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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Easy

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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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391 Easy
391 of 434
Analyzing +4 | -1
x10+10x8a+40x6a2+80x4a3+80x2a4+32a5x^{10}+10 x^8 a+40 x^6 a^2+80 x^4 a^3+80 x^2 a^4+32 a^5 is equal to
View answer and explanation
Correct option: C
(x2+2a)5\left(x^2+2 a\right)^5
Option C ((x2+2a)5(x^2+2 a)^5) is correct because it matches the binomial expansion of the given expression.
392 Easy
392 of 434
Analyzing +4 | -1
If f(x)=∑r=1n{r2(nCr−nCr−1)+(2r+1)nCr}f(x)=\sum\limits_{r=1}^n\left\{r^2\left({ }^n C_r-{ }^n C_{r-1}\right)+(2 r+1)^n C_r\right\} and f(30)=30(2)λf(30)=30(2)^\lambda, then the value of λ\lambda is
View answer and explanation
Correct option: C
5
Option C (5) is correct, but the provided solution steps are incomplete. A full solution is not possible without further information or context.
393 Easy
393 of 434
Understanding +4 | -1
1+13x+1.43.6x2+1.4.73.6.9x3+...1 + \frac{1}{3}x + \frac{{1.4}}{{3.6}}{x^2} + \frac{{1.4.7}}{{3.6.9}}{x^3} + ... is equal to
View answer and explanation
Correct option: D
(1−x)−1/3{\left( {1 - x} \right)^{ - 1/3}}
Option D ((1−x)−1/3{\left( {1 - x} \right)^{ - 1/3}}) is correct because the given series matches the binomial expansion of (1−x)−1/3{\left( {1 - x} \right)^{ - 1/3}}.
394 Easy
394 of 434
Application +4 | -1
For ∣x∣<1\left| x \right| < 1, the constant term in the expansion of 1(x−1)2(x−2)\frac{1}{{{{\left( {x - 1} \right)}^2}\left( {x - 2} \right)}} is
View answer and explanation
Correct option: D
−12- \frac{1}{2}
Option D (−12- \frac{1}{2}) is correct because it represents the constant term obtained after expanding the given expression using partial fractions and binomial expansions.
395 Easy
395 of 434
Application +4 | -1
(23n−1)({2^{3n}} - 1) will be divisible by (∀n∈N)\left( {\forall n \in N} \right)
View answer and explanation
Correct option: C
7
Option C (7) is correct because 23n−1{2^{3n}} - 1 can be rewritten as 8n−1{8^n} - 1, which is always divisible by 7.
396 Easy
396 of 434
Application +4 | -1
In the expansion of (ax+bx)12{\left( {\frac{a}{x} + bx} \right)^{12}},the coefficient of x−10{x^{ - 10}} will be
View answer and explanation
Correct option: C
12a11b12{a^{11}}b
Option C (12a11b12{a^{11}}b) is correct because it is the coefficient of x−10{x^{ - 10}} obtained from the binomial expansion of the given expression.
397 Easy
397 of 434
Application +4 | -1
The 4th term from the end in the expansion of (x32−2x2)9\left(\frac{x^3}{2}-\frac{2}{x^2}\right)^9 is
View answer and explanation
Correct option: C
672x3\frac{672}{x^3}
Option C (672/x3) is correct because it is the 7th term from the beginning, calculated using the binomial theorem.
398 Easy
398 of 434
Application +4 | -1
The coefficients of x2y2,  yzt2{x^2}{y^2},\;yz{t^2} and   xyzt\mathrm{\;}xyzt in the expansion of (x+y+z+t)4{\left( {x + y + z + t} \right)^4} are in the ratio
View answer and explanation
Correct option: B
1:2:41:2:4
Option B (1:2:41:2:4) is correct because the coefficients can be calculated using the multinomial theorem, and their ratio is 6:12:246:12:24 which simplifies to 1:2:41:2:4.
399 Easy
399 of 434
Application +4 | -1
If n−1Cr=(k2−3)⋅nCr+1^{n-1} C_r=\left(k^2-3\right) \cdot{ }^n C_{r+1} then kk belongs to
View answer and explanation
Correct option: D
(3,2](\sqrt{3}, 2]
Option D ((3,2]\sqrt{3}, 2]) is correct because the given condition leads to 3<k2≤43 < k^2 \leq 4, which means kk belongs to the interval (3,2](\sqrt{3}, 2].
400 Easy
400 of 434
Application +4 | -1
The sum of the coefficients of all even degree terms is xx in the expansion of (x+x3−1)6 + (x−x3−1)6 ,{(x + \sqrt {{x^3} - 1} )^6}\, + \,{(x - \sqrt {{x^3} - 1} )^6}\,, (x>1)(x > 1) is equal to
View answer and explanation
Correct option: D
24
Option D (24) is correct because it accurately calculates the sum of the coefficients of all even degree terms in the given expansion.
401 Easy
401 of 434
Application +4 | -1
The number of terms in the expansion of (5+114)124(\sqrt{5}+\sqrt[4]{11})^{124} which are integers, is equal to
View answer and explanation
Correct option: D
32
Option D (32) is correct because there are 32 integer terms in the expansion, determined by the conditions on r.
402 Easy
402 of 434
Application +4 | -1
The coefficient of xm{x^m} in (1+x)p+(1+x)p+1+…+(1+x)n,p≤m≤n{\left( {1 + x} \right)^p} + {\left( {1 + x} \right)^{p + 1}} + \ldots + {\left( {1 + x} \right)^n},p \le m \le n is
View answer and explanation
Correct option: A
  n+1Cm+1{\;^{n + 1}}{C_{m + 1}}
The correct answer is   n+1Cm+1{\;^{n + 1}}{C_{m + 1}} because the coefficient of xm{x^m} can be found by manipulating the sum into a single term and then finding the coefficient of xm+1{x^{m+1}} in (1+x)n+1{\left( {1 + x} \right)^{n + 1}}.
403 Easy
403 of 434
Application +4 | -1
If the ratio of the 7th term from the beginning to the seventh term from the end in the expansion of (23+133)x{\left( {\sqrt[3]{2} + \frac{1}{{\sqrt[3]{3}}}} \right)^x} is 16\frac{1}{6} then xx, is
View answer and explanation
Correct option: A
9
Option A (9) is correct because it satisfies the given ratio condition of the 7th terms from the beginning and end of the expansion.
404 Easy
404 of 434
Application +4 | -1
The coefficient of xn{x^n} in the expansion of (1+x+x2+…)−n{\left( {1 + x + {x^2} + \ldots } \right)^{ - n}} is
View answer and explanation
Correct option: B
(−1)n{\left( { - 1} \right)^n}
Option B ((−1)n{\left( { - 1} \right)^n}) is correct because the given expression simplifies to (1−x)n{\left( {1 - x} \right)^n}, and the coefficient of xn{x^n} is then (−1)n{\left( { - 1} \right)^n}.
405 Easy
405 of 434
Application +4 | -1
Find the coefficient of x13x^{13} in the expansion of (1−x)5(1+x+x2+x3)4(1-x)^5\left(1+x+x^2+x^3\right)^4 is
View answer and explanation
Correct option: A
4
Option A (4) is correct because it is the coefficient of x13x^{13} after simplifying the expression (1−x)5(1+x+x2+x3)4(1-x)^5(1+x+x^2+x^3)^4.
406 Easy
406 of 434
Analyzing +4 | -1
If (1+ax)n=1+10x+40x2+…(1+a x)^n=1+10 x+40 x^2+\ldots, then value of a+na−n\frac{a+n}{a-n} is
View answer and explanation
Correct option: D
−7/3-7 / 3
Option D (-7 / 3) is correct because by comparing coefficients and solving the resulting equations, we find a=2 and n=5, leading to the correct value of (a+n)/(a-n).
407 Easy
407 of 434
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 \ldots+a_{2 n} x^{2 n}. Then value of a0+a3+a6……a_0+a_3+a_6 \ldots \ldots. is equal to
View answer and explanation
Correct option: C
3n−13^{n-1}
Option C (3n−13^{n-1}) is correct because by substituting x=1,ω,ω2x=1, \omega, \omega^2 and adding them, we can isolate the desired sum.
408 Easy
408 of 434
Application +4 | -1
If the (r+1)th\left( {r + 1} \right)th term in the expansion of (a1/3b1/6+b1/2a1/6)21{\left( {\frac{{{a^{1/3}}}}{{{b^{1/6}}}} + \frac{{{b^{1/2}}}}{{{a^{1/6}}}}} \right)^{21}} has equal exponents of both aa and bb, then value of rr is
View answer and explanation
Correct option: B
9
Option 9 is correct. By equating the exponents of a and b in the (r+1)th term, we find r=9.
409 Easy
409 of 434
Analyzing +4 | -1
If X={4n−3n−1:n∈N}X = \left\{ {{4^n} - 3n - 1:n \in N} \right\} and Y={9(n−1):n∈N}Y = \left\{ {9\left( {n - 1} \right):n \in N} \right\} then X∪YX \cup Y is equal to:
View answer and explanation
Correct option: B
Y
Option B (Y) is correct because X⊆YX \subseteq Y, therefore X∪Y=YX \cup Y = Y.
410 Easy
410 of 434
Analyzing +4 | -1
The sum to infinity of the series 1+23+632+10333+1434+….1+\frac{2}{3}+\frac{6}{3^2}+\frac{10^3}{3^3}+\frac{14}{3^4}+\ldots .. , is
View answer and explanation
Correct option: B
3
Option B (3) is correct because the sum to infinity of the series is 3.
411 Easy
411 of 434
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.
412 Easy
412 of 434
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.
413 Easy
413 of 434
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).
414 Easy
414 of 434
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.
415 Easy
415 of 434
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.
416 Easy
416 of 434
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.
417 Easy
417 of 434
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.
418 Easy
418 of 434
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.
419 Easy
419 of 434
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.
420 Easy
420 of 434
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.
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