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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Medium

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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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421 Medium

421 of 468

Application +4 | -1

Value of

S=1 \times 2 \times 3 \times 4+2 \times 3 \times 4 \times 5 +\ldots+n(n+1)(n+2)(n+3)

View answer and explanation

Correct option: A

\frac{1}{5} n(n+1)(n+2)(n+3)(n+4)

Option A (

\frac{1}{5} n(n+1)(n+2)(n+3)(n+4)

) is correct because the summation can be expressed as a telescoping sum of combinations, leading to the given formula.

422 Medium

422 of 468

Understanding +4 | -1

Value of the expression

C_0^2+C_1^2+C_2^2+\cdots+C_n^2 \text { is }

View answer and explanation

Correct option: C

{ }^{2 n} \mathrm{C}_n

Option C (

{ }^{2 n} \mathrm{C}_n

) is correct because the given expression represents the coefficient of

x^n

in the expansion of

(1+x)^n(1+x)^n

423 Medium

423 of 468

Analyzing +4 | -1

(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n

, then

{\sum\limits_{0 \le i} {\sum\limits_{ < j \le n} {({C_i} - {C_j})} } ^2}

View answer and explanation

Correct option: A

(n+1){ }^{2 n} C_n-2^{2 n}

Option A is correct because the given summation simplifies to

(n+1){ }^{2 n} C_n-2^{2 n}

using properties of binomial coefficients.

424 Medium

424 of 468

Analyzing +4 | -1

(1+x)^{10}=a_0+a_1 x+a_2 x^2+\ldots .+a_{10} x^{10}

, then

\left(a_0-a_2+a_4-a_6+a_8-a_{10}\right)^2 +\left(a_1-a_3+a_5-a_7+a_9\right)^2 \text { is equal to }

View answer and explanation

Correct option: B

2^{10}

Option B (

2^{10}

) is correct because by substituting x = i and x = -i into the binomial expansion and multiplying the results, we can simplify the expression to

2^{10}

425 Medium

425 of 468

Analyzing +4 | -1

\sum\limits_{{r=0}}^{2 n} a_r(x-2)^r=\sum\limits_{{r=0}}^{2 n} b_r(x-3)^r

and

a_k=\frac{1}{{ }^{n+k} C_n}

for

0 \leq k \leq n

then

b_n

equals

View answer and explanation

Correct option: D

n+1

Option D (

n+1

) is correct because by substituting

x-3=t

, the coefficient

b_n

is found to be the sum of

a_k { }^{n+k} C_k

from k=0 to n, which simplifies to

n+1

426 Medium

426 of 468

Application +4 | -1

\left(\sqrt[6]{3} \sqrt{2}+\frac{1}{\sqrt[3]{3}}\right)^n, \frac{t_7 \text { from the } 1 \mathrm{st}}{t_7 \text { from the last }}=\frac{1}{6}, n

is equal to

View answer and explanation

Correct option: B

Option B (10) is correct because by equating the ratio of the 7th term from the beginning and the 7th term from the end to 1/6, we can solve for n and find that n=10.

427 Medium

427 of 468

Analyzing +4 | -1

The value of

{ }^n C_0 \cdot{ }^{2 n} C_r-{ }^n C_1,{ }^{2 n-2} C_r

+{ }^n C_2 \cdot{ }^{2 n-4} C_r+\ldots+(-1)^n{ }^n C_n{ }^{2 n-2 n} C_n

View answer and explanation

Correct option: A

\begin{cases}2^{2 n-r}{ }^n C_{r-n}, & \text { if } r>n \\ 0, & \text { if } r < n\end{cases}

Option A is correct because the expression simplifies to the coefficient of

x^{r-n}

(2x+x^2)^n

, which is

2^{2n-r} { }^n C_{r-n}

r>n

and 0 if

r < n

428 Medium

428 of 468

Application +4 | -1

The coefficient of

x^{53}

\sum\limits_{{r=0}}^{100}{ }^{100} C_r(x-3)^{100-r} 2^r

View answer and explanation

Correct option: C

-{ }^{100} C_{53}

Option C (-

{ }^{100} C_{53}

) is correct because simplifying the summation leads to (x-1)¹⁰⁰, and the coefficient of x⁵³ is -

{ }^{100} C_{53}

429 Medium

429 of 468

Application +4 | -1

The sum of the coefficients

x^{2 r}, r=1,2,3, \ldots

, in the expansion of

(1+x)^n

View answer and explanation

Correct option: B

2^{w-1}-1

Option B (

2^{w-1}-1

) is correct because it represents the sum of coefficients of even powers of x in the binomial expansion.

430 Medium

430 of 468

Application +4 | -1

The number of solutions of the equation

x_1+x_2+x_3+x_4=31, x_1, x_2, x_3, x_4 \in \mathrm{W}

View answer and explanation

Correct option: B

\left(\begin{array}{c}34 \\ 3\end{array}\right)

Option C, 120, is correct because it represents the number of non-negative integer solutions to the equation with the given constraints.

431 Medium

431 of 468

Analyzing +4 | -1

\sum\limits_{0 \le r} {\sum\limits_{ < s \le n} {{(^n}{C_r}{ + ^n}Cs)} }

View answer and explanation

Correct option: C

n 2^n

Option C (

n 2^n

) is correct because the summation can be simplified using binomial theorem properties and combinatorial arguments.

432 Medium

432 of 468

Application +4 | -1

The sum of the coefficients in the polynomial expansion of

\left(1+x-3 x^2\right)^{2163}

View answer and explanation

Correct option: B

-1

Option B (-1) is correct because substituting x=1 into the polynomial results in (1+1-3)²¹⁶³ = (-1)²¹⁶³ = -1.

433 Medium

433 of 468

Application +4 | -1

n

is odd, then

C_0^2 - C_1^2 + C_2^2 - C_3^2 + ... + {\left( { - 1} \right)^n}C_n^2

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because when n is odd, the given expression evaluates to zero due to the properties of binomial coefficients.

434 Medium

434 of 468

Application +4 | -1

The term independent of

x

in the expansion of

\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}, x > 0

, is

\alpha

times the corresponding binomial coefficient. Then,

\alpha

is___.

View answer and explanation

Correct option: A

Option A (1) is correct because after finding the value of r for the term independent of x, alpha is found to be 1.

435 Medium

435 of 468

Application +4 | -1

The greatest term (numerically) in the expansion of

(2+3 x)^9

, when

x=\frac{3}{2}

, is

View answer and explanation

Correct option: C

\frac{7 \times 3^{13}}{2}

Option C (

\frac{{7 \times 3^{{13}}}}{{2}}

) is correct because it is the greatest term calculated using the binomial theorem with the given x value.

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436 Medium

436 of 468

Analyzing +4 | -1

The value of

\sum\limits_{{r=0}}^{2 n} r \cdot\left({ }^{2 n} C_r\right) \cdot \frac{1}{(r+2)}

is equal to

View answer and explanation

Correct option: B

\frac{2^{2 n+1}\left(2 n^2-n+1\right)-2}{(2 n+1)(2 n+2)}

Option B (

\frac{{2^{{2n+1}}(2n^2-n+1)-2}}{{(2n+1)(2n+2)}}

) is correct because evaluating the summation leads to this expression.

437 Medium

437 of 468

Analyzing +4 | -1

The value of

\left(\begin{array}{l}30 \\ 0\end{array}\right)\left(\begin{array}{l}30 \\ 10\end{array}\right)-\left(\begin{array}{l}30 \\ 1\end{array}\right)\left(\begin{array}{l}30 \\ 11\end{array}\right)

+\left(\begin{array}{l}30 \\ 2\end{array}\right)\left(\begin{array}{l}30 \\ 12\end{array}\right)+\ldots+\left(\begin{array}{l}30 \\ 20\end{array}\right)\left(\begin{array}{l}30 \\ 30\end{array}\right)

is equal to

View answer and explanation

Correct option: B

{ }^{30} C_{10}

Option B (

{ }^{30} C_{10}

) is correct because the given expression is the coefficient of

x^{20}

in the product of

(1-x)^{30}

and

(x+1)^{30}

, which simplifies to

(1-x^2)^{30}

438 Medium

438 of 468

Understanding +4 | -1

Let

n

be a positive integer such that

\left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n}

, then

\sum\limits_{{r=0}}^{2 n} a_r

View answer and explanation

Correct option: A

3^n

Option A (

3^n

) is correct because substituting x=1 into the given polynomial expression directly yields the sum of the coefficients.

439 Medium

439 of 468

Analyzing +4 | -1

The coefficient of

x^{-5}

in the binomial expansion of

\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}

, where

x \neq 0,1

View answer and explanation

Correct option: A

Option A (1) is correct because after simplifying the expression and finding the general term, the coefficient of

x^{-5}

is found to be 1 when r = 10.

440 Medium

440 of 468

Application +4 | -1

The coefficient of

t^{4}

in the expansion of

\left(\frac{1-t^{6}}{1-t}\right)^{3}

View answer and explanation

Correct option: B

Option B (15) is correct because it is the coefficient of t⁴ in the expansion after simplifying the expression.

441 Medium

441 of 468

Analyzing +4 | -1

Let

n

be a positive integer such that

\left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n}

, then

(n-r) a_r+(2 n-r+1) a_{r-1}, 0 < r < 2 n

View answer and explanation

Correct option: A

(r+1) a_{r+1}

Option A (

(r+1) a_{r+1}

) is correct because it correctly relates the coefficients using differentiation and equating coefficients.

442 Medium

442 of 468

Application +4 | -1

For the natural numbers

m,\,\,n,\,\,\mathrm{if}\,\,{\left( {1 - y} \right)^m}{(1 + y)^n} = 1 + {a_1}y + {a_2}{y^2} + ...... + {a_{m + n}}{y^{m + n}}\mathrm{)}

and

{a_1} = {a_2} = 10

, then the value of

(m + n)

is equal to

View answer and explanation

Correct option: D

Option D (80) is correct because by expanding the given expression and equating coefficients, we can solve for m and n, and find m + n = 80.

443 Medium

443 of 468

Application +4 | -1

The coefficient of

x^4

in the expansion of

\left(1+x-2 x^2\right)^7

View answer and explanation

Correct option: B

-91

Option B (-91) is correct because the coefficient of x⁴ is calculated by expanding the expression using the binomial theorem and summing the relevant terms.

444 Medium

444 of 468

Application +4 | -1

The coefficient of

a^{10} b^7 c^3

in the expansion of

(b c+c a+a b)^{10}

View answer and explanation

Correct option: C

120

Option C, 120, is correct because it is the multinomial coefficient calculated from the given expansion with the correct powers of a, b, and c.

445 Medium

445 of 468

Application +4 | -1

Let

n

be an odd natural number and

A=\sum\limits_{{r=1}}^{\frac{n-1}{2}} \frac{1}{{ }^n C_r}

, then value of

\sum\limits_{{r=1}}^n \frac{r}{{ }^n C_r}

is equal to

View answer and explanation

Correct option: B

n(A+1)

Option B (

n(A+1)

) is correct because by using the property

{ }^n C_r = { }^n C_{n-r}

and summing the series, we arrive at the result

n(A+1)

446 Medium

446 of 468

Analyzing +4 | -1

{}^{20}{C_r}

is the coefficient of

{x^r}

in the expansion of

{\left( {1 + x} \right)^{20}}

, then the value of

\sum\limits_{r = 0}^{20} {{r^2}\,{}^{20}{C_r}}

is equal to

View answer and explanation

Correct option: D

420 \times {2^{18}}

The correct answer is D (

420 \times {2^{18}}

) because it is the result of applying combinatorial identities and the binomial theorem to simplify the summation.

447 Medium

447 of 468

Analyzing +4 | -1

{T_0},\;{T_1},{T_2}, \ldots ,{T_n}

represents the terms in the expansion of

{\left( {x + a} \right)^n}

, then

{\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

{\left( {{x^2} + {a^2}} \right)^n}

Option B (

{\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.

448 Medium

448 of 468

Application +4 | -1

Value of

S=\sum\limits_{{k=1}}^{\infty} \sum\limits_{{r=0}}^k \frac{1}{3^k}\left({ }^k C_r\right)

View answer and explanation

Correct option: A

Option A (2) is correct because the double summation simplifies to a geometric series with a sum of 2.

449 Medium

449 of 468

Application +4 | -1

If coefficient of

a^2 b^3 c^4

(a+b+c)^m

, (where

\left.m \in N\right)

L(L \neq 0)

then in same expansion coefficient of

a^4 b^4 c^1

will be

View answer and explanation

Correct option: D

\frac{\mathrm{L}}{2}

Option D (

\frac{\mathrm{L}}{2}

) is correct because the ratio of the coefficients is determined by the multinomial theorem and the given condition that

L \ne 0

450 Medium

450 of 468

Application +4 | -1

Assuming to be small that

{x^2}

and higher powers of

x

can be neglected, then

\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 - \frac{{305}}{{96}}x

Option B (

1 - \frac{{305}}{{96}}x

) is correct because using the binomial approximation and neglecting higher powers of x leads to this result.

Revision notes and key takeaways

Review BINOMIAL THEOREM in your textbook or class notes, then use this set to check your understanding.

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