Binomial Theorem Test

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Binomial Theorem Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

If the sum of binomial coefficient in the expansion $$(1+x)^{n}$$ is $$256$$, then $$n$$ is

A
$$6$$

B
$$7$$

C
$$8$$

D
$$9$$

Solution

$$(1+x)^n=C_0+C_1x+\cdots+C_nx^n$$
Sum of coefficients is $$C_0+C_1+\cdots+C_n=2^n\\C_0+C_1+\cdots+C_n=256\\2^n=256\\2^n=2^8\\n=8$$
Question 2
1 / -0

The coefficient of $${ x }^{ 3 }$$ in the polynomial $$(x-1) (x-2) (x-4)$$ is

A
1

B
-1

C
7

D
None of these

Solution
Question 3
1 / -0

If the constants term in the expansions of $$(\sqrt{x}-\dfrac{k}{x^2})^{10}$$ is $$405$$, then what can be the value of $$k$$ ?

A
$$\pm 2$$

B
$$\pm 3$$

C
$$\pm 5$$

D
$$\pm 9$$

Solution

Given,

$$\left ( \sqrt{x}-\dfrac{k}{x^2} \right )^{10}$$

$$T_{r+1}={}^{10}C_r(\sqrt{x})^{10-r}\left ( -\dfrac{k}{x^2} \right )^r$$

$$=^{10}C_r(x)^{\frac{1}{2}(10-r)}(-k)^rx^{-2r}$$

$$=^{10}C_r(x)^{\frac {10-5r}{2}}(-k)^r$$

For a constant term, the condition is

$$=x^0$$

$$\Rightarrow \dfrac {10-5r}{2}=0$$

$$\Rightarrow r=2$$

$$T_{2+1}=^{10}C_2(-k)^2$$

Given,

$$405=^{10}C_2(-k)^2$$

$$\dfrac{10 \times 9 \times 8!}{2! \times 8!}(-k)^2=405$$

$$45k^2=405$$

$$k^2=9\Rightarrow k= \pm 3$$
Question 4
1 / -0

The middle term in $${\left( {{x^2} + \dfrac{1}{{{x^2}}} + 2} \right)^n}$$ is

A
$$\dfrac{{n!}}{{{{\left( {\left( {\frac{n}{2}} \right)!} \right)}^2}}}$$

B
$$\dfrac{{\left( {2n} \right)!}}{{{{\left( {\left( {\frac{n}{2}} \right)!} \right)}^2}}}$$

C
$$\dfrac{{1.3.5....\left( {2n + 1} \right)}}{{n!}}{2^n}$$

D
$$\dfrac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}}$$

Solution

$$(x^2+\cfrac{1}{x^2}+2)^n=(x+\cfrac{1}{x})^{2n}$$
Middle term $$=(n+1)th \quad term$$
$$T_{n+1}= { { ^{ 2n }{ C } } }_{ n }(x)^n(\cfrac{1}{x})^n\\ \quad=\cfrac{2n!}{(n!)^2}$$
Question 5
1 / -0

The $$5$$th term in the expansion of $$ \left(2x^2 + \dfrac{3}{x}\right)^5 $$ is

A
$$810.x^{-2}$$

B
$$810.x^{-4}$$

C
$$810$$

D
$$810.x^{-5}$$

Solution

we k.n.t $$(r+1)^{th}$$ term in expansion of $$(a+b)^n$$ is $$^nc_ra^{n-r}b^r$$
$$\Rightarrow T_{r+1}=^nc_ra^{n-r}b^r$$

Here $$a=2x^2$$ and $$b=\dfrac{2}{x}$$ and $$n=5$$

5th term = $$T_5=^5c_4a^{5-4}b^4=5ab^4=5(2x^2)(\dfrac{3}{x})^4=(5*2*3^4)(x^{2-4}=810x^{-2}$$

$$\therefore $$ The $$5$$th term in expansion of $$(2x^2+\dfrac{3}{x})^4$$ is $$810x^{-2}$$.
Question 6
1 / -0

If $$n$$ is a positive integer, then the number of terms in the expansion of $$ (x+a)^n $$ is

A
$$n$$

B
$$n+1$$

C
$$2n$$

D
Infinitely many

Solution

In binomial expansion the terms goes from $$n{ C }_{ 0 }{ x }^{ n }$$ to $$n{ C }_{ n }{ a }^{ n }$$ i.e the base of $$C$$ goes from $$0$$ to $$n$$ and this shows that there must be $$(n+1)$$ terms.
Question 7
1 / -0

The $$ (n+1)^{th} $$ term from the end in $$ \left(x-\dfrac{1}{x}\right)^{3n} $$ is

A
$$ ^{3n}C_n.x^{-n} $$

B
$$ (-1)^n.^{3n}C_n.x^{-n} $$

C
$$ ^{3n}C_n.x^n $$

D
$$ (-1)^n.^{3n}C_n.x^n $$

Solution

$$r^{th}$$ term from end means $${ (n+1 -r+1) }^{ th }$$ term from start.

$$r^{th}$$ term from begning in expansion of $${ (x-a) }^{ n }$$ is given by

$${ T }_{ r }\quad =\quad ({ -1 })^{ r-1 }\binom{n}{ r-1 }{ x }^{ n-r+1 }{ a }^{ r-1 }$$

Here, no. of terms = $$3n+1$$
$$r^{th}$$ term from the end = $$(3n+1) - (n+1) + 1 = {2n+1}^{th}$$ term from the beginning

$$\therefore T_{2n+1} = \quad ^{3n}{C}_{2n}{x}^{n}\dfrac{1}{x^{2n}} = \quad^{3n}{ C }_{ n }{ x }^{ -n }$$
Question 8
1 / -0

In any binomial expansion, the number of terms are

A
$$ \geq 5 $$

B
$$ \geq 2 $$

C
$$ \geq 3 $$

D
$$ \geq 4 $$

Solution

Bi-nomial, involves summation of two terms.
Let the terms be $$x$$ and $$y$$.
Therefore a binomial expansion can be of the form, $$(x+y)^{n}$$. where $$n\geq 1$$
If $$n=1$$, we get only two terms.
If $$n>1$$ where $$n$$ is an integer, then it gives us in total $$(n+1)$$ terms.
Thus, number of terms has to be $$\geq 2$$.
Question 9
1 / -0

The $$4^{th}$$ term in the expansion of $$ \left(\sqrt{x}+\dfrac{1}{x}\right)^{12} $$ is

A
$$110x^{\frac{3}{2}} $$

B
$$ 220x^{\frac{3}{2}} $$

C
$$ 220x^2 $$

D
$$ 110x^2 $$

Solution

Expansion is $${ \left( \sqrt { x } +\dfrac { 1 }{ x } \right) }^{ 12 }\\$$
$$T_{ r+1 }={ { 12 }_{ C } }_{ r }{ \left( \dfrac { 1 }{ x } \right) }^{ r }.{ \left( \sqrt { x } \right) }^{ 12-r }={ { 12 }_{ C } }_{ r }.{ x }^{ 6-1.5r }\\$$
$$4^{ th }$$ term is $$T_{ 4 }={ { 12 }_{ C } }_{ 3 }.{ x }^{ 6-1.5*3 }=220.{ x }^{ \frac { 3 }{ 2 } }\\$$
Correct answer is option B.
Question 10
1 / -0

If $$ T_r $$ denotes the $$ r^{th} $$ term in the expansion of $$\displaystyle \left(x + \frac{1}{x} \right)^{23} $$, then

A
$$ T_{12} = T_{13} $$

B
$$ x^2.T_{13} = T_{12} $$

C
$$ x^2.T_{12} = T_{13} $$

D
$$ T_{12}+T_{13} = 25 $$

Solution

We know that, general term expression for $$(a+b)^n$$
$$T_{r+1} =^{n}C_{r} \times a^{n} \times \left(b \right) ^{n-r} $$

$$T_{12} =^{23}C_{11} \times x^{12} \times \left(\dfrac{1}{x} \right) ^{11} $$
$$T_{12} = ^{23}C_{11} x$$
$$T_{13} =^{23}C_{12} \times x^{11} \times \left(\dfrac{1}{x} \right)^{12} $$
$$T_{13} = ^{23}C_{12} \left(\dfrac{1}{x} \right)$$

$$^{23}C_{12}=^{23}C_{11}$$ ....as $$(11+12=23)$$

$$T_{12} = ^{23}C_{11} x^2 \times \dfrac 1x$$
$$T_{12} =x^2 \times ^{23}C_{12} \times \dfrac 1x$$

$$T_{12} =x^2 \times T_{13}$$

Hence, option B is the correct answer.

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Binomial Theorem Test - 11

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Binomial Theorem Test - 11

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Question 1

$$6$$

$$7$$

$$8$$

$$9$$

Solution

Question 2

1

-1

7

None of these

Solution

Question 3

$$\pm 2$$

$$\pm 3$$

$$\pm 5$$

$$\pm 9$$

Solution

Question 4

$$\dfrac{{n!}}{{{{\left( {\left( {\frac{n}{2}} \right)!} \right)}^2}}}$$

$$\dfrac{{\left( {2n} \right)!}}{{{{\left( {\left( {\frac{n}{2}} \right)!} \right)}^2}}}$$

$$\dfrac{{1.3.5....\left( {2n + 1} \right)}}{{n!}}{2^n}$$

$$\dfrac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}}$$

Solution

Question 5

$$810.x^{-2}$$

$$810.x^{-4}$$

$$810$$

$$810.x^{-5}$$

Solution

Question 6

$$n$$

$$n+1$$

$$2n$$

Infinitely many

Solution

Question 7

$$ ^{3n}C_n.x^{-n} $$

$$ (-1)^n.^{3n}C_n.x^{-n} $$

$$ ^{3n}C_n.x^n $$

$$ (-1)^n.^{3n}C_n.x^n $$

Solution

Question 8

$$ \geq 5 $$

$$ \geq 2 $$

$$ \geq 3 $$

$$ \geq 4 $$

Solution

Question 9

$$110x^{\frac{3}{2}} $$

$$ 220x^{\frac{3}{2}} $$

$$ 220x^2 $$

$$ 110x^2 $$

Solution

Question 10

$$ T_{12} = T_{13} $$

$$ x^2.T_{13} = T_{12} $$

$$ x^2.T_{12} = T_{13} $$

$$ T_{12}+T_{13} = 25 $$

Solution

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