Binomial Theorem Test

Result

Binomial Theorem Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

The number of rational terms in the expansion of $${ \left( 1+\sqrt { 2 } +\sqrt [ 3 ]{ 3 } \right) }^{ 6 }$$ is

A
$$6$$

B
$$7$$

C
$$5$$

D
$$8$$

Solution
Question 2
1 / -0

If $${ x }^{ m }$$ occurs in the expansion of $$(x+\frac { 1 }{ { x }^{ 2 } } )^{ 2n }$$, the coefficient of $${ x }^{ m }$$, is

A
$$\frac { (2n)! }{ m!(2n-m)! } $$

B
$$\frac { (2n)!3!3! }{ (2n-m)! } $$

C
$$\frac { (2n)! }{ (\frac { 2n-m }{ 3 } )!(\frac { 4n+m }{ 3 } )! } $$

D
none of these

Solution
Question 3
1 / -0

The coefficient of x$$^{24}$$ in the expansion of
(1 +3x + 6x$$^2$$ + 10x$$^3$$+ -----------+$$\infty$$)$$^{2/3}$$ =

A
300

B
250

C
25

D
205

Solution

$$\left(1+3 x+6 x^{2}+10 x^{3}+\ldots\infty\right)^{\frac{2}{3}} \\$$

$$\text { We know that, } \\$$

$${ (1+\alpha) }^{n}=1+n \alpha+\frac{n(n-1) \alpha^{2}}{2 !}+\ldots \\$$

$$\text { comparing with given eqn, }$$

$$\frac{n(n-1)}{2} \cdot \alpha^{2}=6 x^{2} \\$$

$$=n(n-1) \frac{9 x^{2}}{n^{2}}=6 x^{2} \\$$

$$\Rightarrow n=-3$$

$$n \alpha=3 x$$

$$\Rightarrow\alpha=\frac{3 x}{n}$$

$$\alpha=-x$$

$$\therefore \text {Expansion is} (1-x)^{-\not 3 \times \frac{2}{\not3}}$$

$$\quad =(1-x)^{-2}$$

$$=1+2 x+\frac{2 \times 3}{2 !} x^{2}+\frac{2 \times 3 \times 4}{3 !} x^{3}+\ldots$$

$$\therefore$$ coeff. of $$x^{24}$$ is 2

$$= \frac{2 \times 3 \times 4 \times\ldots25}{24 !}$$

$$=25$$
Question 4
1 / -0

The number of terms in the expansion of $${ \left[ { a }^{ 3 }+\dfrac { 1 }{ { a }^{ 3 } } +1 \right] }^{ 100 }$$ is

A
$$201$$

B
$$300$$

C
$$200$$

D
$$^{ 100 }{ C }_{ 3 }$$

Solution

$$\left(a^{3}+\frac{1}{a^{3}}+1\right)^{100} \\$$

$$\frac{1}{a^{300}}\left(1+a^{3}+a^{6}\right)^{100}$$

Here, terms will be in the form,

$$a^{0} ; a^{3}, a^{6}\cdots$$

and highest power term will be $$\left(a^{6}\right)^{100}$$

This forms a A.P of powers,

$$\therefore \quad 600=0+(n-1) 3$$

$$\therefore \quad n=201 \\$$

$$\text { Option (a) }$$
Question 5
1 / -0

The middle term in the expansion of $$\left(1-\dfrac{1}{x}\right)^{n}\left(1-x\right)^{n}$$ is

A
$$^{2n}C_{n}$$

B
$$-^{2n}C_{n}$$

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

D
$$none\ of\ these$$

Solution
Question 6
1 / -0

The coefficient of $$a^{3}b^{4}c$$ in the expansion of $$(1+a+b-c)^{9}$$ is

A
$$2\;^{9}C_{7}\cdot ^{7}C_{4}$$

B
$$-2\;^{9}C_{7}\cdot ^{7}C_{3}$$

C
$$^{9}C_{7}\cdot ^{7}C_{3}$$

D
$$-^{9}C_{7}\cdot ^{7}C_{3}$$

Solution
Question 7
1 / -0

If the middle term in the expansion of $$( 1 + x ) ^ { 2 n }$$ is the greatest term, then $$x$$ lies in the interval ___________________.

A
$$\left( \frac { n } { n + 1 } , \frac { n + 1 } { n } \right)$$

B
$$\left( \frac { n + 1 } { n } , \frac { n } { n + 1 } \right)$$

C
$$( n - 2 , n )$$

D
$$( n - 1 , n )$$
Question 8
1 / -0

In the expansion of $$(\dfrac{1+x}{1-x})^2$$ , the coefficient of $$x^n$$ will be

A
$$4n$$

B
$$4n - 3$$

C
$$4n - 1$$

D
none of these

Solution
Question 9
1 / -0

Coefficient of $${ t }^{ 12 }$$ in $$({ 1+t }^{ 2 }{ ) }^{ 6 }{ (1+t }^{ 6 }){ (1+t }^{ 12 })$$ is:

A
24

B
21

C
22

D
23

Solution
Question 10
1 / -0

The coefficient of $$x^{4}$$ in $$\dfrac{3x^{2}+2x}{(x^{2}+2)(x-3)}$$ is

A
$$\dfrac{11}{27}$$

B
$$\dfrac{77}{324}$$

C
$$-\dfrac{77}{324}$$

D
$$\dfrac{11}{27}$$

Solution