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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Mixed difficulty

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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 121-150 of 951 questions

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121 Medium
121 of 951
Application +4 | -1
The sum of the series ∑r=1∞(−1)r−1⋅nCr(a−r)\sum\limits_{{r=1}}^{\infty}(-1)^{r-1} \cdot{ }^n C_r(a-r) is equal to
View answer and explanation
Correct option: C
aa
Option C (aa) is correct because the series simplifies to aa after applying binomial coefficient identities and recognizing the alternating sum.
122 Medium
122 of 951
Application +4 | -1
Suppose k∈Rk \in {R}. If the coefficient of xx in the expansion of (x2+kx)5\left(x^2+\frac{k}{x}\right)^5 is 270, then kk is equal to:
View answer and explanation
Correct option: A
3
Option A (3) is correct because when k=3, the coefficient of x in the expansion of (x2 + k/x)5 is 270.
123 Hard
123 of 951
Application +4 | -1
Let a,ba,b be integers and f(x)f\left( x \right) be a polynomial with integer coefficients such that f(b)−f(a)=1.f\left( b \right) - f\left( a \right) = 1. Then, the value of b−a,b - a, is
View answer and explanation
Correct option: C
1,−11, - 1
Option C (1,−11, -1) is correct because the difference b−ab-a must be a factor of 1, which can only be 1 or -1.
124 Easy
124 of 951
Understanding +4 | -1
Assertion: The rth{r}^{\text {th}} term from the end in the expansion of (x+a)n(x+a)^n is nCn−r+1xr−1an−r+1{ }^n C_{n-r+1} x^{r-1} a^{n-r+1}.

Reason: The rth{r}^{\text {th}} term from the end in the expansion of (x+a)n(x+a)^n is (n−r+2)th(n-r+2)^{\text {th}} term.
View answer and explanation
Correct option: A
Assertion is correct, reason is correct; reason is a correct explanation for assertion
Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because the rth term from the end is indeed the (n-r+2)th term, and the assertion correctly states the value of that term.
125 Medium
125 of 951
Application +4 | -1
If (1+ax)n=1+8x+24x2+⋯(1+a x)^n=1+8 x+24 x^2+\cdots then
View answer and explanation
Correct option: C
a=2a=2
Option C (a=2a=2) is correct because by comparing the coefficients of the given expansion with the binomial expansion, we can solve for aa and nn.
126 Medium
126 of 951
Understanding +4 | -1
The sum of 21C10+21C9+⋯+21C0{ }^{21} C_{10}+{ }^{21} C_9+\cdots+{ }^{21} C_0 is equal to
View answer and explanation
Correct option: A
2202^{20}
Option A (2202^{20}) is correct because it represents half of the total sum of binomial coefficients for 21Ci{ }^{21} C_i, where ii ranges from 0 to 21.
127 Medium
127 of 951
Application +4 | -1
If ∑k=110k⋅nCknCk−1=5n\sum_{k=1}^{10} k \cdot \frac{{ }^n C_k}{{ }^n C_{k-1}}=5 \mathrm{n}, then n\mathrm{n} is equal to
View answer and explanation
Correct option: A
9
Option A (9) is correct because solving the summation equation leads to n = 9.
128 Medium
128 of 951
Understanding +4 | -1
∑0≤i∑<j≤nnCi\sum\limits_{0 \le i} {\sum\limits_{ < j \le n} {^n{C_i}} } is equal to
View answer and explanation
Correct option: A
n2n−1n 2^{n-1}
Option A (n2n−1n 2^{n-1}) is the correct answer because the summation simplifies to n2n−1n 2^{n-1} using combinatorial identities.
129 Medium
129 of 951
Analyzing +4 | -1
The constant term in the expansion of (log⁡10(xlog⁡10x)−log⁡x2100)12\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.
130 Medium
130 of 951
Analyzing +4 | -1
If SS be the sum of coefficients in the expansion of (αx+βy−γz)n{\left( {\mathrm{\alpha }x + \mathrm{\beta }y - \mathrm{\gamma }z} \right)^n}, where (α,β,γ>0)(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma } > 0), then the value of lim⁡n→∞S{S1/n+1}n\lim\limits_{n \to \infty } \frac{S}{{{{\left\{ {{S^{1/n}} + 1} \right\}}^n}}} is
View answer and explanation
Correct option: D
0
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.
131 Easy
131 of 951
Application +4 | -1
The remainder left out when 82n−(62)2n+1{8^{2n}} - {\left( {62} \right)^{2n + 1}} is divided by 9, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the remainder when 8(2n) - (62)(2n+1) is divided by 9 is 2.
132 Medium
132 of 951
Application +4 | -1
The coefticient of x3x^3 in the expansion of (1−x+x2)5\left(1-x+x^2\right)^5 is
View answer and explanation
Correct option: C
-30
Option C (-30) is correct. The coefficient of x3x^3 in the expansion of (1−x+x2)5(1-x+x^2)^5 is found by considering the relevant terms in the multinomial expansion.
133 Easy
133 of 951
Analyzing +4 | -1
Coefficient of the term independent of xx in the expansion of (1+x)2n(x1−x)−2n(1+x)^{2 n}\left(\frac{x}{1-x}\right)^{-2 n} is:
View answer and explanation
Correct option: B
(−1)n  2nCn(-1)^{n} \, \, {}^{2 n} C_n
Option B ( (−1)n  2nCn(-1)^{n} \, \, {}^{2 n} C_n ) is correct because the coefficient of the term independent of xx is found by simplifying the expression and finding the coefficient of x2nx^{2n} in the resulting expansion.
134 Easy
134 of 951
Application +4 | -1
If (1−x+x2)n=a0+a1x+a2x2+…+a2nx2n,{\left( {1 - x + {x^2}} \right)^n} = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_{2n}}{x^{2n}}, then a0+a2+a4+…+a2n{a_0} + {a_2} + {a_4} + \ldots + {a_{2n}} is equal to
View answer and explanation
Correct option: A
3n+12\frac{{{3^n} + 1}}{2}
Option A (3n+12\frac{{{3^n} + 1}}{2}) is correct because it is derived by substituting x = 1 and x = -1 into the given equation and solving for the required sum.
135 Easy
135 of 951
Analyzing +4 | -1
If the coefficient of x7{x^7} in (ax2+1bx)11{\left( {a{x^2} + \frac{1}{{bx}}} \right)^{11}} equal the coefficient of x−7{x^{ - 7}} in (ax−1bx2)11,{\left( {ax - \frac{1}{{b{x^2}}}} \right)^{11}}, then aa and bb satisfy the relation
View answer and explanation
Correct option: A
ab=1ab = 1
Option A (ab=1ab = 1) is correct because equating the coefficients of x7{x^7} and x−7{x^{ - 7}} leads to this relationship between aa and bb.
136 Easy
136 of 951
Application +4 | -1
The remainder when 3373^{37} is divided by 80 is
View answer and explanation
Correct option: B
3
Option B (3) is correct because the remainder is found by using the binomial theorem and focusing on the remainder when divided by 80.
137 Medium
137 of 951
Application +4 | -1
The coefficient of x1012x^{1012} in the expansion of (1+xn+x253)10\left(1+x^{n}+x^{253}\right)^{10}, (where n≤22\mathrm{n} \leq 22 is any positive integer), is
View answer and explanation
Correct option: B
10C4{ }^{10} \mathrm{C}_{4}
Option B (10C4{}^{10} C_{4}) is correct because it correctly identifies the coefficient of x1012 in the given expansion.
138 Easy
138 of 951
Application +4 | -1
If  ∣x∣<1,  then   1+n(2x1+x)+n(n+1)2!(2x1+x)2+\,\left| x \right| < 1,\;\mathrm{then\;}\,1 + n\left( {\frac{{2x}}{{1 + x}}} \right) + \frac{{n\left( {n + 1} \right)}}{{2!}}{\left( {\frac{{2x}}{{1 + x}}} \right)^2} + …is equal to
View answer and explanation
Correct option: D
(1+x1−x)n{\left( {\frac{{1 + x}}{{1 - x}}} \right)^n}
The correct answer is D ((1+x1−x)n{(\frac{{1 + x}}{{1 - x}})^n}) because it is the simplified form of the given infinite series using the binomial theorem for negative exponents.
139 Medium
139 of 951
Application +4 | -1
If the fourth term in the expansion of (ax+1x)n\left(a x+\frac{1}{x}\right)^n is 52\frac{5}{2}, then the value of a×n{a} \times {n} is
View answer and explanation
Correct option: A
3
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.
140 Medium
140 of 951
Understanding +4 | -1
The coefficient of x9x^9 in the expansion ∑r=070(70r)(x−5)70−r4r\sum\limits_{{r=0}}^{70}\left(\begin{array}{c}70 \\ r\end{array}\right)(x-5)^{70-r} 4^r is
View answer and explanation
Correct option: C
−(7061)-\left(\begin{array}{l}70 \\ 61\end{array}\right)
Option C (−(7061)-\left(\begin{array}{l}70 \\ 61\end{array}\right)) is correct because after simplifying the summation, the coefficient of x9x^9 is found using binomial expansion.
141 Easy
141 of 951
Analyzing +4 | -1
If nn is the number of irrational terms in the expansion of (314+518)60{\left( {{3^{\frac{1}{4}}} + {5^{\frac{1}{8}}}} \right)^{60}}, then (n−1)(n - 1) is divisible by:
View answer and explanation
Correct option: B
26
Option B (26) is correct because after calculating the number of irrational terms, subtracting 1 gives a number divisible by 26.
142 Medium
142 of 951
Application +4 | -1
If coefficient of x2y3z4x^2 y^3 z^4 in (x+y+z)n(x+y+z)^n is AA, then coefficient of x4y4zx^4 y^4 z is
View answer and explanation
Correct option: C
A2\frac{A}{2}
Option C (A2\frac{A}{2}) is correct because the coefficient is determined by the multinomial coefficient formula, and swapping exponents changes the factorials in the denominator.
143 Easy
143 of 951
Application +4 | -1
The greatest term in the expansion of (3+5x)15(3+5 x)^{15}, when x=1/5x=1 / 5, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the greatest term is 15C4(3)11{ }^{15} C_4(3)^{11}, which is not present in the given options.
144 Medium
144 of 951
Application +4 | -1
If the third term in the expansion of [x+xlog⁡10x]5\left[x+x^{\log _{10} x}\right]^5 is 10610^6, then xx must be
View answer and explanation
Correct option: C
10
Option C (10) is correct because substituting x=10 into the third term of the expansion results in 106, satisfying the given condition.
145 Medium
145 of 951
Analyzing +4 | -1
If (1−x+x2)n=a0+a1x+a2x2+…+a2nx2n\left(1-x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n}, where a0,a1,a2,…,a2na_0, a_1, a_2, \ldots, a_{2 n} are in AP, then ana_n is equal to
View answer and explanation
Correct option: A
12n+1\frac{1}{2 n+1}
Option A (12n+1\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 an=12n+1a_n = \frac{{1}}{{2n+1}}.
146 Easy
146 of 951
Analyzing +4 | -1
If {p}\{ p\} denotes the fractional part of the number p, then {38200}\left\{ {{{\frac{3}{8}}^{200}}} \right\} is equal to
View answer and explanation
Correct option: D
18\frac{1}{8}
Option D is correct because by using the binomial theorem and properties of fractional parts, the fractional part of the given expression is 18\frac{1}{8}.
147 Easy
147 of 951
Application +4 | -1
If in the expansion of (1+ax)n,n∈N,{\left( {1 + ax} \right)^n},n \in N, the coefficients of xx and x2{x^2} are 8 and 24 respectively, then
View answer and explanation
Correct option: A
a=2,n=4a = 2,n = 4
Option A (a=2,n=4a = 2, n = 4) is correct because the coefficients of x and x² in the binomial expansion of (1+ax)n(1 + ax)^n are given as 8 and 24 respectively, leading to the solution a = 2 and n = 4.
148 Hard
148 of 951
Analyzing +4 | -1
If the coefficients of three consecutive terms in the expansion of (1+x)n(1+x)^{n} are in the ratio 1:5:201: 5: 20, then the coefficient of the fourth term is
View answer and explanation
Correct option: D
3654
Option C (0) is correct because after simplifying the expression for y, it becomes a constant, and the second derivative of a constant is zero.
149 Medium
149 of 951
Application +4 | -1
The value of ∑r=0nnCr(sin⁡rx)\sum\limits_{{r=0}}^n{ }^n C_r(\sin r x) is equal to
View answer and explanation
Correct option: A
2n⋅cos⁡nx2⋅sin⁡nx22^n \cdot \cos ^n \frac{x}{2} \cdot \sin \frac{n x}{2}
Option A (2n⋅cos⁡nx2⋅sin⁡nx22^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+eix)n(1+e^{ix})^n simplified using trigonometric identities.
150 Medium
150 of 951
Application +4 | -1
If xx is very small in magnitude compared with aa, then (aa+x)12+(aa−x)12\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+34x2a22+\frac{3}{4} \frac{x^2}{a^2}
Option D (2+34x2a22+\frac{3}{4} \frac{x^2}{a^2}) is correct because using the binomial approximation for small x/a leads to this result.
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