Binomial Theorem Test

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Binomial Theorem Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Find the coefficient of $$\cfrac { 1 }{ { y }^{ 2 } } $$ in $${ \left( y+\cfrac { { c }}{ { y }^{ 2 } } \right) }^{ 10 }$$.

A
$$210{ c }^{ 4 }$$

B
$$210{ c }^{ 5 }$$

C
$$120{ c }^{ 3 }$$

D
$$120{ c }^{ 4 }$$

Solution

$$(y+\frac{c}{y^2})^{10}$$
$$T_{r+1}=\:^{10}C_{r}y^{10-3r}c^r$$
Hence for $$y^{-2}$$
$$10-3r=-2$$
$$12=3r$$
$$r=4$$
Coefficient will be
$$\:^{10}C_{4}c^4$$
$$=210c^4$$
Question 2
1 / -0

If $${ x }^{ 4r }$$ occurs in the expansion of $${ \left( x+\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ 4n }$$ , then its coefficient is

A
$$\cfrac { 4n! }{ \left( \cfrac { 4 }{ 3 } (n-r) \right) !\left( \cfrac { 4 }{ 3 } (2n+r) \right) ! } $$

B
$$\cfrac { 4n! }{ \left( \cfrac { 3}{ 4 } (n-r) \right) !\left( \cfrac { 3 }{ 4 } (2n+r) \right) ! } $$

C
$$\cfrac { 4n! }{ \left( \cfrac { 4 }{ 3 } (2n-r) \right) !\left( \cfrac { 4 }{ 3 } (2n+r) \right) ! } $$

D
None of these

Solution

The $$(k+1)^{th}$$ term of the given expression is
$$T_{k+1} = {^{4n}\textrm{C}_{k}} x^{(4n-k)}( \displaystyle \frac{1}{x^{2}})^{k}$$

We know the coefficient of $$x$$ is $$4r$$, so, $$4n - k- 2k = 4r$$
$$k = \displaystyle \frac{4(n-r)}{3}$$

The $$(k+1)^{th}$$ term will be,
$${ \displaystyle ^{4n}\textrm{C}_{\dfrac{4(n-r)}{3}}} (x)^ { \left(4n- \frac{4(n-r)}{3} \right)} ( \displaystyle \frac{1}{x^{2}})^{ \frac{4(n-r)}{3}}$$

The coeffecient will be $${ \displaystyle ^{4n}\textrm{C}_{\frac{4(n-r)}{3}}}$$
That is,
= $$  \displaystyle \frac{4n!}{(4n-\dfrac{4n}{3}+ \displaystyle \frac{4r}{3})!(\frac{4(n-r)}{3})!}$$

= $$  \displaystyle \frac{4n!}{( \displaystyle \frac{4(2n+r)}{3})!( \displaystyle \frac{4(n-r)}{3})!}$$
Question 3
1 / -0

Determine the value of $$x$$ in the expression of $${ ( 2+x) }^{ 5 }$$, if the second term in the expansion is $$240$$

A
$$(24)^\frac{1}{4}$$

B
$$6$$

C
$$3$$

D
None of the above

Solution

The general term of $$(a+x)^n$$ is $$^nC_r a^{(n-r)}x^r$$

Here, $$n=5$$
we will get second term if we put $$r=1$$

The second term will be $$^5C_1(2^{(5-1)})x$$
$$=5(2^{4})x$$

$$5(2^{4})x=240$$ ...... given

Therefore
$$5(16)x=240$$
$$80(x)=240$$
$$x=3$$
Question 4
1 / -0

Find the $$(n+1)^{th}$$ term from the end in the expansion of $${ \left( x-\cfrac { 1 }{ x } \right) }^{ 2n }$$

A
$${ \left( -1 \right) }^{ n }. { _{ }^{ 2n }{ C } }_{ n }$$

B
$${ \left( -1 \right) }^{ n+1 }. { _{ }^{ 2n }{ C } }_{ n-1 }$$

C
$${ \left( -1 \right) }^{ n }. { _{ }^{ 2n }{ C } }_{ n -1}$$

D
None of these

Solution

There are total $$2n+1$$ terms in the given expansion.
Hence $$(n+1)^{th}$$ term, from the end is the $$(n+1)^{th}$$ term form beginning the beginning also since, it is the middle term.
$$T_{n+1}$$$$=\:^{2n}C_{n}x^{n}(-\dfrac{1}{x^{n}})^{n}$$

$$=(-1)^{n}\:^{2n}C_{n}$$
Question 5
1 / -0

Find the middle term in the expansion of $${ \left( \cfrac { 2x }{ 3 } +\cfrac { 3 }{ 2x } \right) }^{ 10 }$$.

A
$$210$$

B
$$630$$

C
$$252$$

D
$$756$$

Solution

The middle term will be the 6th term. It will also be the only term independent of $$x$$.
Hence the coefficient will be
$$T_{5+1}=\:^{10}C_{5}$$
$$=\displaystyle\frac{10!}{5!(5!)}$$
$$=252$$
Question 6
1 / -0

If $${ \left( 1+2x+x^{ 2 } \right) }^{ n }=\sum _{ r=0 }^{ 2n }{ { a }_{ r }{ x }^{ r } } $$, then $${ a }_{ r }=$$

A
$${ \left( _{ }^{ n }{ { C }_{ r }^{ } } \right) }^{ 2 }$$

B
$$^{ n }{ { C }_{ r }^{ } }.^{ n }{ { C }_{ r+1 }^{ } }$$

C
$$^{ 2n }{ { C }_{ r }^{ } }$$

D
$$^{ 2n }{ { C }_{ r+1 }^{ } }$$

Solution

We have $$\displaystyle { \left( 1+2x+{ x }^{ 2 } \right) }^{ n }=\sum _{ r=0 }^{ 2n }{ { a }_{ r }{ x }^{ r } } $$
$$\displaystyle \Rightarrow { \left( 1+x \right) }^{ 2n }=\sum _{ r=0 }^{ 2n }{ { a }_{ r }{ x }^{ r } } $$
$$\displaystyle \Rightarrow \sum _{ r=0 }^{ 2n }{ ^{ 2n }{ { C }_{ r } }{ x }^{ r } } =\sum _{ r=0 }^{ 2n }{ { a }_{ r }{ x }^{ r } } \Rightarrow { a }_{ r }=^{ 2n }{ { C }_{ r } }$$
Question 7
1 / -0

The middle term in the expansion of $${ \left( 1+x \right) }^{ 2n }$$ is , $$n$$ being a positive integer is

A
$$\cfrac { \left\{ 1.3.5....(2n) \right\} { 2 }^{ n } }{ n! } { x }^{ n }\\ \quad \quad \quad \quad \quad \quad \quad \quad $$

B
$$\cfrac { \left\{1.3.5....(2n) \right\} { 2 }^{ n }n! }{ n! } { x }^{ n }\\ \quad \quad \quad \quad \quad \quad \quad \quad $$

C
$$\cfrac { \left\{ 1.3.5....(2n-1) \right\} { 2 }^{ n }n! }{ n! } { x }^{ n }\\ \quad \quad \quad \quad \quad \quad \quad \quad $$

D
$$\cfrac { \left\{ 1.3.5....(2n-1) \right\} { 2 }^{ n } }{ n! } { x^n }\\ \quad \quad \quad \quad \quad \quad \quad \quad $$

Solution

The number of terms in the expansion of $${ \left( 1+x \right) }^{ 2n } $$ is $$(2n+1)$$ (odd), its middle term is $$\dfrac{2n+1+1}2=(n+1)^{th}$$ term.$$\therefore$$ Required term $$={ T }_{ n+1 }={ _{ }^{ 2n }{ C } }_{ n }{ x }^{ n }\\ =\cfrac { 2n! }{ n!n! } { x }^{ n }=\cfrac { \left\{ 1.2.3.4.5.6...(2n-1)2n \right\} }{ n!n! } { x }^{ n }\\ =\cfrac { \left\{ 1.3.5....(2n-1) \right\} \left\{ 2.4.6.....2 \right\} n }{ n!n! } { x }^{ n }\\ =\cfrac { \left\{ 1.3.5....(2n-1) \right\} { 2 }^{ n }(1.2.3....n) }{ n!n! } { x }^{ n }\\ =\cfrac { \left\{ 1.3.5....(2n-1) \right\} { 2 }^{ n } }{ n! } { x }^{ n }\\ $$
Question 8
1 / -0

If $$n>2$$, then find the value of $${ C }_{ 1 }{ \left( a-1 \right) }^{ 2 }-{ C }_{ 2 }{ \left( a-2 \right) }^{ 2 }+{ C }_{ 3 }{ \left( a-3 \right) }^{ 2 }-.....+{ \left( -1 \right) }^{ n-1 }{ C }_{ n }{ \left( a-n \right) }^{ 2 }$$ where $${ C }_{ r }$$ stands for $$\quad { _{ }^{ n }{ C } }_{ r }$$

A
$${ a }^{ 3}$$

B
$${ a }$$

C
$$\dfrac{ a }{2}$$

D
$${ a }^{ 2 }$$

Solution

Let $$n=3$$
Hence the above expression reduces to
$$\:^{3}C_{1}(a-1)^{2}-\:^{3}C_{2}(a-2)^{2}+\:^{3}C_{3}(a-3)^{3}$$
$$=3(a-1)^{2}-3(a-2)^{2}+(a-3)^{3}$$
$$=3a^2-6a+3-(3a^2-12a+12)+(a^2-6a+9)$$
$$=a^2$$
Hence answer is Option D
Question 9
1 / -0

find the 7th term in the expansion of $${ \left( 4x-\frac { 1 }{ 2\sqrt { x } } \right) }^{ 13 }$$

A
$$439296{ x }^{ 7 }$$

B
$$439296{ x }^{ 4 }$$

C
$$439396{ x }^{ 7 }$$

D
$$43396{ x }^{ 4 }$$

Solution

The 7th term of the given binomial expansion will be,
$$=$$ $$ \displaystyle _{6}^{13}\textrm{C} \; (4x)^{7} \;(-\frac{1}{2x^{\frac{1}{2}}})^{6}$$

$$=$$ $$ \displaystyle\frac{13! \times 2^{14} \times x^{7}}{7! . 6! \times 2^{6}.x^{3}}$$

$$=$$ $$ \displaystyle\frac{13! \times 2^{14}}{7! \times 6! \times 2^{6}}x^{4}$$

$$=439296x^{4}$$
Question 10
1 / -0

The $$8^{th}$$ term of $$\displaystyle { \left( 3x+\frac { 2 }{ 3{ x }^{ 2 } } \right) }^{ 12 }$$, when expanded ina scending power of $$x$$, is

A
$$\displaystyle \frac { 228096 }{ { x }^{ 3 } } $$

B
$$\displaystyle \frac { 228096 }{ { x }^{ 9 } } $$

C
$$\displaystyle \frac { 328179 }{ { x }^{ 3 } } $$

D
None of these

Solution

When $$\displaystyle { \left( 3x+\frac { 2 }{ 3{ x }^{ 2 } } \right) }^{ 12 }$$ is expanded, the power of $$x$$ goes on decreasing as the terms proceed.
hence, it is expanded in descending power of $$x$$.
So $$\displaystyle { \left( \frac { 2 }{ 3{ x }^{ 2 } } +3x \right) }^{ 12 }$$, when expanded will be ascending power of $$x$$.
Now $${ t }_{ 8 }$$ in $$\displaystyle { \left( \frac { 2 }{ 3{ x }^{ 2 } } +3x \right) }^{ 12 }$$ $$\displaystyle =_{ }^{ 12 }{ { C }_{ 7 }^{ } }{ \left( \frac { 2 }{ 3{ x }^{ 2 } } \right) }^{ 12-7 }.{ \left( 3x \right) }^{ 7 }$$
$$\displaystyle =\frac { 12! }{ 7!5! } .{ \left( \frac { 2 }{ 3{ x }^{ 2 } } \right) }^{ 5 }.{ \left( 3x \right) }^{ 7 }=\frac { 12\times 11\times 10\times 9\times 8 }{ 5\times 4\times 3\times 2 } .\frac { { 2 }^{ 5 }.{ 3 }^{ 2 } }{ { x }^{ 3 } } $$ $$\displaystyle =\frac { 228096 }{ { x }^{ 3 } } $$

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Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 21

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Binomial Theorem Test - 21

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Question 1

$$210{ c }^{ 4 }$$

$$210{ c }^{ 5 }$$

$$120{ c }^{ 3 }$$

$$120{ c }^{ 4 }$$

Solution

Question 2

$$\cfrac { 4n! }{ \left( \cfrac { 4 }{ 3 } (n-r) \right) !\left( \cfrac { 4 }{ 3 } (2n+r) \right) ! } $$

$$\cfrac { 4n! }{ \left( \cfrac { 3}{ 4 } (n-r) \right) !\left( \cfrac { 3 }{ 4 } (2n+r) \right) ! } $$

$$\cfrac { 4n! }{ \left( \cfrac { 4 }{ 3 } (2n-r) \right) !\left( \cfrac { 4 }{ 3 } (2n+r) \right) ! } $$

None of these

Solution

Question 3

$$(24)^\frac{1}{4}$$

$$6$$

$$3$$

None of the above

Solution

Question 4

$${ \left( -1 \right) }^{ n }. { _{ }^{ 2n }{ C } }_{ n }$$

$${ \left( -1 \right) }^{ n+1 }. { _{ }^{ 2n }{ C } }_{ n-1 }$$

$${ \left( -1 \right) }^{ n }. { _{ }^{ 2n }{ C } }_{ n -1}$$

None of these

Solution

Question 5

$$210$$

$$630$$

$$252$$

$$756$$

Solution

Question 6

$${ \left( _{ }^{ n }{ { C }_{ r }^{ } } \right) }^{ 2 }$$

$$^{ n }{ { C }_{ r }^{ } }.^{ n }{ { C }_{ r+1 }^{ } }$$

$$^{ 2n }{ { C }_{ r }^{ } }$$

$$^{ 2n }{ { C }_{ r+1 }^{ } }$$

Solution

Question 7

$$\cfrac { \left\{ 1.3.5....(2n) \right\} { 2 }^{ n } }{ n! } { x }^{ n }\\ \quad \quad \quad \quad \quad \quad \quad \quad $$

$$\cfrac { \left\{1.3.5....(2n) \right\} { 2 }^{ n }n! }{ n! } { x }^{ n }\\ \quad \quad \quad \quad \quad \quad \quad \quad $$

$$\cfrac { \left\{ 1.3.5....(2n-1) \right\} { 2 }^{ n }n! }{ n! } { x }^{ n }\\ \quad \quad \quad \quad \quad \quad \quad \quad $$

$$\cfrac { \left\{ 1.3.5....(2n-1) \right\} { 2 }^{ n } }{ n! } { x^n }\\ \quad \quad \quad \quad \quad \quad \quad \quad $$

Solution

Question 8

$${ a }^{ 3}$$

$${ a }$$

$$\dfrac{ a }{2}$$

$${ a }^{ 2 }$$

Solution

Question 9

$$439296{ x }^{ 7 }$$

$$439296{ x }^{ 4 }$$

$$439396{ x }^{ 7 }$$

$$43396{ x }^{ 4 }$$

Solution

Question 10

$$\displaystyle \frac { 228096 }{ { x }^{ 3 } } $$

$$\displaystyle \frac { 228096 }{ { x }^{ 9 } } $$

$$\displaystyle \frac { 328179 }{ { x }^{ 3 } } $$

None of these

Solution

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