

Step 2: Applying the Given Ratio

Given $5α = 6β \Rightarrow (β)/(α) = (5)/(6)$ .

(β)/(α) = (1)/(n+1) ²ⁿ⁺¹Cₙ₊₂(1)/(n+1) ²ⁿ⁺¹Cₙ₊₁ = ²ⁿ⁺¹Cₙ₊₂²ⁿ⁺¹Cₙ₊₁

Using the ratio property $ⁿCᵣⁿCᵣ₋₁ = (n-r+1)/(r)$ : Here, $n \to 2n+1$ $n \to 2n+1$ and $r \to n+2$ $r \to n+2$.

(β)/(α) = (2n+1 - (n+2) + 1)/(n+2) = (2n+1 - n - 2 + 1)/(n+2) = (n)/(n+2)

Step 3: Solving for n

Equate and solve:

(n)/(n+2) = (5)/(6)

6n = 5n + 10 \Rightarrow n = 10

Pattern Recognition

Any fractional binomial coefficient like $Ck / (k+x)$ $C_k / (k+x)$ triggers an absorption identity to step up the index. Then, multiplying terms turns into a simple combinatorial Vandermonde convolution.

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

Let $α = Σk=0ⁿ( (ⁿCk)²k+1)$ and $β = Σk=0ⁿ⁻¹( ⁿCkⁿCk+1k+2)$ . If $5α = 6β$ $5\alpha = 6\beta$ , then $n$ $n$ equals

Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluating Beta

Step 2: Applying the Given Ratio

Step 3: Solving for n

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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Let α = Σk=0ⁿ( (ⁿCk)²k+1)$\alpha = \sum_{k=0}^{n}\left(\frac{({}^nC_k)^2}{k+1}\right)$ and β = Σk=0ⁿ⁻¹( ⁿCkⁿCk+1k+2)$\beta = \sum_{k=0}^{n-1}\left(\frac{{}^nC_k{}^nC_{k+1}}{k+2}\right)$ . If 5α = 6β$5\alpha = 6\beta$ , then n$n$ equals

Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluating Beta

Step 2: Applying the Given Ratio

Step 3: Solving for n

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_30_january_evening

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Let $α = Σk=0ⁿ( (ⁿCk)²k+1)$ and $β = Σk=0ⁿ⁻¹( ⁿCkⁿCk+1k+2)$ . If $5α = 6β$ $5\alpha = 6\beta$ , then $n$ $n$ equals