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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Medium

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91 Medium
91 of 468
Application +4 | -1
Coefficient of x7x^7 in the expansion of (1−2x+x3)6\left(1-2 x+x^3\right)^6 is
View answer and explanation
Correct option: B
360
Option B (360) is correct because it is the sum of the coefficients of x7 obtained from the multinomial expansion of the given expression.
92 Medium
92 of 468
Analyzing +4 | -1
If x+y=1x+y=1, then ∑r=0nr2nC,xryn−r\sum_{r=0}^n r^{2 n} C, x^r y^{n-r} equals:
View answer and explanation
Correct option: C
nx(nx+y)n x(n x+y)
Option C (n x(n x+y)) is correct because the given summation can be simplified using binomial theorem properties and the condition x+y=1.
93 Medium
93 of 468
Application +4 | -1
The coefficient of xnx^n in (1−x+x22!−x33!+⋯+(−1)nxnn!)2\left(1-x+\frac{x^2}{2 !}-\frac{x^3}{3 !}+\cdots+\frac{(-1)^n x^n}{n !}\right)^2 is
View answer and explanation
Correct option: B
(−2)nn!\frac{(-2)^n}{n !}
Option B ((−2)nn!\frac{(-2)^n}{n !}) is the correct answer because it is the coefficient of xnx^n in the expansion of the given expression.
94 Medium
94 of 468
Application +4 | -1
Value of ∑k=0nnCksin⁡(kx)cos⁡(n−k)x\sum\limits_{{k=0}}^n{ }^n C_k \sin (k x) \cos (n-k) x is
View answer and explanation
Correct option: A
2n−1sin⁡(nx)2^{n-1} \sin (n x)
Option A (2n−1sin⁡(nx)2^{n-1} \sin (n x)) is correct because the summation can be simplified using trigonometric identities and the binomial theorem.
95 Medium
95 of 468
Application +4 | -1
Let xx be the 7th 7^{\text {th }} term from the beginning and yy be the 7th 7^{\text {th }} term from the end in the expansion of (31/3+141/3)n\left(3^{1 / 3}+\frac{1}{4^{1 / 3}}\right)^n. If y=12xy=12 x then the value of nn is :
View answer and explanation
Correct option: A
9
Option A (9) is correct because by equating the expressions for the 7th term from the beginning and end, and using the given relation y=12x, we can solve for n.
96 Medium
96 of 468
Application +4 | -1
The value of 47C4+∑j=0350−jC3+∑k=0556−kC53−k is { }^{47} C_4+\sum\limits_{{j=0}}^3{ }^{50-j} C_3+\sum\limits_{{k=0}}^5{ }^{56-k} C_{53-k} \text { is }
View answer and explanation
Correct option: A
57C4{ }^{57} C_4
Option A (57C4{ }^{57} C_4) is correct because the given expression simplifies to 57C4{ }^{57} C_4 using combinatorial identities.
97 Medium
97 of 468
Analyzing +4 | -1
The coefficient of x2m+1{x^{{2^{m + 1}}}} in the expansion of E=1(1+x)(1+x2)(1+x4)(1+x8)…(1+x2m)  ,  ∣x∣<1E = \frac{1}{{\left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^4}} \right)\left( {1 + {x^8}} \right) \ldots \left( {1 + {x^{{2^m}}}} \right)}}\;,\;\left| x \right| < 1 is
View answer and explanation
Correct option: C
1
Option C (1) is correct because after simplifying the expression, the coefficient of x^(2(m+1)) in the resulting series is 1.
98 Medium
98 of 468
Application +4 | -1
14C7+∑i=1317−iC6={ }^{14} C_7+\sum\limits_{{i=1}}^3{ }^{17-i} C_6=
View answer and explanation
Correct option: B
17C7{ }^{17} {C}_7
Option B (17C7{ }^{17} {C}_7) is correct because the summation can be simplified using the identity (nk)+(nk+1)=(n+1k+1){n \choose k} + {n \choose k+1} = {n+1 \choose k+1} repeatedly.
99 Medium
99 of 468
Application +4 | -1
If   nC0,    nC1,    nC2….  nCn  {\;^n}{C_0},\;{\;^n}{C_1},\;{\;^n}{C_2} \ldots .{\;^n}{C_n}\; denote the coefficient of the binomial expansion (1+x)n{\left( {1 + x} \right)^n}, then the value of C1+3C3+5C5+...{C_1} + 3{C_3} + 5{C_5} + ... is
View answer and explanation
Correct option: A
n2n−2n{2^{n - 2}}
Option A (n2(n-1)) is correct because the value of the series C1 + 3C3 + 5C5 + ... is equal to n2(n-2).
100 Medium
100 of 468
Application +4 | -1
If (1+x)n=C0+C1x+C2x2+…+CnXn(1+\mathrm{x})^n=\mathrm{C}_0+\mathrm{C}_1 \mathrm{x}+\mathrm{C}_2 \mathrm{x}^2+\ldots+\mathrm{C}_{\mathrm{n}} \mathrm{X}^n, then C0+2C1+3C2+…+(n+1)Cn\mathrm{C}_0+2 \mathrm{C}_1+3 \mathrm{C}_2+\ldots+(\mathrm{n}+1) \mathrm{C}_{\mathrm{n}} is equal to-
View answer and explanation
Correct option: A
2n−1(n+2)2^{n-1}(n+2)
Option A (2n−1(n+2)2^{n-1}(n+2)) is correct because differentiating the binomial expansion and adding it to the original expansion gives the desired sum.
101 Medium
101 of 468
Application +4 | -1
The greatest value of the term independent of xx in the expansion of (xsin⁡α+x−1cos⁡α)10,  α∈R{\left( {x\sin \mathrm{\alpha } + {x^{ - 1}}\cos \mathrm{\alpha }} \right)^{10}},\;\mathrm{\alpha } \in R is
View answer and explanation
Correct option: C
125  (10C5)\frac{1}{{{2^5}}}\;{(^{10}}{C_5})
Option C (125  (10C5)\frac{1}{{{{2^5}}}}\;{(^{10}}{C_5})) is correct because the greatest value of the term independent of x is obtained when sin(2α) = 1.
102 Medium
102 of 468
Analyzing +4 | -1
The coefficient of x256{x^{256}} in the expansion of (1−x)101(x2+x+1)100{\left( {1 - \mathrm{ }x} \right)^{101}}{\left( {\mathrm{ }{x^2} + \mathrm{ }x + 1} \right)^{100}} is
View answer and explanation
Correct option: B
100C15^{100}{C_{15}}
The correct answer is B (100C15^{100}{C_{15}}) because it is the coefficient of x256{x^{256}} obtained after simplifying the expression using the identity a3−b3=(a−b)(a2+ab+b2){a^3} - {b^3} = (a - b)({a^2} + ab + {b^2}).
103 Medium
103 of 468
Application +4 | -1
If II is integral part of (2+3)n(2+\sqrt{3})^n and ff is its fractional part. Then (I+f)(1−f)(I+f)(1-f) is
View answer and explanation
Correct option: B
1
Option B (1) is correct because (I+f)(1−f)=(2+3)n(2−3)n=1(I+f)(1-f) = (2+\sqrt{3})^n(2-\sqrt{3})^n = 1, based on the properties of the integral and fractional parts.
104 Medium
104 of 468
Application +4 | -1
If (1+x−2x2)6=∑r=012Arxr\left(1+x-2 x^2\right)^6=\sum\limits_{{r=0}}^{12} A_r x^r, then value of A2+A4A_2+A_4 +A6+…+A12+A_6+\ldots+A_{12}, is
View answer and explanation
Correct option: B
31
Option B (31) is correct because by substituting x=1 and x=-1 into the given expression and adding the resulting equations, we can isolate the sum of even-powered coefficients.
105 Medium
105 of 468
Application +4 | -1
∑r=1n(−1)r+1⋅nCrr+1\sum\limits_{{r=1}}^n(-1)^{r+1} \cdot \frac{{ }^n C_r}{r+1} is equal to
View answer and explanation
Correct option: D
nn+1\frac{n}{n+1}
Option D (nn+1\frac{n}{n+1}) is correct because by manipulating the summation and using binomial coefficient identities, we can simplify the expression to nn+1\frac{n}{n+1}.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
106 Medium
106 of 468
Application +4 | -1
The value of
(21C1−10C1)+(21C2−10C2)+(21C3−10C3)+(21C4−10C4)+….+(21C10−10C10) is : \begin{array}{l} \left({ }^{21} \mathrm{C}_{1}-{ }^{10} \mathrm{C}_{1}\right)+\left({ }^{21} \mathrm{C}_{2}-{ }^{10} \mathrm{C}_{2}\right)+\left({ }^{21} \mathrm{C}_{3}-{ }^{10} \mathrm{C}_{3}\right)+\left({ }^{21} \mathrm{C}_{4}-{ }^{10} \mathrm{C}_{4}\right) \\ +\ldots .+\left({ }^{21} \mathrm{C}_{10}-{ }^{10} \mathrm{C}_{10}\right) \text { is : } \end{array}
View answer and explanation
Correct option: A
220−2102^{20}-2^{10}
Option A (220−2102^{20}-2^{10}) is correct because it involves summing binomial coefficients and using properties of binomial expansions.
107 Medium
107 of 468
Application +4 | -1
Middle term in the expansion of (1+3x+3x2+x3)6\left(1+3 x+3 x^2+x^3\right)^6 is
View answer and explanation
Correct option: C
10th10 \mathrm{th}
Option C (10th) is correct because the given expression simplifies to (1+x)18, and the middle term of this expansion is the 10th term.
108 Medium
108 of 468
Analyzing +4 | -1
The greatest value of the term independent of xx, as α\alpha varies over RR, in the expansion of (xcos⁡α+sin⁡αx)20{\left( {x\cos \mathrm{\alpha } + \frac{{\sin \mathrm{\alpha }}}{x}} \right)^{20}} is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the greatest value of the term independent of x is actually   20C10(12)10{\;^{20}}{C_{10}}{\left( {\frac{1}{2}} \right)^{10}}, which is not among the given options.
109 Medium
109 of 468
Analyzing +4 | -1
If CrC_r stands for nCr{ }^n C_r, then the sum of the series
2(n2)!(n2)!n![C02−2C12+3C22……+(−1)n(n+1)Cn2]\frac{2\left(\frac{n}{2}\right) !\left(\frac{n}{2}\right) !}{n !}\left[C_0^2-2 C_1^2+3 C_2^2 \ldots \ldots+(-1)^n(n+1) C_n^2\right] where 'n{n}' is an even positive integer is equal to
View answer and explanation
Correct option: C
(−1)n/2(n+2)(-1)^{n / 2}(n+2)
Option C ((−1)n/2(n+2)(-1)^{n / 2}(n+2)) is correct because after simplifying the series, the sum equals this expression.
110 Medium
110 of 468
Analyzing +4 | -1
If xx is nearly equal to one, then the approximate value of mxm−nxnm−n\frac{m x^m-n x^n}{m-n} is
View answer and explanation
Correct option: C
xm+nx^{m+n}
Option C (xm+nx^{m+n}) is correct because using the approximation x=1+hx = 1 + h and neglecting higher powers of hh simplifies the expression.
111 Medium
111 of 468
Application +4 | -1
The greatest term in the expansion of (1+x)10(1+x)^{10}, when x=23x=\frac{2}{3}, is
View answer and explanation
Correct option: A
210(23)4210\left(\frac{2}{3}\right)^4
Option A (210(2/3)4) is correct because it is found by determining the greatest term in the binomial expansion using the ratio test and calculating the corresponding term.
112 Medium
112 of 468
Application +4 | -1
If (1+x)(1+x+x2)(1+x+x2+x3)(1+x)\left(1+x+x^2\right)\left(1+x+x^2+x^3\right).(1+x+x2+x3+…+xn)=a0+a1x+a2x2+…+amxm\left(1+x+x^2+x^3+\ldots+x^n\right)=a_0+a_1 x+a_2 x^2+\ldots+a_m x^m then ∑r=0mar=\sum_{r=0}^m a_r=
View answer and explanation
Correct option: D
nn !
Option D (nn !) is correct because substituting x=1 into the given equation results in the factorial of n.
113 Medium
113 of 468
Understanding +4 | -1
1+13x+1×43×6x2+1×4×73×6×9x3+…1+\frac{1}{3} x+\frac{1 \times 4}{3 \times 6} x^2+\frac{1 \times 4 \times 7}{3 \times 6 \times 9} x^3+\ldots is equal to
View answer and explanation
Correct option: D
(1−x)−1/3(1-x)^{-1 / 3}
Option D ((1−x)−1/3(1-x)^{-1 / 3}) is correct because the given series is the binomial expansion of (1−x)−1/3(1-x)^{-1 / 3}.
114 Medium
114 of 468
Application +4 | -1
The coefficient of x6⋅y−2x^6 \cdot y^{-2} in the expansion of (x2y−yx)12\left(\frac{x^2}{y}-\frac{y}{x}\right)^{12} is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the conditions for the powers of x and y to be x6 and y-2 cannot be simultaneously satisfied.
115 Medium
115 of 468
Analyzing +4 | -1
If (3+x2008+x2009)2010=a0+a1x+a2x2\left(3+x^{2008}+x^{2009}\right)^{2010}=a_0+a_1 x+a_2 x^2 +…+anxn+\ldots+a_n x^n, the value of a0−a12−a22+a3−a42−a52+a6−…a_0-\frac{a_1}{2}-\frac{a_2}{2}+a_3-\frac{a_4}{2}-\frac{a_5}{2}+a_6-\ldots is
View answer and explanation
Correct option: B
220102^{2010}
Option B (220102^{2010}) is correct because substituting complex cube roots of unity into the given polynomial and summing the results leads to the desired expression.
116 Medium
116 of 468
Application +4 | -1
If the rth r^{\text {th }} term in the expansion of (x3−2x2)10\left(\frac{x}{3}-\frac{2}{x^2}\right)^{10} contains x4x^4, then the value of rr is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the rth r^{\text {th }} term in the expansion of (x3−2x2)10\left(\frac{x}{3}-\frac{2}{x^2}\right)^{10} containing x4x^4 leads to the equation 13−3r=413-3r = 4, which gives r=3r=3.
117 Medium
117 of 468
Application +4 | -1
(200)−(201)+(202)−(203)+…+(2010)=\left(\begin{array}{c} 20 \\ 0 \end{array}\right)-\left(\begin{array}{c} 20 \\ 1 \end{array}\right)+\left(\begin{array}{c} 20 \\ 2 \end{array}\right)-\left(\begin{array}{c} 20 \\ 3 \end{array}\right)+\ldots+\left(\begin{array}{l} 20 \\ 10 \end{array}\right)=
View answer and explanation
Correct option: B
12(2010)\frac{1}{2}\left(\begin{array}{l}20 \\ 10\end{array}\right)
Option B is correct because by substituting x = -1 into the binomial expansion and using the property that C(n,r) = C(n, n-r), the given sum simplifies to half of C(20, 10).
118 Medium
118 of 468
Application +4 | -1
If (1+x−2x2)8=a0+a1x+a2x2+…+a16x16\left(1+x-2 x^2\right)^8=a_0+a_1 x+a_2 x^2+\ldots+a_{16} x^{16}, then the sum a1+a3+a5+…+a15a_1+a_3+a_5+\ldots+a_{15} is equal to
View answer and explanation
Correct option: A
−27-2^7
Option A (-27) is correct because it calculates the sum of odd-powered coefficients using the given polynomial and substituting x=1 and x=-1.
119 Medium
119 of 468
Analyzing +4 | -1
Let nn be a positive integer such that (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}, then a0+a1+a2+…+an−1a_0+a_1+a_2+\ldots+a_{n-1} is
View answer and explanation
Correct option: C
12(3n−an)\frac{1}{2}\left(3^n-a_n\right)
Option C (12(3n−an)\frac{1}{2}(3^n-a_n)) is correct because it utilizes the symmetry of the coefficients and the total sum to isolate the desired partial sum.
120 Medium
120 of 468
Application +4 | -1
Let mm and nn be the coefficients of seventh and thirteenth terms respectively in the expansion of (13x13+12x23)18\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}. Then (nm)13\left(\frac{n}{m}\right)^{\frac{1}{3}} is
View answer and explanation
Correct option: C
94\frac{9}{4}
Option C (94\frac{9}{4}) is correct because it is the simplified ratio of the coefficients of the seventh and thirteenth terms in the binomial expansion, raised to the power of 13\frac{1}{3}.
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