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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 1-30 of 951 questions
1 Medium
1 of 951
Application +4 | -1
(10C0)2−(10C1)2+…−(10C9)2+(10C10)2\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2 equals
View answer and explanation
Correct option: B
−10C5-{ }^{10} C_5
Option B (-10C5{ }^{10} C_5) is correct because the given expression is the coefficient of x10x^{10} in the expansion of (1−x2)10(1-x^2)^{10}, which can be found using the binomial theorem.
2 Easy
2 of 951
Analyzing +4 | -1
If aa and dd are two complex numbers, then the sum to (n+1)\left( {n + 1} \right) terms of the following series aC0−(a+d)C1+(a+2d)C2−...+...a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ... is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the given series can be expressed as a combination of binomial coefficients with alternating signs, which sums to zero.
3 Medium
3 of 951
Analyzing +4 | -1
The number of different terms in the expansion of (1−x)201(1+x+x2)200(1-x)^{201}\left(1+x+x^2\right)^{200} is
View answer and explanation
Correct option: D
402
Option D (402) is correct because the expression simplifies to (1-x)(1-x3)200, which has terms from x0 to x601, and considering the (1-x) factor, the number of terms is 2*201 = 402.
4 Medium
4 of 951
Application +4 | -1
The sum of the coefficient of the binomial expansion of (1x+2x)n\left(\frac{1}{x}+2 x\right)^n is equal to 6561 . The constant term in the expansion is
View answer and explanation
Correct option: B
16⋅8C416 \cdot{ }^8 C_4
Option B (16⋅8C416 \cdot{ }^8 C_4) is correct because it calculates the constant term in the binomial expansion after finding the value of n.
5 Easy
5 of 951
Application +4 | -1
The coefficient of x5{x^5} in the expansion of 1+x21+x,∣x∣<1,\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1, is
View answer and explanation
Correct option: D
−2- 2
Option D (-2) is correct because the coefficient of x5 in the expansion of (1+x2)/(1+x) is found by multiplying the series expansion of (1+x2) and (1+x)-1 and identifying the x5 term.
6 Medium
6 of 951
Application +4 | -1
If (2n+10)+(2n+13)+(2n+16)+…=170\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170, then nn equals
View answer and explanation
Correct option: B
4
Option B (4) is correct because substituting n=4 into the given equation yields 170, satisfying the condition.
7 Easy
7 of 951
Analyzing +4 | -1
Let (1+x)n=∑r=0nCr  xr{\left( {1 + x} \right)^n} = \sum \limits_{r = 0}^n {C_r}\;{x^r} and C1C0+2C2C1+3C3C2+…+nCnCn−1=1kn(n+1),\frac{{{C_1}}}{{{C_0}}} + 2\frac{{{C_2}}}{{{C_1}}} + 3\frac{{{C_3}}}{{{C_2}}} + \ldots + n\frac{{{C_n}}}{{{C_{n - 1}}}} = \frac{1}{k}n\left( {n + 1} \right), then the value of kk, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because by simplifying the summation and comparing it to the given expression, we find that k = 2.
8 Medium
8 of 951
Analyzing +4 | -1
If S=∑n=1∞(nC0+nC1+nC2+…+nCnnPn)S=\sum\limits_{{n=1}}^{\infty}\left(\frac{{ }^n C_0+{ }^n C_1+{ }^n C_2+\ldots+{ }^n C_n}{{ }^n P_n}\right), then SS is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the sum evaluates to e2−1e^2 - 1, which is not among the given options.
9 Easy
9 of 951
Application +4 | -1
The sum of the coefficients in the expansion of (1+2x+3x2+...+nxn)2{\left( {1 + 2x + 3{x^2} + ... + n{x^n}} \right)^2} is
View answer and explanation
Correct option: D
∑n3\sum {n^3}
Option D (∑n3\sum {n^3}) is correct because substituting x=1 into the given expression directly yields the sum of the coefficients, which simplifies to (∑n)2=∑n3(\sum n)^2 = \sum n^3.
10 Hard
10 of 951
Application +4 | -1
For natural numbers m,nm,n if (1−y)m(1+y)n=1+a1y2+{\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n} = 1 + {a_1}{y^2} +….and a1=a2=10,  {a_1} = {a_2} = 10,\; then (m,n)\left( {m,n} \right) is
View answer and explanation
Correct option: C
(35,45)\left( {35,45} \right)
Option C ((35,45)) is correct because these values satisfy the equations derived from the given conditions on the binomial expansion.
11 Easy
11 of 951
Analyzing +4 | -1
The coefficient of the term independent of xx in the expansion of [(x+1)x2/3−x1/3+1−(x−1)x−x1/2]10is{\left[ {\frac{{\left( {x + 1} \right)}}{{{x^{2/3}} - {x^{1/3}} + 1}} - \frac{{\left( {x - 1} \right)}}{{x - {x^{1/2}}}}} \right]^{10}}\mathrm{is}
View answer and explanation
Correct option: A
210
Option A (210) is correct because the coefficient of the term independent of x is found by simplifying the expression and using binomial theorem.
12 Easy
12 of 951
Application +4 | -1
If term independent of xx in the expansion of (x2−3x)6\left(x^2-\frac{3}{x}\right)^6 is λ35\lambda 3^5, then find the value of λ\lambda.
View answer and explanation
Correct option: A
5
Option A (5) is correct because the term independent of x is found by setting the power of x to zero and solving for the binomial coefficient.
13 Easy
13 of 951
Application +4 | -1
aC0+(a+b)C1+(a+2b)C2+…+(a+nb)Cna C_0+(a+b) C_1+(a+2 b) C_2+\ldots+(a+n b) C_n is equal to
View answer and explanation
Correct option: B
(2a+nb)2n−1(2 a+n b) 2^{n-1}
Option B ((2a+nb)2n−1(2 a+n b) 2^{n-1}) is correct because the summation can be split into two parts, one involving 'a' and the other involving 'b', and then simplified using binomial coefficient identities.
14 Medium
14 of 951
Analyzing +4 | -1
The coefficient of xr{0≤r≤(n−1)}x^r\{0 \leq r \leq(n-1)\} in the expansion of (x+2)n−1+(x+2)n−2(x+1)+(x+2)n−3(x+2)^{n-1}+(x+2)^{n-2}(x+1)+(x+2)^{n-3} (x+1)2+…+(x+1)n−1(x+1)^2+\ldots+(x+1)^{n-1} is equal to
View answer and explanation
Correct option: C
(2n−r−1)nCr\left(2^{n-r}-1\right)^n C_r
Option C ((2n−r−1)nCr\left(2^{{n-r}}-1\right)^n C_r) is correct because the given series can be simplified to (x+2)n−(x+1)n(x+2)^n - (x+1)^n, and the coefficient of xrx^r in this expression is (2n−r−1)nCr\left(2^{{n-r}}-1\right)^n C_r.
15 Medium
15 of 951
Application +4 | -1
If (1+x)n=C0+C1x+C2x2+…+Cnxn(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n, then (C0+C1)(C1+C2)…(Cn−1+Cn)C1C2…Cn\frac{\left(C_0+C_1\right)\left(C_1+C_2\right) \ldots\left(C_{n-1}+C_n\right)}{C_1 C_2 \ldots C_n} equals-
View answer and explanation
Correct option: B
(n+1)nn!\frac{(n+1)^n}{n !}
Option B ((n+1)nn!\frac{{(n+1)^n}}{{n!}}) is correct because simplifying the expression by using the binomial coefficient formula leads to this result.
16 Easy
16 of 951
Application +4 | -1
For a positive integer nn, if the mean of the binomial coefficients in the expansion of (a+b)2n−3(a+b)^{2 n-3} is 16, then nn is equal to:
View answer and explanation
Correct option: A
5
Option A (5) is correct because the mean of the binomial coefficients is given as 16, and solving the equation derived from this information yields n=5n=5.
17 Medium
17 of 951
Application +4 | -1
The greatest integer less than or equal to (2+1)6(\sqrt{2}+1)^6 is
View answer and explanation
Correct option: B
197
Option B (197) is correct because the greatest integer less than or equal to (2+1)6(\sqrt{2}+1)^6 is found by considering its conjugate and using binomial expansion.
18 Easy
18 of 951
Application +4 | -1
Coefficient of x′′x^{\prime \prime} in the expression (1−2x+3x2−4x3+⋯ )−n\left(1-2 x+3 x^2-4 x^3+\cdots\right)^{-n} equals
View answer and explanation
Correct option: A
2nCn{ }^{2 n} C_n
Option A (2nCn{ }^{2 n} C_n) is correct because the given expression simplifies to (1+x)(2n), and the coefficient of x^n is 2nCn{ }^{2 n} C_n.
19 Medium
19 of 951
Analyzing +4 | -1
If the coefficient of x15x^{15} in the expansion of (ax3+1bx1/3)15\left(a x^3+\frac{1}{b x^{1 / 3}}\right)^{15} is equal to the coefficient of x−15x^{-15} in the expansion of (ax1/3−1bx3)15\left(a x^{1 / 3}-\frac{1}{b x^3}\right)^{15}, where aa and bb are positive real numbers, then for each such ordered pair (a,b)(a, b)
View answer and explanation
Correct option: A
ab=1a b=1
The correct answer is ab=1. By equating the coefficients of x15 and x-15 in the given expansions and simplifying, we arrive at the relationship ab=1.
20 Easy
20 of 951
Analyzing +4 | -1
The sum ∑i=0m(10i)(20m−i)\sum\limits_{i=0}^m\left(\begin{array}{c}10 \\ i\end{array}\right)\left(\begin{array}{c}20 \\ m-i\end{array}\right), (where pq=0\frac{p}{q}=0, if p<q)p < q) is maximum when mm is
View answer and explanation
Correct option: C
15
Option C (15) is correct because the sum represents the coefficient of xmx^m in (1+x)30(1+x)^{30}, which is maximum when m=30/2=15m = 30/2 = 15.
21 Medium
21 of 951
Application +4 | -1
The number of irrational terms in the expansion of (51/6+21/8)100{({5^{1/6}} + {2^{1/8}})^{100}} is
View answer and explanation
Correct option: B
97
Option B (97) is correct because there are 4 rational terms in the expansion, and thus 101 - 4 = 97 irrational terms.
22 Hard
22 of 951
Understanding +4 | -1
nC0−12nC1+13nC2−…….+(−1)nnCnn+1={ }^{\mathrm{n}} \mathrm{C}_0-\frac{1}{2}{ }^{\mathrm{n}} \mathrm{C}_1+\frac{1}{3}{ }^{\mathrm{n}} \mathrm{C}_2-\ldots \ldots .+(-1)^{\mathrm{n}} \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}}{\mathrm{n}+1}=
View answer and explanation
Correct option: C
1n+1\frac{1}{n+1}
Option C (1n+1\frac{{1}}{{n+1}}) is correct because the given expression is the result of integrating the binomial expansion of (1+x)n(1+x)^n and then substituting x=−1x = -1.
23 Easy
23 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 the coefficient of x5{x^5} can be found by multiplying the relevant terms in the expansions of (1+x2)5{\left( {1 + {x^2}} \right)^5} and (1+x)4{\left( {1 + x} \right)^4}.
24 Medium
24 of 951
Application +4 | -1
The remainder when (2021)2022+(2022)2021(2021)^{2022}+(2022)^{2021} is divided by 7 is
View answer and explanation
Correct option: A
0
Option A is correct because when we divide (2021)2022+(2022)2021(2021)^{2022}+(2022)^{2021} by 7, the remainder is 0.
25 Medium
25 of 951
Analyzing +4 | -1
The value of 1+222!+243!+264!+…1+12!+23!+224!+…\frac{1+\frac{2^2}{2 !}+\frac{2^4}{3 !}+\frac{2^6}{4 !}+\ldots}{1+\frac{1}{2 !}+\frac{2}{3 !}+\frac{2^2}{4 !}+\ldots} is
View answer and explanation
Correct option: C
e2−1e^2-1
Option C (e2−1e^2-1) is correct because the numerator and denominator can be expressed in terms of exponential series.
26 Medium
26 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+\cdots+a_{2 n} x^{2 n}, then a0+a2+a4+⋯+a2na_0+a_2+a_4+\cdots+a_{2 n} 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 the correct answer because it is the calculated sum of the even-indexed coefficients.
27 Easy
27 of 951
Analyzing +4 | -1
The maximum value of the term independent of ′t′'t' in the expansion of (tx1/5+(1−x)1/10t)10{\left( {t{x^{1/5}} + \frac{{{{\left( {1 - x} \right)}^{1/10}}}}{t}} \right)^{10}}, where x∈(0,  1)x \in \left( {0,\,\,1} \right) is
View answer and explanation
Correct option: B
2.10!33(5!)2\frac{{2.10!}}{{3\sqrt 3 {{\left( {5!} \right)}^2}}}
Option B is correct. The maximum value of the term independent of ′t′'t' is found by using the binomial theorem and maximizing the resulting expression using calculus.
28 Easy
28 of 951
Application +4 | -1
The sum of coefficients of the expansion   (1x+2x)n  \mathrm{\;}{\left( {\frac{1}{x} + 2x} \right)^n}\mathrm{\;} in 6561. The coefficient of term independent of xx is
View answer and explanation
Correct option: A
16    8C416\;{\;^8}{C_4}
Option A (16  8C416{\;^8}{C_4}) is correct because it represents the coefficient of the term independent of x in the expansion, calculated by finding the value of r for which the power of x is zero.
29 Medium
29 of 951
Application +4 | -1
The greatest term in the expansion of (10+3x)12(10+3 x)^{12}, when x=4x=4 is _____ and its value is _____
View answer and explanation
Correct option: A
T8,(125)(120)5×144\mathrm{T}_8,\left(\begin{array}{c}12 \\ 5\end{array}\right)(120)^5 \times 144
Option A is correct because calculating the ratio of consecutive terms and finding where it becomes less than 1 determines the greatest term, which is T_8.
30 Easy
30 of 951
Application +4 | -1
If x>0,y>0x>0, y>0 then ∑r=1∞(x2x+3y)r−1=\sum\limits_{{r=1}}^{\infty}\left(\frac{x}{2 x+3 y}\right)^{r-1}=
View answer and explanation
Correct option: B
2x+3yx+3y\frac{2 x+3 y}{x+3 y}
Option B (2x+3yx+3y\frac{{2 x+3 y}}{{x+3 y}}) is correct because it is derived from summing the infinite geometric series.
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