Integral part of $(2+\sqrt{5})^{2 n+1}$ is $(n \in N)$

an even or an odd number depending upon the value of $n$

$\left( {\begin{array}{*{20}{c}} n \\ 0 \end{array}} \right)\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n \\ 1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} n \\ {r + 1} \end{array}} \right) + ... + \left( {\begin{array}{*{20}{c}} n \\ {n - r} \end{array}} \right)\left( {\begin{array}{*{20}{c}} n \\ n \end{array}} \right) = $

$\left( {\begin{array}{*{20}{c}} {2n} \\ {n - r} \end{array}} \right)$

$\left( {\begin{array}{*{20}{c}} {2n} \\ {r - 1} \end{array}} \right)$

$\left( {\begin{array}{*{20}{c}} {2n} \\ r \end{array}} \right)$

$\left( {\begin{array}{*{20}{c}} {2n} \\ {n - r + 1} \end{array}} \right)$

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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Medium

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121 Medium

121 of 468

Analyzing +4 | -1

The value of

\frac{{{{18}^3} + {7^3} + 3.18.7.25}}{{{3^6} + 6.243.2 + 15.181.4 + 20.27.8 + 15.9.16 + 6.3.32 + 64}},

View answer and explanation

Correct option: B

The correct answer is B (1) because both the numerator and denominator can be simplified using the binomial theorem, resulting in the same expression.

122 Medium

122 of 468

Analyzing +4 | -1

Larger of

99^{50}+100^{50}

and

101^{50}

View answer and explanation

Correct option: A

101^{50}

Option A (

101^{50}

) is correct because

101^{50}

is larger than

99^{50} + 100^{50}

123 Medium

123 of 468

Application +4 | -1

If the 4 th term in the expansion of

\left(p x+x^{-1}\right)^m

is 2.5 for all

x \in R

, then

View answer and explanation

Correct option: B

p=\frac{1}{2}, m=6

Option B (

p=\frac{1}{2}, m=6

) is correct because by equating the power of x to zero and comparing the constant term, we can find the values of p and m.

124 Medium

124 of 468

Application +4 | -1

The coefficient of

x^5

in the expansion of

\left(1+x+x^2+x^3+x^4\right)^3

View answer and explanation

Correct option: B

Option B,

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

, is correct because it represents the number of non-negative integer solutions to the given equation.

125 Medium

125 of 468

Application +4 | -1

Let

\left[ x \right]

denote the greatest integer less than or equal to

x

. If

x = {\left( {\sqrt 3 + 1} \right)^5}

, then

\left[ x \right]

is equal to

View answer and explanation

Correct option: D

152

Option D (152) is correct because it is the greatest integer less than or equal to the given expression after expansion and approximation.

126 Medium

126 of 468

Application +4 | -1

(1+\mathrm{x})^n=\mathrm{C}_0+\mathrm{C}_1 \mathrm{x}+\mathrm{C}_2 \mathrm{x}^2+\ldots+\mathrm{C}_n \mathrm{x}^{\mathrm{n}}

, then

\mathrm{C}_0 \mathrm{C}_1+\mathrm{C}_1 \mathrm{C}_2+\mathrm{C}_2 \mathrm{C}_3+\ldots .+\mathrm{C}_{n-1} \mathrm{C}_n

is equal to-

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the given expression can be interpreted as the coefficient of

x^{n-1}

in the expansion of

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

, which simplifies to

{}^{2n}C_{n-1}

127 Medium

127 of 468

Application +4 | -1

If it is known that the third term of the binomial expansion

\left(x+x^{\log _{10} x}\right)^5

10^6

then

x

is equal to

View answer and explanation

Correct option: A

Option A (10) is correct because solving the equation formed by the third term of the binomial expansion leads to x=10.

128 Medium

128 of 468

Application +4 | -1

If the third term in the expansion of

\left(\frac{1}{x}+x^{\log _{10} x}\right)^5

x > 1

, is 1000, then value of

x

View answer and explanation

Correct option: C

100

Option C (100) is correct because solving the equation derived from the third term of the binomial expansion yields x=100.

129 Medium

129 of 468

Application +4 | -1

If the coefficient of the

x^7

and

x^8

in the expansion of

\left(2+\frac{x}{3}\right)^n

are equal then the value of

{n}

is equal to

View answer and explanation

Correct option: C

Option C (55) is correct because setting the coefficients of

x^7

and

x^8

equal in the binomial expansion of

\left(2+\frac{{x}}{{3}}\right)^n

and solving for n yields 55.

130 Medium

130 of 468

Application +4 | -1

Let

n \in N

. If

(1+x)^n=a_0+a_1 x+a_2 x^2+\ldots+a_n x^n

and

a_{n-3}, a_{n-2}, a_{n-1}

are in

\mathrm{AP}

, then

View answer and explanation

Correct option: A

a_1, a_2, a_3

are in

\mathrm{AP}

Option A (

a_1, a_2, a_3

are in AP) is correct because binomial coefficients equidistant from the beginning and the end are equal, implying

a_1, a_2, a_3

are in AP.

131 Medium

131 of 468

Analyzing +4 | -1

\sum\limits_{{r=0}}^n(-1)^r \frac{{ }^n C_r}{{ }^{r+3} C_r}

is equal to

View answer and explanation

Correct option: B

\frac{3 !}{2(n+3)}

Option C (

\frac{{3 !}}{{(n+3)}}

) is the correct answer. After simplifying the summation, the expression reduces to this form.

132 Medium

132 of 468

Application +4 | -1

(1+a x)^n=1+8 x+24 x^2+\ldots \ldots

; then

\frac{a-n}{a+n}

is equal to

View answer and explanation

Correct option: C

-1 / 3

Option C (-1/3) is correct because by comparing coefficients in the binomial expansion, we can solve for a and n and then calculate (a-n)/(a+n).

133 Medium

133 of 468

Application +4 | -1

The value of

\sqrt{e}

will be

View answer and explanation

Correct option: A

1.648

Option A (1.648) is correct because it is the approximate value of

\sqrt{e}

calculated using the Taylor series expansion of

e^x

134 Medium

134 of 468

Application +4 | -1

Let

{S}_{{n}}={ }^{{n}} {C}_0+{ }^{{n}} {C}_3+{ }^{{n}} {C}_6+{ }^{{n}} {C}_9+\ldots \ldots

, then the value of

{S}_{{n}}

, if

{n}

is not a multiple of 3, is

View answer and explanation

Correct option: B

\frac{2^n-(-1)^n}{3}

Option B is correct because the value of S_n, when n is not a multiple of 3, is (2^n - (-1)^n) / 3, which is derived using complex roots of unity.

135 Medium

135 of 468

Application +4 | -1

The value of

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

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the summation simplifies to an arithmetic series after simplifying the binomial coefficients.

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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

136 Medium

136 of 468

Analyzing +4 | -1

Integral part of

(2+\sqrt{5})^{2 n+1}

(n \in N)

View answer and explanation

Correct option: A

an even number

Option A (an even number) is correct because the integral part of the expression is always an even number.

137 Medium

137 of 468

Analyzing +4 | -1

y = \frac{1}{3} + \frac{{1.3}}{{3.6}} + \frac{{1.3.5}}{{3.6.9}} + \ldots ,

then the value of

{y^2} + 2\;y

View answer and explanation

Correct option: A

Option A (2) is correct because by comparing the given series with the binomial expansion and simplifying, we find that

{y^2} + 2y = 2

138 Medium

138 of 468

Application +4 | -1

2^{60}

when divided by 7 leave the remainder

View answer and explanation

Correct option: A

Option A (1) is correct because the remainder is found by rewriting the expression and applying the binomial theorem.

139 Medium

139 of 468

Analyzing +4 | -1

The sum of

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

View answer and explanation

Correct option: D

\frac{2}{n+2}

Option D (

\frac{{2}}{{(n+2)}}

) is correct because after simplifying the summation, the expression reduces to this form.

140 Medium

140 of 468

Application +4 | -1

If the coefficients of

x^2

and

x^3

in the expansion of

(3+k x)^9

are equal, then the value of

k

View answer and explanation

Correct option: B

9 / 7

Option B (9 / 7) is correct because equating the coefficients of x² and x³ leads to the equation

{ }^9 C_2 3^7 k^2 = { }^9 C_3 3^6 k^3

, which simplifies to k = 9/7.

141 Medium

141 of 468

Application +4 | -1

The value of

\left( {\frac{{{\;^{50}}{C_0}}}{1} + \frac{{{\;^{50}}{C_2}}}{3} + \frac{{{\;^{50}}{C_4}}}{5} + ... + \frac{{{\;^{50}}{C_{50}}}}{{51}}} \right)

View answer and explanation

Correct option: A

\frac{{{2^{50}}}}{{51}}

Option A is correct. The given sum can be simplified using binomial coefficient identities and properties of series.

142 Medium

142 of 468

Analyzing +4 | -1

\frac{1}{n+1}{ }^{n} C_{n}+\frac{1}{n}{ }^{n} C_{n-1}+\ldots+\frac{1}{2}{ }^{n} C_{1}+{ }^{n} C_{0}=\frac{1023}{10}

then

n

is equal to

View answer and explanation

Correct option: A

Option A (9) is correct because solving the given summation involving binomial coefficients leads to n = 9.

143 Medium

143 of 468

Application +4 | -1

For

n > 3, a, b \in \mathbf{R}

, let

S(n, a, b)

=\sum\limits_{{r=0}}^n(-1)^r \frac{n(n-1) \cdots(n-r+1)}{r !}(a-r)(b-r)

then

S\left(5, \frac{1}{2}, \frac{1}{7}\right)

View answer and explanation

Correct option: A

Option A (0) is correct because the given summation simplifies to zero due to the properties of binomial coefficients and alternating sums.

144 Medium

144 of 468

Analyzing +4 | -1

Suppose

[x]

denote the greatest integer

\leq x

, and

n \in {N}

, then

\lim \limits_{{n \rightarrow \infty}} \frac{\left[{ }^n C_0 x^2\right]+\left[{ }^n C_1 x^2\right]+\cdots+\left[{ }^n C_n x^2\right]}{2^n-2}

is equal to

View answer and explanation

Correct option: B

x^2

Option B (

x^2

) is correct because using the sandwich theorem, the limit of the given expression as n approaches infinity is

x^2

145 Medium

145 of 468

Application +4 | -1

The coefficient of

x^5

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

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the given expression simplifies to (1-x)³, and the coefficient of x⁵ in this polynomial is 0.

146 Medium

146 of 468

Application +4 | -1

The value of

{ }^{n-1} C_0 \cdot{ }^n C_2+{ }^{n-1} C_1 \cdot{ }^n C_3+{ }^{n-1} C_2 \cdot{ }^n C_4 +\ldots+{ }^{n-1} C_{n-2} \cdot{ }^n C_n \text { is }

View answer and explanation

Correct option: C

{ }^{2 n-1} C_{n-2}

Option C (

{ }^{2 n-1} C_{n-2}

) is correct because it represents the coefficient of

x^{n-2}

in the expansion of

(1+x)^n(1+x)^{n-1}

147 Medium

147 of 468

Application +4 | -1

\left( {\begin{array}{cccccccccccccccccccc} n \\ 0 \end{array}} \right)\left( {\begin{array}{cccccccccccccccccccc} n \\ r \end{array}} \right) + \left( {\begin{array}{cccccccccccccccccccc} n \\ 1 \end{array}} \right)\left( {\begin{array}{cccccccccccccccccccc} n \\ {r + 1} \end{array}} \right) + ... + \left( {\begin{array}{cccccccccccccccccccc} n \\ {n - r} \end{array}} \right)\left( {\begin{array}{cccccccccccccccccccc} n \\ n \end{array}} \right) =

View answer and explanation

Correct option: C

\left( {\begin{array}{cccccccccccccccccccc} {2n} \\ {n - r} \end{array}} \right)

Option C

\left( {\begin{array}{cccccccccccccccccccc} {2n} \\ {n - r} \end{array}} \right)

is correct because it represents the coefficient of

x^{n-r}

in the expansion of

(1+x)^n(1+x)^n = (1+x)^{2n}

148 Medium

148 of 468

Application +4 | -1

The value of

\sum\limits_{{r=0}}^{22}{ }^{22} C_{r}{ }^{23} C_{r}

View answer and explanation

Correct option: D

{ }^{45} \mathrm{C}_{23}

Option D (

{ }^{45} \mathrm{C}_{23}

) is correct because the sum can be identified as the coefficient of

x^{23}

in the expansion of

(1+x)^{22}(1+x)^{23}

149 Medium

149 of 468

Analyzing +4 | -1

For

0 \leq r < 2 n,\left({ }^{2 n+r} C_n\right)\left({ }^{2 n-r} C_n\right)

cannot exceed

View answer and explanation

Correct option: B

{ }^{4 n} C_{2 n}

Option B (

{}^{4 n} C_{2 n}

) is correct because the product of binomial coefficients cannot exceed the largest binomial coefficient in the expansion of

(1+x)^{4n}

150 Medium

150 of 468

Analyzing +4 | -1

The value of

\sum\limits_{0 \le i} {\sum\limits_{ < j \le n} {i{.^n}{C_j}} }

is equal to

View answer and explanation

Correct option: C

n(n-1) \cdot 2^{n-3}

Option C (

n(n-1) \cdot 2^{n-3}

) is correct because it is derived by simplifying the double summation involving binomial coefficients.

Revision notes and key takeaways

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