For some n neq 10, let the coefficients of the 5th, 6th and 7th terms in the binomial expansion of (1 + x)n+4 be in…

Q1 jee_main_2024_27_jan_morning Properties of Binomial Coefficients

Given

^n-1C_r=(k^2-8)^nC_r+1

, this holds true if and only if:

A. $2sqrt2 lt k le 3$ $2\sqrt{2} \lt k \le 3$
B. $2sqrt3 lt k le 3sqrt2$ $2\sqrt{3} \lt k \le 3\sqrt{2}$
C. $2sqrt3 lt k lt 3sqrt3$ $2\sqrt{3} \lt k \lt 3\sqrt{3}$
D. $2sqrt2 lt k lt 2sqrt3$ $2\sqrt{2} \lt k \lt 2\sqrt{3}$

Solution

### Related Formula

frac^n-1C_r^nC_r+1 = fracr+1n

### Core Logic From the given equation:

(k^2 - 8) = frac^n-1C_r^nC_r+1

Applying the combination property, we get:

k^2 - 8 = fracr+1n

Since

n, r ge 0

and

n ge r+1

for the combination to be valid, the ratio

fracr+1n

must satisfy:

0 < fracr+1n le 1

Substituting this bound into our expression:

0 < k^2 - 8 le 1

### Step 1: Solving the Inequalities First inequality:

k^2 - 8 > 0 Rightarrow k^2 > 8 Rightarrow k in (-infty, -2sqrt2) cup (2sqrt2, infty)

Second inequality:

k^2 - 8 le 1 Rightarrow k^2 - 9 le 0 Rightarrow k^2 le 9 Rightarrow -3 le k le 3

Taking the intersection of both intervals:

k in [-3, -2sqrt2) cup (2sqrt2, 3]

### Step 2: Final Conclusion Looking at the options, we consider the positive domain interval:

2sqrt2 < k le 3

### Pattern Recognition Combinatorics identities often reduce to bounds on variables. Memorize the ratio

frac^n-1C_r^nC_r+1 = fracr+1n

and use the strict combinatorial limit

0 < fracr+1n le 1

to form algebraic inequalities. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem Class 11 Maths: Linear Inequalities

Q5 jee_main_2024_27_jan_morning Sum of Binomial Coefficients

If

A

denotes the sum of all the coefficients in the expansion of

(1-3x+10x^2)^n

and

B

denotes the sum of all the coefficients in the expansion of

(1+x^2)^n

, then :

A. $A=B^3$ $A=B^{3}$
B. $3A=B$ $3A=B$
C. $B=A^3$ $B=A^{3}$
D. $A=3B$ $A=3B$

Solution

### Related Formula

textSum of all coefficients in f(x)^n = f(1)^n

### Core Logic To find the sum of all coefficients in a polynomial expansion, substitute the variable

x = 1

. For

A

(Sum of coefficients of

(1-3x+10x^2)^n

):

A = (1 - 3(1) + 10(1)^2)^n

A = (1 - 3 + 10)^n = (8)^n

For

B

(Sum of coefficients of

(1+x^2)^n

):

B = (1 + (1)^2)^n

B = (1 + 1)^n = (2)^n

### Step 1: Establishing the Relation We have

A = 8^n

and

B = 2^n

. Observe that

8 = 2^3

.

A = (2^3)^n = (2^n)^3

Substituting

B = 2^n

into the expression:

A = B^3

### Pattern Recognition Summing coefficients of *any* multi-nomial expansion is trivially simple: just plug in

x=1

(or

x=y=z=1

for multivariable expressions). It collapses the variables, leaving only the arithmetic sum of the coefficients. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem

Q29 jee_main_2024_29_jan_morning Sum of Binomial Coefficients

If

frac^11C_12+frac^11C_23+dots+frac^11C_910=fracnm

with

gcd(n,m)=1

, then

n+m

is equal to

Numerical Answer. Answer: 2041 to 2041

Solution

### Related Formula

frac^nC_rr+1 = frac^n+1C_r+1n+1

sum_k=0^n ^nC_k = 2^n

### Core Logic The given series can be rewritten using summation notation:

S = sum_r=1^9 frac^11C_rr+1

Applying the coefficient shifting identity

frac^nC_rr+1 = frac^n+1C_r+1n+1

:

S = sum_r=1^9 frac^12C_r+112

S = frac112 sum_r=1^9 ^12C_r+1

### Step 1: Expand and Complete the Series Expand the internal sum by shifting the index bounds:

S = frac112 left( ^12C_2 + ^12C_3 + dots + ^12C_10 right)

We know the complete sum of binomial coefficients for

n=12

is

2^12

. We just need to subtract the missing boundary terms:

k = 0, 1, 11, 12

.

2^12 = sum_k=0^12 ^12C_k

The missing terms evaluate to:

^12C_0 = 1

^12C_1 = 12

^12C_11 = 12

^12C_12 = 1

Sum of missing terms

= 1 + 12 + 12 + 1 = 26

. ### Step 2: Calculate Final Fraction Substitute this back into the series equation:

S = frac112 left[ 2^12 - 26 right]

S = frac112 [ 4096 - 26 ]

S = frac407012

Simplify the fraction by dividing by 2 to achieve the coprime structure

fracnm

:

S = frac20356

Thus,

n = 2035

and

m = 6

, and they are coprime (

gcd(2035, 6) = 1

). Calculate

n + m

:

n + m = 2035 + 6 = 2041

### Pattern Recognition Whenever you see binomial coefficients divided by their sequential index (

r+1

), always use the absorption identity

frac1n+1 binomn+1r+1

to bump the top index up by 1. Then fill the array to force the complete

2^n+1

sum. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

Q17 jee_main_2024_30_january_evening Binomial Coefficients

Suppose

2 - p, p, 2 - alpha, alpha

are the coefficient of four consecutive terms in the expansion of

(1 + x)^n

. Then the value of

p^2 - alpha^2 + 6alpha + 2p

equals

A. $4$ $4$
B. $10$ $10$
C. $8$ $8$
D. $6$ $6$

Solution

### Related Formula

textCoefficient of (r+1)^textth text term in (1+x)^n text is ^nC_r

^nC_r + ^nC_r+1 = ^n+1C_r+1

### Core Logic Let the four consecutive binomial coefficients be

^nC_r, ^nC_r+1, ^nC_r+2, ^nC_r+3

. Given these correspond to

2 - p, p, 2 - alpha, alpha

respectively. From the properties of combinations:

^nC_r + ^nC_r+1 = (2 - p) + p = 2

Rightarrow ^n+1C_r+1 = 2 quad dots(1)

Similarly:

^nC_r+2 + ^nC_r+3 = (2 - alpha) + alpha = 2

Rightarrow ^n+1C_r+3 = 2 quad dots(2)

### Step 1: Identifying the Inconsistency From equations (1) and (2):

^n+1C_r+1 = ^n+1C_r+3 = 2

By combination properties, if

^nC_x = ^nC_y

and

x neq y

, then

x + y = n

. So,

(r+1) + (r+3) = n+1 Rightarrow 2r + 4 = n+1 Rightarrow n = 2r + 3

. Substitute

n

back into the equality:

^2r+4C_r+1 = 2

The binomial coefficient must be

ge 2

. However, for any valid integer

r ge 0

,

^2r+4C_r+1

grows very rapidly. Let's test small values: If

r = 0

,

^4C_1 = 4 neq 2

. If

r = 1

,

^6C_2 = 15 neq 2

. Hence, no integer values satisfy this condition, making the given data inherently inconsistent. ### Step 2: Conclusion Due to inconsistent data resulting in a mathematically impossible scenario, this question was treated as a Bonus/Dropped question. ### Pattern Recognition Adding consecutive binomial coefficients yields the sum from Pascal's triangle identity

^nC_r + ^nC_r+1 = ^n+1C_r+1

. If identical sums yield small integer invariants like 2, they typically break bounding limits for

^nC_k

combinations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem

Q24 jee_main_2024_30_january_evening Properties of Binomial Coefficients

Let

alpha = sum_k=0^nleft(frac(^nC_k)^2k+1right)

and

beta = sum_k=0^n-1left(frac^nC_k^nC_k+1k+2right)

. If

5alpha = 6beta

, then

n

equals

Numerical Answer. Answer: 10 to 10

Solution

### Related Formula

frac^nC_rr+1 = frac^n+1C_r+1n+1

sum_k=0^n ^n+1C_k+1 cdot ^nC_n-k = ^2n+1C_n+1

### Core Logic Simplify

alpha

by absorbing the denominator:

alpha = sum_k=0^n frac^nC_k cdot ^nC_kk + 1

Multiply and divide by

(n+1)

:

alpha = frac1n+1 sum_k=0^n fracn+1k+1 ^nC_k cdot ^nC_n-k

alpha = frac1n+1 sum_k=0^n ^n+1C_k+1 cdot ^nC_n-k

This corresponds to choosing

k+1

items from a set of

n+1

, and

n-k

items from a set of

n

. Total items chosen =

(k+1) + (n-k) = n+1

from a total pool of

(n+1) + n = 2n+1

.

alpha = frac1n+1 cdot ^2n+1C_n+1

### Step 1: Evaluating Beta Now for

beta

:

beta = sum_k=0^n-1 frac^nC_k cdot ^nC_k+1k + 2

Convert

^nC_k

to

^nC_n-k

and absorb the denominator into

^nC_k+1

:

beta = frac1n+1 sum_k=0^n-1 ^nC_n-k cdot fracn+1k+2 ^nC_k+1

beta = frac1n+1 sum_k=0^n-1 ^nC_n-k cdot ^n+1C_k+2

Here, total items chosen is

(n-k) + (k+2) = n+2

from a total pool of

n + (n+1) = 2n+1

.

beta = frac1n+1 cdot ^2n+1C_n+2

### Step 2: Applying the Given Ratio Given

5alpha = 6beta Rightarrow fracbetaalpha = frac56

.

fracbetaalpha = fracfrac1n+1 ^2n+1C_n+2frac1n+1 ^2n+1C_n+1 = frac^2n+1C_n+2^2n+1C_n+1

Using the ratio property

frac^nC_r^nC_r-1 = fracn-r+1r

: Here,

n to 2n+1

and

r to n+2

.

fracbetaalpha = frac2n+1 - (n+2) + 1n+2 = frac2n+1 - n - 2 + 1n+2 = fracnn+2

### Step 3: Solving for n Equate and solve:

fracnn+2 = frac56

6n = 5n + 10 Rightarrow n = 10

### Pattern Recognition Any fractional binomial coefficient like

C_k / (k+x)

triggers an absorption identity to step up the index. Then, multiplying terms turns into a simple combinatorial Vandermonde convolution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem

Solution & Explanation

Reference Study Guides

More Binomial Theorem Previous-Year Questions — Page 5

Solution

Solution

Solution

Solution

Solution

More Binomial Theorem Questions — jee_main_2025_24_jan_morning

We Map Every Repeating Question in Competitive Exams.

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

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

For some n neq 10$n \neq 10$, let the coefficients of the 5^textth$5^{\text{th}}$, 6^textth$6^{\text{th}}$ and 7^textth$7^{\text{th}}$ terms in the binomial expansion of (1 + x)^n+4$(1 + x)^{n+4}$ be in A.P. Then the largest coefficient in the expansion of (1 + x)^n+4$(1 + x)^{n+4}$ is :

Solution & Explanation

Reference Study Guides

More Binomial Theorem Previous-Year Questions — Page 5

Solution

Solution

Solution

Solution

Solution

More Binomial Theorem Questions — jee_main_2025_24_jan_morning

We Map Every Repeating Question in Competitive Exams.

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

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back