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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Mixed difficulty

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Practice format and chapter quiz

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 901-930 of 951 questions
901 Easy
901 of 951
Application +4 | -1
The coefficient of x5{x^5} in the expansion of (1+x2)5(1+x)4,{\left( {1 + {x^2}} \right)^5}{\left( {1 + x} \right)^4}, is
View answer and explanation
Correct option: B
60
Option B (60) is correct because by expanding the product of binomials and finding the terms that contribute to x5, the coefficient is 60.
902 Easy
902 of 951
Application +4 | -1
If a,  b,  ca,\;b,\;c are in AP, then the sum of the coefficients of {1+(ax2−2bx+c)2}1973{\left\{ {1 + {{\left( {a{x^2} - 2bx + c} \right)}^2}} \right\}^{1973}} is
View answer and explanation
Correct option: D
1
Option D (1) is correct because when a,b,ca, b, c are in AP, a−2b+c=0a - 2b + c = 0, and substituting x=1x = 1 into the expression results in 1.
903 Easy
903 of 951
Application +4 | -1
C0Cr+C1Cr+1+C2Cr+2+...+Cn−rCn{C_0}{C_r} + {C_1}{C_{r + 1}} + {C_2}{C_{r + 2}} + ... + {C_{n - r}}{C_n} is equal to
View answer and explanation
Correct option: A
(2n)!(n−r)!(n+r)!\frac{{\left( {2n} \right)!}}{{\left( {n - r} \right)!\left( {n + r} \right)!}}
Option A ((2n)!(n−r)!(n+r)!\frac{{\left( {2n} \right)!}}{{\left( {n - r} \right)!\left( {n + r} \right)!}}) is correct because it represents the coefficient of x(n+r) in the expansion of (1+x)(2n).
904 Medium
904 of 951
Application +4 | -1
Let nn be an odd natural number and A=∑r=1n−121nCrA=\sum\limits_{{r=1}}^{\frac{n-1}{2}} \frac{1}{{ }^n C_r}, then value of ∑r=1nrnCr\sum\limits_{{r=1}}^n \frac{r}{{ }^n C_r} is equal to
View answer and explanation
Correct option: B
n(A+1)n(A+1)
Option B (n(A+1)n(A+1)) is correct because by using the property nCr=nCn−r{ }^n C_r = { }^n C_{n-r} and summing the series, we arrive at the result n(A+1)n(A+1).
905 Easy
905 of 951
Application +4 | -1
The number of non zero terms in the expansion of (1+32x)9+(1−32x)9(1+3 \sqrt{2} x)^9+(1-3 \sqrt{2} x)^9 is
View answer and explanation
Correct option: C
5
Option C (5) is correct because only the even-powered terms survive after adding the two expansions.
906 Medium
906 of 951
Analyzing +4 | -1
If 20Cr{}^{20}{C_r} is the coefficient of xr{x^r} in the expansion of (1+x)20{\left( {1 + x} \right)^{20}}, then the value of ∑r=020r2 20Cr\sum\limits_{r = 0}^{20} {{r^2}\,{}^{20}{C_r}} is equal to
View answer and explanation
Correct option: D
420×218420 \times {2^{18}}
The correct answer is D (420×218420 \times {2^{18}}) because it is the result of applying combinatorial identities and the binomial theorem to simplify the summation.
907 Easy
907 of 951
Application +4 | -1
The first 3 terms in the expansion of (1+ax)n  (n≠0){\left( {1 + ax} \right)^n}\;\left( {n \ne 0} \right) are 1,  6x,1,\;6x, and 16x216{x^2}. Then, the value of aa and nn are respectively
View answer and explanation
Correct option: C
23\frac{2}{3} and 9
Option C (23\frac{2}{3} and 9) is correct because these values satisfy the equations derived from the given expansion.
908 Easy
908 of 951
Application +4 | -1
If the expansion of (1+x)20,{\left( {1 + x} \right)^{20}}, the coefficients of rth{r^{th}} and (r+4)th{\left( {r + 4} \right)^{th}} terms are equal, then the value of rr, is
View answer and explanation
Correct option: C
9
Option C (9) is correct because the equality of binomial coefficients implies either the indices are equal or their sum equals the upper index.
909 Easy
909 of 951
Application +4 | -1
The value of xx in the expansion [x+xlog⁡10x]5{\left[ {x + {x^{{{\log }_{10}}x}}} \right]^5}, if the third term in the expansion is 1000000, is
View answer and explanation
Correct option: A
10
Option A (10) is correct because substituting x=10 into the third term of the expansion satisfies the given condition that the third term is 1000000.
910 Easy
910 of 951
Analyzing +4 | -1
Statement-1 ∑r=0n(r+1)nCr=(n+2)⋅2n−1\sum\limits_{r=0}^n(r+1){ }^n C_r=(n+2) \cdot 2^{n-1}
Statement - 2 ∑r=0n(r+1)nCrxr\sum\limits_{{r=0}}^n(r+1)^n C_r x^r=(1+x)n+nx(1+x)n−1=(1+x)^n+n x(1+x)^{n-1}
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true; Statement-2 is a
correct explanation for Statement-1
Option A is correct because both Statement-1 and Statement-2 are true, and Statement-2 provides a general formula that, when evaluated at x=1, proves Statement-1.
911 Medium
911 of 951
Analyzing +4 | -1
If T0,  T1,T2,…,Tn{T_0},\;{T_1},{T_2}, \ldots ,{T_n} represents the terms in the expansion of (x+a)n{\left( {x + a} \right)^n}, then (T0−  T2+T4−...)2+(T1−  T3+T5−...)2{\left( {{T_0} - \;{T_2} + {T_4} - ...} \right)^2} + {\left( {{T_1} - \;{T_3} + {T_5} - ...} \right)^2} is equal to
View answer and explanation
Correct option: B
(x2+a2)n{\left( {{x^2} + {a^2}} \right)^n}
Option B ((x2+a2)n{\left( {{x^2} + {a^2}} \right)^n}) is correct because by replacing 'a' with 'ai' and '-ai' in the binomial expansion and multiplying the resulting equations, we obtain the given expression.
912 Easy
912 of 951
Application +4 | -1
The number of terms with integral coefficients in the expansion of (71/3+51/2x)600,{\left( {{7^{1/3}} + {5^{1/2}}x} \right)^{600}}, is
View answer and explanation
Correct option: C
101
Option C (101) is correct because the number of terms with integral coefficients corresponds to the number of multiples of 6 between 0 and 600, inclusive.
913 Easy
913 of 951
Application +4 | -1
In the expansion of (x+x2−1)6+(x−x2−1)6,{\left( {x + \sqrt {{x^2} - 1} } \right)^6} + {\left( {x - \sqrt {{x^2} - 1} } \right)^6}, the number of terms, is
View answer and explanation
Correct option: D
4
Option D (4) is correct because after simplification and combining like terms, the expansion results in a polynomial with only 4 terms.
914 Easy
914 of 951
Application +4 | -1
In the expansion of (x3−1x2)15\left(x^3-\frac{1}{x^2}\right)^{15}, the constant term is
View answer and explanation
Correct option: B
−15C6-{ }^{15} C_6
Option B (−15C6-{ }^{15} C_6) is correct because the constant term is found by determining the value of r that makes the exponent of x equal to zero, and then calculating the corresponding binomial coefficient.
915 Medium
915 of 951
Application +4 | -1
Value of S=∑k=1∞∑r=0k13k(kCr)S=\sum\limits_{{k=1}}^{\infty} \sum\limits_{{r=0}}^k \frac{1}{3^k}\left({ }^k C_r\right) is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the double summation simplifies to a geometric series with a sum of 2.

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916 Medium
916 of 951
Application +4 | -1
If coefficient of a2b3c4a^2 b^3 c^4 in (a+b+c)m(a+b+c)^m, (where m∈N)\left.m \in N\right) is L(L≠0)L(L \neq 0) then in same expansion coefficient of a4b4c1a^4 b^4 c^1 will be
View answer and explanation
Correct option: D
L2\frac{\mathrm{L}}{2}
Option D (L2\frac{\mathrm{L}}{2}) is correct because the ratio of the coefficients is determined by the multinomial theorem and the given condition that L≠0L \ne 0.
917 Easy
917 of 951
Application +4 | -1
In the expansion of (2x2−1x)12{\left( {2{x^2} - \frac{1}{x}} \right)^{12}}, the term independent of xx is
View answer and explanation
Correct option: C
9th
Option C (9th) is correct because the term independent of xx occurs when the power of xx is 0, which happens at the 9th term in the expansion.
918 Easy
918 of 951
Application +4 | -1
If the term independent of xx in the expansion of (x−kx2)10\left(\sqrt{x}-\frac{k}{x^2}\right)^{10} is 405 , then kk equals
View answer and explanation
Correct option: B
3,−33,-3
Option B (3,−33, -3) is correct because the term independent of x is found by setting the power of x to zero and solving for k.
919 Medium
919 of 951
Application +4 | -1
Assuming to be small that x2{x^2} and higher powers of xx can be neglected, then (1+34x)−4(16−3x)1/2(8+x)2/3\frac{{{{\left( {1 + \frac{3}{4}x} \right)}^{ - 4}}{{\left( {16 - 3x} \right)}^{1/2}}}}{{{{\left( {8 + x} \right)}^{2/3}}}} is approximately equal to
View answer and explanation
Correct option: B
1−30596x1 - \frac{{305}}{{96}}x
Option B (1−30596x1 - \frac{{305}}{{96}}x) is correct because using the binomial approximation and neglecting higher powers of x leads to this result.
920 Medium
920 of 951
Application +4 | -1
If (1+x)n=∑r=0ncrxr(1+x)^n=\sum\limits_{{r=0}}^n c_r x^r, then ∑r=0200crr+1=\sum\limits_{{r=0}}^{200} \frac{c_r}{r+1}=
View answer and explanation
Correct option: D
2201−1201\frac{2^{201}-1}{201}
Option D (2201−1201\frac{{2^{201}-1}}{{201}}) is correct because it is derived from integrating the binomial expansion and evaluating the sum.
921 Hard
921 of 951
Application +4 | -1
If n∈N−{1}n \in \mathrm{N}-\{1\}, then ∑r=0n{(−1)r(100−r)(nr)}=\sum\limits_{{r=0}}^n\left\{(-1)^r(100-r)\left(\begin{array}{l}n \\ r\end{array}\right)\right\}=
View answer and explanation
Correct option: D
0
Option D (0) is correct because the given summation simplifies to zero when n>1n > 1 due to the properties of binomial coefficients.
922 Easy
922 of 951
Application +4 | -1
Sum of the series
2C0+C1222+C2323+C3424+……⋅Cnn+12n+12 \mathrm{C}_0+\frac{\mathrm{C}_1}{2} 2^2+\frac{\mathrm{C}_2}{3} 2^3+\frac{\mathrm{C}_3}{4} 2^4+\ldots \ldots \cdot \frac{\mathrm{C}_{\mathrm{n}}}{\mathrm{n}+1} 2^{\mathrm{n}+1} equals-
View answer and explanation
Correct option: B
3n+1−1n+1\frac{3^{\mathrm{n}+1}-1}{\mathrm{n}+1}
The correct answer is B (3n+1−1n+1\frac{{3^{n+1}-1}}{{n+1}}) because the given series can be expressed as a summation involving binomial coefficients, which simplifies to the given expression.
923 Easy
923 of 951
Analyzing +4 | -1
If (2x2−x−1)5=a0+a1x+a2x2+...+a10x10,  then  a2+a4+a6+a8+a10{(2{x^2} - x - 1)^5} = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_{10}}{x^{10}},\;\mathrm{then\;}{a_2} + {a_4} + {a_6} + {a_8} + {a_{10}} is equal to
View answer and explanation
Correct option: A
15
Option A (15) is correct because by substituting x = 1 and x = -1 into the given equation and adding the resulting equations, we can isolate and find the value of a2+a4+a6+a8+a10{a_2} + {a_4} + {a_6} + {a_8} + {a_{10}}.
924 Medium
924 of 951
Application +4 | -1
If ' nn ' is positive integer and three consecutive coefficient in the expansion of (1+x)n(1+x)^n are in the ratio 6:33:1106: 33: 110, then nn is equal to :
View answer and explanation
Correct option: C
12
Option C (12) is correct because solving the ratio of consecutive coefficients in the binomial expansion leads to n=12.
925 Medium
925 of 951
Analyzing +4 | -1
The value of ∑r=0       r≤sn∑s=1nnCssCr\sum\limits_{\mathop {r = 0}\limits_{\,\,\,\,\,\,\,r \le s} }^n {\sum\limits_{s = 1}^n {^n} {C_s}^s{C_r}} is
View answer and explanation
Correct option: A
3n−13^n-1
Option A (3n−13^n-1) is correct because the double summation simplifies to a binomial expansion minus the first term.
926 Easy
926 of 951
Application +4 | -1
The sum of the magnitudes of the coefficients in the expansion of (1−x+x2−x3)n,{\left( {1 - x + {x^2} - {x^3}} \right)^n}, is
View answer and explanation
Correct option: D
4n{4^n}
Option D is correct. The sum of the magnitudes of the coefficients is found by substituting x = -1 into the expression.
927 Hard
927 of 951
Analyzing +4 | -1
If (1+x)n=∑r=0nnCrxn(1+x)^n=\sum\limits_{{r=0}}^n{ }^n C_r x^n, then C01⋅222+C12⋅323\frac{C_0}{1 \cdot 2} 2^2+\frac{C_1}{2 \cdot 3} 2^3 +C23⋅424+…+Cn(n+1)(n+2)2n+2+\frac{C_2}{3 \cdot 4} 2^4+\ldots+\frac{C_n}{(n+1)(n+2)} 2^{n+2} is equal to
View answer and explanation
Correct option: C
3n+2−2n−5(n+1)(n+2)\frac{3^{n+2}-2 n-5}{(n+1)(n+2)}
Option C (3n+2−2n−5(n+1)(n+2)\frac{{3^{n+2}-2 n-5}}{{(n+1)(n+2)}}) is correct because it is the sum of the series after applying binomial theorem and simplifying the expression.
928 Medium
928 of 951
Application +4 | -1
If in the expansion of (1+x)m(1−x)n(1+x)^m(1-x)^n, the coefficients of xx and x2x^2 are 3 and -6 respectively, then mm is
View answer and explanation
Correct option: C
12
Option C (12) is correct because solving the system of equations derived from the coefficients of x and x² yields m=12.
929 Easy
929 of 951
Analyzing +4 | -1
Sum of the series ∑k=0n(2n)!(k!)2(n−k!)2\sum\limits_{{k=0}}^n \frac{(2 n) !}{(k !)^2(n-k !)^2} equals
View answer and explanation
Correct option: B
(2nCn)2\left({ }^{2 n} C_n\right)^2
Option B ((2nCn)2\left({ }^{2 n} C_n\right)^2) is correct because the given sum can be simplified to (2nCn)∑k=0n(nCk)2\left({ }^{2 n} C_n\right) \sum_{k=0}^n\left({ }^n C_k\right)^2, which equals (2nCn)2\left({ }^{2 n} C_n\right)^2.
930 Medium
930 of 951
Application +4 | -1
The greatest term in the expansion of (2x+5y)13(2 x+5 y)^{13} when x=10x=10 and y=2y=2
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated greatest term does not match any of the provided options.
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