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All questions of Binomial Theorem & its Simple Applications for NDA Exam

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The number of terms in the expansion of (x – y + 2z)7 are:
  • a)
    35
  • b)
    38
  • c)
    36
  • d)
    37
Correct answer is option 'C'. Can you explain this answer?

Tejas Verma answered
Here the number of terms can be calculated by:
= ((n+ 1) * (n+2)) /2
where, n =7
∴ Number of terms = 36

In the expansion of (a+b)n, nN the number of terms is:
  • a)
    n + 1
  • b)
    n
  • c)
    n – 1
  • d)
    an
Correct answer is option 'A'. Can you explain this answer?

Sruthi Manoj answered
For example if it is (a b)², then the expansion is a² 2ab b² which consist of 3 terms. Similarly (a b)¹= a b which consists of 2 terms. therefore that implies (a b) raised to n, then the no. of terms is n 1.

The expansion of  , in powers of x, is valid if
  • a)
    |x| > 2
  • b)
    x < 2
  • c)
    x > 2
  • d)
    |x| < 2
Correct answer is option 'D'. Can you explain this answer?

Knowledge Hub answered
In case of negative or fractional power, expansion (1+x)^n is valid only when |x| < 1
(6 - 3x)-1/2
= (6-1/2 (1 - x/2)-1/2)
So, this equation exists only when |x/2| < 1
|x| < 2

If n is a +ve integer, then the binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are
  • a)
    additive inverse of each other
  • b)
    multiplicative inverse of each other
  • c)
    equal
  • d)
    nothing can be said
Correct answer is option 'C'. Can you explain this answer?

Hansa Sharma answered
(x+a)n = nC0 xn + nC1 x(n-1) a1 + nC2 x(n-2) a2 + ..........+ nC(n-1) xa(n-1) + nCn  an
Now, nC0 = nCn, nC1 = nCn-1,    nC2 = nCn-2,........
therefore, nCr = nCn-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.

The value of 1261/3 upto three decimals is
  • a)
    5.013
  • b)
    5.014
  • c)
    5.012
  • d)
    5.011
Correct answer is option 'A'. Can you explain this answer?

Hiral Choudhary answered
To find the value of 1261/3 upto three decimals, we can use the long division method or a calculator to obtain the quotient.

Long division method:

- We first write 1261 as the dividend and 3 as the divisor.
- We then perform the division as follows:

4.203666...

- We stop at three decimal places, which gives us the final answer as 5.013.

Therefore, the correct option is A) 5.013.

the coefficient of xn in the expansion of 
  • a)
    n
  • b)
    2n
  • c)
    4n
  • d)
    zero
Correct answer is option 'C'. Can you explain this answer?

Suresh Reddy answered
[(1+x)/(1−x)]2
⇒ (1+x)2(1−x)-2
⇒ (1 + x2 + 2x)[1 + 2x + 3x2 +..... +(n−1)xn-2 + nxn-1 + (n+1)xn +...]
coeff of xn will be given by
(I) When 1 will be multiplied by (n+1)xn
(II) When x2 will be multiplied by (n−1)xn-1
(III) When 2x will be multiplied by nxn-1
∴ coeff. of xn = n + 1 + n − 1 + 2n
= 4n

What is the coefficient of x5 in the expansion of (1-x)-6 ?
  • a)
    252
  • b)
    250
  • c)
    -252
  • d)
    251
Correct answer is option 'A'. Can you explain this answer?

Preeti Iyer answered
(1-x)-6 
=> (1-x)(-6/1)
It is in the form of (1-x)(-p/q), p =6, q=1
(1-x)(-p/q) = 1+p/1!(x/q)1 + p(p+q)/2!(x/q)2 + p(p+q)(p+2q)/3!(x/q)3 + p(p+q)(p+2q)(p+3q)/4!(x/q)4........
= 1+6/1!(x/1)1 + 6(7)/2!(x/1)2 + 6(7)(8)/3!(x/1)3 + 6(7)(8)(9)/4!(x/1)4 +.......................
So, coefficient of x5 is (6*7*8*9*10)/120
= 252

 If in the expansion of (1+x)20, the coefficients of rth and (r+4)th terms are equal, then the value of r is equal to:
  • a)
    9
  • b)
    7
  • c)
    10
  • d)
    8
Correct answer is option 'A'. Can you explain this answer?

Raghav Bansal answered
Coefficients of the rth and (r+4)th terms in the given expansion are Cr−120  and 20Cr+3.
Here,Cr−120  = 20Cr+3
⇒ r−1+r+3 = 20 
[∵ if nCx = nCy  ⇒ x = y or x+y = n]
⇒ r = 2 or 2r = 18
⇒ r = 9  

The coefficient of y in the expansion of (y² + c/y)5 is 
  • a)
    10c 
  • b)
    10c² 
  • c)
    10c³ 
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Varun Kapoor answered
Given, binomial expression is (y² + c / y)5 
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r 
= 5Cr × y10-3r × Cr 
Now, 10 – 3r = 1 
⇒ 3r = 9 
⇒ r = 3 
So, the coefficient of y = 5C3 × c³ = 10c³

 What is the general term in the expansion of (2y-4x)44?
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Om Desai answered
 n = 44, p = 2y q = -4x
General term of (p+q)n is given by
T(r+1) = nCr . pr . q(n-r)
= 44Cr . (2y)r . (-4x)(44-r)

The coefficient of x4 in the expansion of (1 + x + x2 + x3)n is:
  • a)
    nC4 + nC2
  • b)
    nC4 + nC2+ nC4.nC2
  • c)
    nC4 + nC2+ nC1.nC2
  • d)
    nC4
Correct answer is option 'C'. Can you explain this answer?

Gaurav Kumar answered
x4 can be achieved in the following ways: 
x4 . 1(n-4) . (x2)0 . (x3)0
Hence, coefficient will be  nC4 .
x2 . 1(n-3) . (x2)1 . (x3)0
Hence, coefficient will be 3nC3.
x1 . 1(n-2) . (x2)0 . (x3)1
Hence, coefficient will be 2nC2.
x0 . 1(n-2) .(x2)2 .(x3)0
Hence, coefficient will be nC2 .
Hence, the required coefficient will be 
nC4 + 3nC3 + 3nC2
= nC4 + 3(nC3 + nC2).
= nC4  + 3(n+1C3).
= nC4  + nC2 + nC1 . nC2

Which of the following is divisible by 25:
  • a)
    6n - 5n + 1
  • b)
    6n + 5n
  • c)
    6n - 5n
  • d)
    6n - 5n - 1
Correct answer is option 'D'. Can you explain this answer?

Riya Banerjee answered
we can write (6ⁿ ) = (1 + 5)ⁿ
we know, according to binomial theorem,
(1 + x)ⁿ = 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! +.............∞ use this here,
(6)ⁿ = (1 + 5)ⁿ = 1 + 5n + n(n-1)5²/2! + n(n-1)(n-2)5³/3! +...........∞
= 1 + 5n + 5²{ n(n-1)/2! + n(n-1)(n-2)5/3! +.......∞}
Let P = n(n-1)/2! + n(n-1)(n-2)5/3! +.........∞
6ⁿ = 1 + 5n + 25P
6ⁿ - 5n = 1 + 25P -------(1)
but we know, according to Euclid algorithm ,
dividend = divisor × quotient + remainder ---(2)
compare eqn (1) to (2)
we observed that 6ⁿ -5 n always leaves the remainder 1 when divided by 25

The general term in the expansion of (a - b)n is
  • a)
    Tr+1 = (-1)r nCr an-r br
  • b)
    Tr+1 = nCr an-r br
  • c)
    Tr+1 = nCr ar br
  • d)
    Tr+1 = (-1)r nCr ar br
Correct answer is option 'A'. Can you explain this answer?

Poonam Reddy answered
If a and b are real numbers and n is a positive integer, then:
(a - b)n = nC0 an + nC1 a(n – 1) b1 + nC2 a(n – 2) b2+ ...... + nCr a(n – r) br+ ... + nCnbn,
The general term or (r + 1)th term in the expansion is given by:
Tr + 1 = (-1)Cr a(n–r) br

If 2nd, 3rd and 4th terms in the expansion of (x+a)n are 240, 720 and 1080 respectively, then the value of n is
  • a)
    15
  • b)
    20
  • c)
    5
  • d)
    10
Correct answer is option 'C'. Can you explain this answer?

Shiksha Academy answered
General term Tr+1 of (x+y)n is given by 
Tr+1 = nCr xn-r yr
T2 = nC2 xn-2 y = 240
T3 = nC3 xn-3 y2 = 720
T4 = nC4 xn-4 y3 = 1080
T3/T2 = [(n-1)/2] * [y/x] = 3......(1)
T4/T2 = {[(n-1)(n-2)]/(3*2)} * x2/y2 = 9/2
T4/T3 = [(n-2)/3] * [y/x] = 3/2...(2)
Dividing 1 by 2
[(n-1)/2] * [3/(n-2)] = 2
⇒ 3n−3 = 4n−8
⇒ 5 = n

 n-1Cr = (k2 - 3). nCr + 1 [JEE 2004 (Scr.)]
  • a)
     
  • b)
  • c)
     
  • d)
     
Correct answer is option 'D'. Can you explain this answer?

Rohit Jain answered
(n - 1)!/{r! × (n - 1 - r)!} = (k² - 3) × n!/(r + 1)!(n - r - 1)!
or, (n - 1)!/r! = (k² - 3) × n!/(r + 1)!
or, (n - 1)!/r! = (k² - 3) × n(n - 1)!/(r + 1)r!
or, 1/1 = (k² - 3) × n/(r + 1)
or, (r + 1)/n = (k² - 3)
we know, r and n are integers so, (r + 1)/n  (0, 1]
so, 0 < (r + 1)/n ≤ 1
or, 0 < k² - 3 ≤ 1
or, 3 < k² ≤ 4
or, √3 < k ≤ 2 , -2 ≤ k < -√3
hence, k  (√3, 2]

The coefficient of x3 in the binomial expansion of   
  • a)
    792m5
  • b)
    942m7
  • c)
    330m4
  • d)
    792m6
Correct answer is option 'C'. Can you explain this answer?

Geetika Shah answered
T(r+1) = 11Cr x(11-2r) (m/r)r (-1)r
= 11Cr x(11-2r) mr (-1)r
Coefficient of x3 = 11 - 2r = 3
8 = 2r 
r = 4
T5 = 11C4 x3 m4 (-1)
Coefficient of x3 = 11C4 m4
= 330 m4

The coefficient of x17 in the expansion of (x- 1) (x- 2) …..(x – 18) is
  • a)
    - 171
  • b)
    342
  • c)
    171/2
  • d)
    684
Correct answer is option 'A'. Can you explain this answer?

Infinity Academy answered
The coefficient of x17 is given by 
−1 + (−2) + (−3) + ….. (−18)
= −1 − 2 − 3….. − 18
= − (18(18+1))/2
= − 9(19)
= − 171

In the expansion of (1+x)60, the sum of coefficients of odd powers of x is
  • a)
    261
  • b)
    260
  • c)
    2
    59
  • d)
    none of these.
Correct answer is option 'C'. Can you explain this answer?

Mohit Mittal answered
C is correct as
sum of ( 1+ x )^60 gives 2^60 which includes both even and odd powers of x
so for only one type of power ( odd power) of x we divide 2^ 60 by 2 so we get 2^ 59

Find the largest co-efficient in the expansion of (1 + x)n, given that the sum of co-efficients of the terms in its expansion is 4096.     [REE 2000 (Mains)]
Correct answer is '924'. Can you explain this answer?

Ameya Iyer answered
We know that, the coefficients in a binomial expansion is obtained by replacing each variable by unit in the given expression.
Therefore, sum of the coefficients in (a+b)^n
= 4096=(1+1)n
⇒ 4096=(2)n
⇒ (2)12=(2)n
⇒ n=12
Here n is even, so the greatest coefficient is nCn/2  i.e., 12C6 = 924

The middle term in the expansion of (2x+3y)12 is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Arun Khanna answered
There will be 13 terms, so the middle term is term #7
Term(7) = C(12,6)(2x)^6 (3y)^6
= 924(64x^6)(729y^6)
= 43110144 x^6 y^6
so the correct option is B

The coefficient of second, third and fourth terms in the binomial expansion of (1+x)n(‘n’, a + ve integer) are in A.P.., if n is equal to
  • a)
    4
  • b)
    7
  • c)
    5
  • d)
    6
Correct answer is option 'B'. Can you explain this answer?

Pallabi Patel answered
Explanation:

To find the coefficient of the second, third, and fourth terms in the binomial expansion of (1+x)^n, we need to understand the general formula for the binomial expansion and the concept of the arithmetic progression (A.P.).

Binomial Expansion:
The binomial expansion of (1+x)^n can be calculated using the binomial theorem. According to the binomial theorem, the expansion is given by:

(1+x)^n = C(n,0) + C(n,1)x + C(n,2)x^2 + C(n,3)x^3 + ... + C(n,n)x^n

where C(n,r) represents the binomial coefficient, given by:

C(n,r) = n! / (r!(n-r)!)

Arithmetic Progression (A.P.):
An arithmetic progression (A.P.) is a sequence of numbers in which the difference between any two consecutive terms is constant. The common difference between the terms is denoted by 'd'.

For an A.P., the nth term (Tn) can be calculated using the formula:
Tn = a + (n-1)d

where 'a' is the first term and 'd' is the common difference.

Coeficients in A.P.:
Now, to determine the coefficients of the second, third, and fourth terms in the binomial expansion, we need to find the values of r for which the binomial coefficients form an arithmetic progression.

From the binomial expansion formula, we can see that the coefficients C(n,1), C(n,2), and C(n,3) correspond to the second, third, and fourth terms respectively.

Let's calculate the difference between the consecutive binomial coefficients:

d1 = C(n,2) - C(n,1) = (n! / (2!(n-2)!)) - (n! / (1!(n-1)!))
= (n(n-1)(n-2)! / (2(n-2)!)) - (n(n-1)(n-2)! / (1(n-1)!))
= n(n-1)(n-2)! / (2(n-2)!) - n(n-1)(n-2)! / (1(n-1)!)
= n(n-1) / 2 - n(n-1)
= -n(n-1) / 2

d2 = C(n,3) - C(n,2) = (n! / (3!(n-3)!)) - (n! / (2!(n-2)!))
= (n(n-1)(n-2)! / (3(n-3)!)) - (n(n-1)(n-2)! / (2(n-2)!))
= n(n-1) / 3 - n(n-1) / 2
= -n(n-1) / 6

We can observe that the differences d1 and d2 are both negative and have a common factor of -n(n-1). This implies that the coefficients C(n,1), C(n,2), and C(n,3) form an arithmetic progression with a common difference of -n(n-1)/2.

Conclusion:
Since the coefficients of the second,

 If in the expansion of (1 + x)m (1 – x)n, the co-efficients of x and x2 are 3 and – 6 respectively, then m is [JEE 99,2 ]
  • a)
    6
  • b)
    9
  • c)
    12
  • d)
    24
Correct answer is option 'C'. Can you explain this answer?

Sushil Kumar answered
(1+x)m (1−x)n = (1 + mx + m(m−1)x2/2!)(1 − nx + n(n−1)x2/2! ) 
= 1 + (m−n)x + [(n2 − n)/2 − mn + (m2 − m)/2] x2 
Given m − n = 3 or n = m − 3
Hence (n2 − n)/2 − mn + (m2 − m)/2 = 6
⇒ [(m−3)(m−4)]/2 − m(m−3) + (m2 − m)/2 = −6
⇒ m2 − 7m + 12 − 2m2 + 6m + m2 − m + 12 = 0
⇒ − 2m + 24 = 0
⇒ m = 12.

If the second, third, and fourth terms in the expansion of (x + y)n are 135, 30, and 10/3, respectively, then 6 (n3 + x2 + y) is equal to ________.
Correct answer is '806'. Can you explain this answer?

Nipuns Institute answered
T2 = nC1 y1 x(n-1) = 135
T3 = nC2 y2 x(n-2) = 30
T4 = nC3 y3 x(n-3) = 10/3
⇒ 135/30 = (x/y) * n * 2 / n(n-1) = (2 / n-1) * (x/y) ... (i)
30 / (10/3) = n(n-1) / 2 / n(n-1)(n-2) * 3! * (x/y)
9 = (3 / n-2) * (x/y)
3(n - 2) = 135 / 60 (n - 1) ⇒ n = 5
⇒ x = 9y ... (i)
y * x4 = 27 ⇒ x / 9 * x4 = 33
⇒ x5 = 35 ⇒ x = 3y = 1/3
⇒ 6 (53 + 32 + 1/3) = 6 (125 + 9 + 1/3)
= 6(134) + 2 = 806

Number of integral terms in the expansion of (71/2 + 111/6)824 is equal to ________.
Correct answer is '138'. Can you explain this answer?

Sai Kulkarni answered
General term in expansion of (71/2 + 111/6)824 is T(r+1) = 824Cr * 7((824-r)/2) * 11(r/6)
For integral term, r must be a multiple of 6.
Hence r = 0, 6, 12, ..., 822

n-1Cr = (k² - 8) nCr+1 if and only if:
  • a)
    2√2 < k < 2√3
  • b)
    2√2 ≤ k ≤ 3
  • c)
    2√3 < k < 3√3
  • d)
    2√3 ≤ k ≤ 3√2
Correct answer is option 'B'. Can you explain this answer?

KP Classes answered
n-1Cr = (k² - 8) nCr+1

(n-1Cr) / (nCr+1) = k² - 8
(r + 1) / n = k² - 8
⇒ k² - 8 > 0
(k - 2√2)(k + 2√2) > 0
k ∈ (-∞, -2√2) ∪ (2√2, ∞) .... (I)
∴ n ≥ r + 1, (r + 1) / n ≤ 1
⇒ k² - 8 ≤ 1
k² - 9 ≤ 0
-3 ≤ k ≤ 3 .... (II)
From equation (I) and (II) we get
k ∈ [-3, -2√2) ∪ (2√2, 3]

The greatest coefficient in the expansion of (1+x)12 is
  • a)
    C (12, 4)
  • b)
    C (12, 6)
  • c)
    C (12, 5)
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?

Naina Bansal answered
The binomial expansion formula states that:
(a + b)^n = a^n + n*a^(n-1)*b + C(n,2)*a^(n-2)*b^2 + ... + b^n

where C(n,k) is the binomial coefficient, given by:

C(n,k) = n! / (k! * (n-k)!)

In the expansion of (1+x)^12, the greatest coefficient will be the one with the largest value of k. Since we are given that n = 12, the greatest coefficient will be the one with the largest value of k such that k <= n. The largest value of k that satisfies this condition is 6, which corresponds to the coefficient C(12,6).
Therefore, the greatest coefficient in the expansion of (1+x)^12 is C(12,6). The correct answer is therefore (b) C (12, 6).

If x = 9950+1005 and y = (101)50, then
  • a)
    x < y
  • b)
    x = y
  • c)
    x > y
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

Nitya Yadav answered
The given equations are:
Ifx = 9950
1005andy = (101)50

To solve for x, we divide both sides of the first equation by 100:
Ifx/100 = 9950/100
x = 99.5

To solve for y, we divide both sides of the second equation by 1005:
andy = (101)50/1005
andy = 50

Therefore, the values of x and y are:
x = 99.5
y = 50

Chapter doubts & questions for Binomial Theorem & its Simple Applications - Mathematics for NDA 2025 is part of NDA exam preparation. The chapters have been prepared according to the NDA exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for NDA 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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