If $S$ be the sum of coefficients in the expansion of ${\left( {{\rm{\alpha }}x + {\rm{\beta }}y - {\rm{\gamma }}z} \right)^n}$, where $({\rm{\alpha }},{\rm{\beta }},{\rm{\gamma }} > 0)$, then the value of $\mathop {\lim }\limits_{n \to \infty } \frac{S}{{{{\left\{ {{S^{1/n}} + 1} \right\}}^n}}}$ is

${e^{\left( {\frac{{{\rm{\alpha \beta }}}}{{\rm{\gamma }}}} \right)}}$

${e^{\left( {\frac{{{\rm{\alpha }} + {\rm{\beta }} - {\rm{\gamma }}}}{{{\rm{\alpha }} + {\rm{\beta }} - {\rm{\gamma }} + 1}}} \right)}}$

$\frac{{{\rm{\alpha \beta }}}}{{\rm{\gamma }}}$

The coefficient of ${x^{100}}$ in the expansion of $\mathop \sum \limits_{j = 0}^{200} {(1 + x)^j}$ is

$\left( {_{100}^{200}} \right)$

$\left( {_{102}^{201}} \right)$

$\left( {_{101}^{200}} \right)$

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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Medium

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Showing 61-90 of 468 questions

61 Medium

61 of 468

Application +4 | -1

The sum of the series

\sum\limits_{{r=1}}^{\infty}(-1)^{r-1} \cdot{ }^n C_r(a-r)

is equal to

View answer and explanation

Correct option: C

a

Option C (

a

) is correct because the series simplifies to

a

after applying binomial coefficient identities and recognizing the alternating sum.

62 Medium

62 of 468

Application +4 | -1

Suppose

k \in {R}

. If the coefficient of

x

in the expansion of

\left(x^2+\frac{k}{x}\right)^5

is 270, then

k

is equal to:

View answer and explanation

Correct option: A

Option A (3) is correct because when k=3, the coefficient of x in the expansion of (x² + k/x)⁵ is 270.

63 Medium

63 of 468

Application +4 | -1

(1+a x)^n=1+8 x+24 x^2+\cdots

then

View answer and explanation

Correct option: C

a=2

Option C (

a=2

) is correct because by comparing the coefficients of the given expansion with the binomial expansion, we can solve for

a

and

n

64 Medium

64 of 468

Understanding +4 | -1

The sum of

{ }^{21} C_{10}+{ }^{21} C_9+\cdots+{ }^{21} C_0

is equal to

View answer and explanation

Correct option: A

2^{20}

Option A (

2^{20}

) is correct because it represents half of the total sum of binomial coefficients for

{ }^{21} C_i

, where

i

ranges from 0 to 21.

65 Medium

65 of 468

Application +4 | -1

\sum_{k=1}^{10} k \cdot \frac{{ }^n C_k}{{ }^n C_{k-1}}=5 \mathrm{n}

, then

\mathrm{n}

is equal to

View answer and explanation

Correct option: A

Option A (9) is correct because solving the summation equation leads to n = 9.

66 Medium

66 of 468

Understanding +4 | -1

\sum\limits_{0 \le i} {\sum\limits_{ < j \le n} {^n{C_i}} }

is equal to

View answer and explanation

Correct option: A

n 2^{n-1}

Option A (

n 2^{n-1}

) is the correct answer because the summation simplifies to

n 2^{n-1}

using combinatorial identities.

67 Medium

67 of 468

Analyzing +4 | -1

The constant term in the expansion of

\left(\log _{10}\left(x^{\log _{10} x}\right)-\log _{x^2} 100\right)^{12}

is:

View answer and explanation

Correct option: A

495

The correct answer is A (495) because the constant term in the binomial expansion is found by setting the power of x to zero and calculating the corresponding binomial coefficient.

68 Medium

68 of 468

Analyzing +4 | -1

S

be the sum of coefficients in the expansion of

{\left( {\mathrm{\alpha }x + \mathrm{\beta }y - \mathrm{\gamma }z} \right)^n}

, where

(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma } > 0)

, then the value of

\lim\limits_{n \to \infty } \frac{S}{{{{\left\{ {{S^{1/n}} + 1} \right\}}^n}}}

View answer and explanation

Correct option: D

Option D (0) is correct because the limit of the given expression as n approaches infinity is 0, given the condition that α + β - γ + 1 > α + β - γ.

69 Medium

69 of 468

Application +4 | -1

The coefticient of

x^3

in the expansion of

\left(1-x+x^2\right)^5

View answer and explanation

Correct option: C

-30

Option C (-30) is correct. The coefficient of

x^3

in the expansion of

(1-x+x^2)^5

is found by considering the relevant terms in the multinomial expansion.

70 Medium

70 of 468

Application +4 | -1

The coefficient of

x^{1012}

in the expansion of

\left(1+x^{n}+x^{253}\right)^{10}

, (where

\mathrm{n} \leq 22

is any positive integer), is

View answer and explanation

Correct option: B

{ }^{10} \mathrm{C}_{4}

Option B (

{}^{10} C_{4}

) is correct because it correctly identifies the coefficient of x¹⁰¹² in the given expansion.

71 Medium

71 of 468

Application +4 | -1

If the fourth term in the expansion of

\left(a x+\frac{1}{x}\right)^n

\frac{5}{2}

, then the value of

{a} \times {n}

View answer and explanation

Correct option: A

Option A (3) is correct because by equating the power of x to zero and using the given information about the fourth term, we can find the values of a and n, and their product is 3.

72 Medium

72 of 468

Understanding +4 | -1

The coefficient of

x^9

in the expansion

\sum\limits_{{r=0}}^{70}\left(\begin{array}{c}70 \\ r\end{array}\right)(x-5)^{70-r} 4^r

View answer and explanation

Correct option: C

-\left(\begin{array}{l}70 \\ 61\end{array}\right)

Option C (

-\left(\begin{array}{l}70 \\ 61\end{array}\right)

) is correct because after simplifying the summation, the coefficient of

x^9

is found using binomial expansion.

73 Medium

73 of 468

Application +4 | -1

If coefficient of

x^2 y^3 z^4

(x+y+z)^n

A

, then coefficient of

x^4 y^4 z

View answer and explanation

Correct option: C

\frac{A}{2}

Option C (

\frac{A}{2}

) is correct because the coefficient is determined by the multinomial coefficient formula, and swapping exponents changes the factorials in the denominator.

74 Medium

74 of 468

Application +4 | -1

If the third term in the expansion of

\left[x+x^{\log _{10} x}\right]^5

10^6

, then

x

must be

View answer and explanation

Correct option: C

Option C (10) is correct because substituting x=10 into the third term of the expansion results in 10⁶, satisfying the given condition.

75 Medium

75 of 468

Analyzing +4 | -1

\left(1-x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n}

, where

a_0, a_1, a_2, \ldots, a_{2 n}

are in AP, then

a_n

is equal to

View answer and explanation

Correct option: A

\frac{1}{2 n+1}

Option A (

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

) is correct because by substituting x=1 into the given equation and using the properties of an arithmetic progression (AP), we can derive that

a_n = \frac{{1}}{{2n+1}}

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76 Medium

76 of 468

Application +4 | -1

The value of

\sum\limits_{{r=0}}^n{ }^n C_r(\sin r x)

is equal to

View answer and explanation

Correct option: A

2^n \cdot \cos ^n \frac{x}{2} \cdot \sin \frac{n x}{2}

Option A (

2^n \cdot \cos ^n \frac{x}{2} \cdot \sin \frac{n x}{2}

) is correct because it is the imaginary part of the complex number

(1+e^{ix})^n

simplified using trigonometric identities.

77 Medium

77 of 468

Application +4 | -1

x

is very small in magnitude compared with

a

, then

\left(\frac{a}{a+x}\right)^{\frac{1}{2}}+\left(\frac{a}{a-x}\right)^{\frac{1}{2}}

can be approximately equal to

View answer and explanation

Correct option: D

2+\frac{3}{4} \frac{x^2}{a^2}

Option D (

2+\frac{3}{4} \frac{x^2}{a^2}

) is correct because using the binomial approximation for small x/a leads to this result.

78 Medium

78 of 468

Application +4 | -1

Let

{(x + 10)^{50}} + {(x - 10)^{50}} = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_{50}}{x^{50}}

, for all

x \in R;

then

\frac{{{a_2}}}{{{a_0}}}

is equal to

View answer and explanation

Correct option: A

12.25

Option A (12.25) is correct because the ratio of coefficients a2/a0 is calculated by expanding the binomial expressions and comparing coefficients.

79 Medium

79 of 468

Application +4 | -1

If the

r

th term is the middle term in the expansion of

\left(x^2-\frac{1}{2 x}\right)^{20}

, then the

(r+3)

th term is

View answer and explanation

Correct option: C

-\frac{1}{2^{13}},{ }^{20} C_{13} \cdot x

Option C is correct because the (r+3)th term, which is the 14th term, is calculated using the binomial theorem and simplified to the given expression.

80 Medium

80 of 468

Application +4 | -1

The 14th term from the end in the expansion of

{\left( {\sqrt x - \sqrt y } \right)^{17}}

View answer and explanation

Correct option: C

{\;^{17}}{C_4}{x^{13/2}}{y^2}

Option C (

{\;^{17}}{C_4}{x^{13/2}}{y^2}

) is correct because the 14th term from the end is the 5th term from the beginning in the binomial expansion.

81 Medium

81 of 468

Analyzing +4 | -1

\left[\lim \limits_{{n \rightarrow \infty}} \sum\limits_{{r=0}}^n\left({ }^n C_r \cdot \frac{1}{(r+4) \cdot n^r}\right)\right]

is equal to
(where [.] denotes greatest integer function):

View answer and explanation

Correct option: A

Option A (0) is correct because the determinant becomes zero as two rows are identical after summation.

82 Medium

82 of 468

Analyzing +4 | -1

\alpha=\frac{5}{2 ! 3}+\frac{5 \cdot 7}{3 ! 3^2}+\frac{5 \cdot 7 \cdot 9}{4 ! 3^3}+\ldots

, then

\alpha^2+4 \alpha

is equal to

View answer and explanation

Correct option: B

Option B (23) is correct because after simplifying the series and using the binomial theorem, the expression

\alpha^2 + 4\alpha

evaluates to 23.

83 Medium

83 of 468

Application +4 | -1

The coefficient of

{x^{100}}

in the expansion of

\sum \limits_{j = 0}^{200} {(1 + x)^j}

View answer and explanation

Correct option: D

\left( {_{100}^{201}} \right)

Option D (

\left( {_{100}^{201}} \right)

) is correct because the sum of binomial coefficients simplifies to this after applying the hockey-stick identity.

84 Medium

84 of 468

Analyzing +4 | -1

Let

P(x)

be a polynomial with real coefficients such that

\int\limits_{0}^1 x^m P(1-x) d x=0 \, \forall \, m \in {N} \cup\{0\}

, then

View answer and explanation

Correct option: D

P(x)=0

Option D (

P(x)=0

) is correct because the given integral condition implies that the polynomial must be identically zero.

85 Medium

85 of 468

Understanding +4 | -1

The total number of different terms in the expansion of

\left(x_1+x_2+x_3+\ldots+x_n\right)^m

is equal to

View answer and explanation

Correct option: B

{ }^{n+m-1} C^{n-1}

Option B (

{ }^{n+m-1} C^{n-1}

) is correct because it represents the number of non-negative integer solutions to the equation

y_1 + y_2 + ... + y_n = m

86 Medium

86 of 468

Application +4 | -1

The constant term in the expansion of

{\left( {1 + x} \right)^{10}}{\left( {1 + \frac{1}{x}} \right)^{12}}

View answer and explanation

Correct option: A

{\;^{22}}{C_{10}}

Option A (

{{\;^{22}}}{C_{10}}

) is correct because the constant term in the expansion of

{\left( {1 + x} \right)^{10}}{\left( {1 + \frac{1}{x}} \right)^{12}}

is the coefficient of

x^{12}

{\left( {1 + x} \right)^{22}}

, which is

{{\;^{22}}}{C_{12}} = {{\;^{22}}}{C_{10}}

87 Medium

87 of 468

Application +4 | -1

Value of '

a

', if

17^{\text {th}}

and

18^{\text {th}}

terms in the expansion of

(2+a)^{50}

are equal, is

View answer and explanation

Correct option: A

Option A (1) is correct because equating the 17th and 18th terms in the expansion of

(2+a)^{50}

leads to the equation

\frac{{ }^{50} {C}_{16}}{{ }^{50} {C}_{17}} \times 2={a}

, which gives

a=1

88 Medium

88 of 468

Analyzing +4 | -1

The coefficient of

x^{301}

(1+x)^{500}+x(1+x)^{499}

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

View answer and explanation

Correct option: B

{ }^{501} C_{200}

The coefficient of x³⁰¹ is

{ }^{501} C_{200}

. By summing the geometric series and using the identity

{ }^n C_r + { }^n C_{r+1} = { }^{n+1} C_{r+1}

, the coefficient can be simplified to

{ }^{501} C_{200}

89 Medium

89 of 468

Analyzing +4 | -1

The coefficients of

{x^6}

in the expansion of

\log \left\{ {{{\left( {1 + x} \right)}^{1 + x}}{{\left( {1 - x} \right)}^{1 - x}}} \right\},

View answer and explanation

Correct option: A

\frac{1}{{15}}

Option A (

\frac{1}{{15}}

) is correct because the coefficient of x⁶ in the expansion is calculated correctly using the logarithmic properties and series expansion.

90 Medium

90 of 468

Application +4 | -1

The coefficient of

a^8 b^4 c^9 d^9

(a b c+a b d+a c d+b c d)^{10}

View answer and explanation

Correct option: C

2520

Option C (2520) is correct because the coefficient is found by using multinomial theorem and simplifying the expression.

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