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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-60 of 468 questions

31 Medium

31 of 468

Application +4 | -1

In how many ways 4 coins can be chosen out of 20 coins of ₹1, 15 coins of ₹2 and 10 coins of ₹5?

View answer and explanation

Correct option: D

The correct answer is 15 because it represents the number of ways to choose 4 coins from the given set of coins.

32 Medium

32 of 468

Application +4 | -1

Let

(x+10)^{50}+(x-10)^{50}=\mathrm{a}_{0}+\mathrm{a}_{1} x+\mathrm{a}_{2} x^{2}+\ldots .+\mathrm{a}_{50} x^{50}

, for all

x \in \mathbf{R}

; then

\frac{\mathrm{a}_{2}}{\mathrm{a}_{0}}

is equal to :

View answer and explanation

Correct option: C

12.25

The correct answer is C (12.25) because it correctly calculates the ratio of a₂ to a₀ using the binomial theorem expansion of the given expression.

33 Medium

33 of 468

Application +4 | -1

C_0, C_1, C_2, \ldots, C_n

denote the binomial coefficients in the expansion of

(1+x)^n

, then

\sum\limits_{{r=0}}^n(-1)^{r n} C_r \frac{1+r \log _e 10}{\left(1+\log _e 10^n\right)^r}

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the given summation simplifies to 0 using binomial theorem and algebraic manipulation.

34 Medium

34 of 468

Analyzing +4 | -1

C_1, C_2, \ldots, C_n

are coefficients in the expansion of

(1+x)^n=1+C_1 x+C_2 x^2+\ldots+C_n x^n

, then the value of

C_1^2-2 C_2^2+3 C_3^2-\ldots-2 n C_{2 n}^2

View answer and explanation

Correct option: C

2(-1)^{n-1} n^{2 n-1} C_n

Option C (

2(-1)^{n-1} n^{2 n-1} C_n

) is correct because it is derived by differentiating the binomial expansion and comparing coefficients.

35 Medium

35 of 468

Understanding +4 | -1

The number of terms in the expansion of the multinomial

(x+y+z)^n

View answer and explanation

Correct option: A

\frac{1}{2}(n+1)(n+2)

Option A (

\frac{1}{2}(n+1)(n+2)

) is correct because it represents the number of non-negative integral solutions of the equation p+q+r=n, which corresponds to the number of terms in the multinomial expansion.

36 Medium

36 of 468

Analyzing +4 | -1

(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n

, then

\sum\limits_{0 \le r} {\sum\limits_{ < s \le n} {(r + s)} } {C_r}{C_s}

is equal to

View answer and explanation

Correct option: A

n\left[2^{2 n-1}-{ }^{2 n-1} C_{n-1}\right]

Option A (

n[2^{2 n-1}-{ }^{2 n-1} C_{n-1}]

) is correct because it correctly uses the symmetry and properties of binomial coefficients to simplify the summation.

37 Medium

37 of 468

Application +4 | -1

The coefficient of

x^7

in the expansion of

(1-2 x)^{-2}

View answer and explanation

Correct option: B

1024

Option B (1024) is correct because it is the coefficient of

x^7

obtained by applying the binomial theorem to

(1-2x)^{-2}

38 Medium

38 of 468

Application +4 | -1

The number of integral terms in the expansion of

(\sqrt{3}+\sqrt[8]{5})^{256}

View answer and explanation

Correct option: C

Option C (33) is correct because the number of integral terms is found by determining the values of r for which both exponents are integers.

39 Medium

39 of 468

Analyzing +4 | -1

The expression

\frac{1}{{\sqrt {4x + 1} }}\left\{ {{{\left( {1 + \frac{{\sqrt {4x + 1} }}{2}} \right)}^7} - {{\left( {1 - \frac{{\sqrt {4x + 1} }}{2}} \right)}^7}} \right\}

is a polynomial in

x

of degree

View answer and explanation

Correct option: D

Option D (3) is correct because after simplification, the highest power of x in the polynomial is 3.

40 Medium

40 of 468

Application +4 | -1

A student is taking examination of three subjects and each paper has maximum of

n

marks. In how many ways, can he score

2 n

marks?

View answer and explanation

Correct option: A

\frac{(n+1)(n+2)}{2}

Option A,

\frac{(n+1)(n+2)}{2}

, is correct because it represents the number of ways the student can score 2n marks in three subjects, each with a maximum of n marks.

41 Medium

41 of 468

Analyzing +4 | -1

The expression

\left(x+\sqrt{x^3-1}\right)^6+\left(x-\sqrt{x^3-1}\right)^6

is a polynomial of degree

View answer and explanation

Correct option: C

Option C (9) is correct because after simplification using the binomial theorem, the highest power of x is 9.

42 Medium

42 of 468

Application +4 | -1

The value of

\sum\limits_{{r=0}}^{50}\left({ }^{100} C_r,{ }^{200} C_{150+r}\right)

is equal to

View answer and explanation

Correct option: A

{ }^{300} C_{50}

Option A (

{ }^{300} C_{50}

) is correct because the summation represents the coefficient of

x^{50}

in the product of two binomial expansions, which can be simplified using the Vandermonde identity.

43 Medium

43 of 468

Understanding +4 | -1

For each

n \in N, 2^{3n}-1

is divisible by

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because

2^{3n}-1

is divisible by 7, not 8, 16, or 32.

44 Medium

44 of 468

Application +4 | -1

\left(x-2+\frac{1}{x}\right)^{30}=n_0 x^{30}+n_1 x^{29}+\ldots \ldots+n_{29} x+n_{30}+n_{31} x^{-1}+\ldots \ldots+n_{60} x^{-30}

and

C=n_0+n_1+n_2+\ldots \ldots+n_{60}

. Find the value of

(a+b)

C-n_{30}=-\left(\begin{array}{l}a \\ b\end{array}\right)

\left[{\left(\begin{array}{l}n \\ r\end{array}\right) \text{ denotes }{ }^n C_r}\right]

View answer and explanation

Correct option: A

Option A (90) is correct because the sum of coefficients C is 0, and C - n_30 = - (a choose b), leading to a = 60, b = 30, and a+b = 90.

45 Medium

45 of 468

Application +4 | -1

The interval in which

x

must lie so that the numerically greatest term in the expansion of

{\left( {1 - x} \right)^{21}}

has the numerically greatest coefficient, is

View answer and explanation

Correct option: B

\left( {\frac{5}{6},\frac{6}{5}} \right)

Option B

\left( {\frac{5}{6},\frac{6}{5}} \right)

is correct because it represents the interval where the numerically greatest term has the numerically greatest coefficient.

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46 Medium

46 of 468

Application +4 | -1

\frac{1}{\left(\begin{array}{c}101 \\ 0\end{array}\right)}-\frac{1}{\left(\begin{array}{c}101 \\ 1\end{array}\right)}+\frac{1}{\left(\begin{array}{c}101 \\ 2\end{array}\right)}-\frac{1}{\left(\begin{array}{c}101 \\ 3\end{array}\right)}+\ldots+\frac{(-1)^{101}}{\left(\begin{array}{l}101 \\ 101\end{array}\right)}=

View answer and explanation

Correct option: D

Option D (0) is correct because the terms cancel out in pairs due to the symmetry of binomial coefficients and alternating signs.

47 Medium

47 of 468

Analyzing +4 | -1

{\left( {1 + x} \right)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ... + {C_n}{x^n}

, then the value of

\sum \limits_{0 \le r < s \le n} \sum \left( {r + s} \right)\left( {{C_r} + {C_s}} \right)

View answer and explanation

Correct option: A

{n^2}{.2^n}

Option A is correct because the value of the summation is found to be n² * 2^n by using properties of binomial coefficients and summations.

48 Medium

48 of 468

Application +4 | -1

S_n=\sum_{r=0}^n \frac{1}{{ }^n C_r}

and

t_n=\sum_{r=0}^n \frac{r}{{ }^n C_r}

, then

\frac{t_n}{S_n}

is equal to-

View answer and explanation

Correct option: A

\frac{1}{2} \mathbf{n}

Option A (

\frac{1}{2}n

) is correct because by using the property

{ }^n C_r = { }^n C_{n-r}

and summing the series, we can find the ratio

\frac{t_n}{S_n}

49 Medium

49 of 468

Application +4 | -1

The sum of coefficients in the expansion of

\left(1+x-3 y^2\right)^{2163}

View answer and explanation

Correct option: B

-1

Option B (-1) is correct because the sum of coefficients is found by substituting x=1 and y=1 into the expression, resulting in (1+1-3)⁽²¹⁶³⁾ = (-1)⁽²¹⁶³⁾ = -1.

50 Medium

50 of 468

Application +4 | -1

The remainder when

4^{101}

is divided by 101 is

View answer and explanation

Correct option: A

Option A (4) is correct because, according to Fermat's Little Theorem, a^(p-1) is congruent to 1 mod p, where p is a prime number.

51 Medium

51 of 468

Application +4 | -1

The fractional part of

\frac{2^{4 n}}{15}

View answer and explanation

Correct option: A

\frac{1}{15}

Option A (

\frac{{1}}{{15}}

) is correct because the fractional part of

\frac{{2^{{4n}}}}{{15}}

\frac{{1}}{{15}}

, which is found using binomial expansion.

52 Medium

52 of 468

Application +4 | -1

The coefficient of

{x^n}

in the expansion of

\frac{1}{{\left( {1 - x} \right)\left( {3 - x} \right)}},

View answer and explanation

Correct option: A

\frac{{{3^{n + 1}} - 1}}{{{{2.3}^{n + 1}}}}

Option A (

\frac{{{3^{n + 1}} - 1}}{{{{2.3}^{n + 1}}}}

) is correct because it is the coefficient of

{x^n}

obtained by partial fraction decomposition and binomial expansion.

53 Medium

53 of 468

Analyzing +4 | -1

{\left( {1 + x + {x^2}} \right)^n} = \sum \limits_{r = 0}^{2n} {a_r}{x^r}

then,

{a_1} - 2{a_2} + 3{a_3} \ldots - 2n{a_{2n}}\;

is equal to

View answer and explanation

Correct option: B

- n

The correct answer is B (-n) because by differentiating the given equation and substituting x = -1, we can find the value of the expression.

54 Medium

54 of 468

Application +4 | -1

Given that positive integers

r>1, n>2

and the coefficient of

(3r)

th and

(r+2)

nd terms in the expansion of

(1+x)^{2 n}

are equal, then

n=

View answer and explanation

Correct option: A

n=2 r

Option A is correct because equating the coefficients and using the properties of binomial coefficients leads to n = 2r.

55 Medium

55 of 468

Application +4 | -1

a, b, p \in {N}

and

p

is prime, then

(a+b)^p-a^p-b^p

is divisible by

View answer and explanation

Correct option: A

p

Option A (

p

) is correct because

(a+b)^p-a^p-b^p

can be expressed as a sum of terms, each of which is divisible by

p

due to the properties of binomial coefficients and prime numbers.

56 Medium

56 of 468

Analyzing +4 | -1

Coefficient of

x^{301}

(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+\cdots+x^{500}

is equal to

View answer and explanation

Correct option: C

{ }^{501} C_{301}

Option C (

{ }^{501} C_{301}

) is correct because the coefficient of

x^{301}

can be simplified using combinatorial identities to

{ }^{501} C_{301}

57 Medium

57 of 468

Application +4 | -1

(1+x)^{100}=c_0+c_1 x+c_2 x^2+\ldots+c_{100} x^{100}

, then

2^1 \frac{c_0}{1}+2^2 \frac{c_1}{2}+2^3 \frac{c_2}{3}+\ldots+2^{101} \frac{c_{100}}{101}=

View answer and explanation

Correct option: B

\frac{3^{101}-1}{101}

Option B is the correct answer because the given expression can be evaluated using integration of the binomial expansion.

58 Medium

58 of 468

Analyzing +4 | -1

Sum of the first 20 terms of the series

\frac{1}{(2)(4)}+\frac{(1)(3)}{(2)(4)(6)}+\frac{(1)(3)(5)}{(2)(4)(6)(8)}+\ldots

View answer and explanation

Correct option: C

\frac{1}{2}-\frac{1}{2^{42}}\left({ }^{42} C_{21}\right)

Option C,

\frac{{1}}{{2}}-\frac{{1}}{{2^{42}}}\left({ }^{{42}} C_{{21}}\right)

, is correct because the series can be expressed as a telescoping sum, and the sum of the first 20 terms is equal to

b_1 - b_{21}

59 Medium

59 of 468

Application +4 | -1

The value of

{ }^{n-1} C_0 \cdot{ }^n C_1+{ }^{n-1} C_1 \cdot{ }^n C_2+{ }^{n-1} C_2 \cdot{ }^n C_3 +\ldots+{ }^{n-1} C_{n-1} \cdot{ }^n C_n \text { is equal to }

View answer and explanation

Correct option: B

{ }^{2 n-1} C_{n-1}

Option B (

{ }^{2 n-1} C_{n-1}

) is correct because the sum represents the coefficient of x^(n-1) in the product of two binomial expansions.

60 Medium

60 of 468

Application +4 | -1

If terms independent of

x

in the expansion of

\left(3 x-\frac{1}{x}\right)^{20}

and

\left(x+\frac{\sqrt[3]{3^{10}}}{x}\right)^{18}

are

A

and

B

respectively then

\left(\frac{9}{38} A+B\right)

equals:

View answer and explanation

Correct option: B

3^{10} \cdot { }^{19} C_9

Option B is correct because after calculating A and B, substituting into the expression yields

3^{{10}} \cdot {{}^{{19}}}C_9

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