Binomial Theorem Test

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Binomial Theorem Test - 19

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Weekly Quiz Competition

Question 1
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Find the sum $$\displaystyle\sum _{ r=1 }^{ n }{ \cfrac { { r }^{ n }{ C }_{ r } }{ { C }_{ r-1 } } } $$

A
$$\displaystyle\cfrac { n(n+1) }{ 3! } $$

B
$$\displaystyle\cfrac { n(n-1) }{ 2 } $$

C
$$\displaystyle\cfrac { n(n+1) }{ 2 } $$

D
None of these

Solution

$$\displaystyle \sum _{ r=1 }^{ n }{ \frac { { r }^{ n }{ C }_{ r } }{ ^{ n }{ C }_{ r-1 } } } =\sum _{ r=1 }^{ n }{ r\frac { n! }{ r!\left( n-r \right) ! } \times \frac { \left( n-\left( r-1 \right) \right) !\left( r-1 \right) ! }{ n! } } $$
$$\displaystyle =\sum _{ r=1 }^{ n }{ \left( n-r+1 \right) } =\left( { n }^{ 2 }-\frac { n\left( n-1 \right) }{ 2 } +n \right) =\frac { n\left( n+1 \right) }{ 2 } $$
Question 2
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The middle term in the expansion of $$\displaystyle \left ( 1-\frac{1}{x} \right )^{n}\left ( 1-x \right )^{n},$$ is

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

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

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

D
none of these

Solution

$$(1-\displaystyle \frac{1}{x})^{n}(1-x)^{n}$$
$$(\displaystyle \frac{x-1}{x})^{n}(1-x)^{n}$$
Case I - $$n$$ is even.
Hence the above expression reduces to
$$\displaystyle \frac{1}{x^n}(1-x)^{2n}$$
The middle term will be $$(n+1)^{th}$$ term.
Hence the coefficient will be.
$$T_{n+1}=\:^{2n}C_{n}$$
Case II - $$n$$ is odd.
Then the above expression reduces to.
$$-\displaystyle \frac{1}{x^{n}}(1-x)^{2n}$$
Middle term will be $$(n+1)^{th}$$ term.
Hence coefficient will be
$$T_{n+1}=-((-1)^{n}\:^{2n}C_{n})=\:^{2n}C_{n}$$
Question 3
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If A is the coefficient of the middle term in the expansion of $$\displaystyle (1+x)^{2n}$$ and B and C are the coefficients of two middle terms in the expansion of $$\displaystyle(1+x)^{2n-1}$$, then

A
$$\displaystyle A+B=C$$

B
$$\displaystyle B+C=A$$

C
$$\displaystyle C+A=B$$

D
$$\displaystyle A+B+C=0$$

Solution

For $$(1+x)^{2n}$$:
The middle term will be $$(\frac{N}{2}+1)$$ th term.
$$=\frac{2n}{2}+1$$ th term
$$=(n+1)^{th}$$ term
Hence coefficient of the middle term will be
$$\:^{2n}C_{n}$$
$$=A$$.
For $$(1+x)^{2n-1}$$,
The middle terms will be $$(\frac{N+1}{2}+1)$$ and $$(\frac{N+1}{2})$$ terms
$$=(\frac{2n}{2}+1)^{th}$$ term and $$\frac{2n}{2}^{th}$$ term.
$$=(n+1)^{th}$$ term and $$(n)^{th}$$ term.
Hence coefficient of the middle terms will be
$$\:^{2n-1}C_{n}=B$$ and $$\:^{2n-1}C_{n-1}=C$$
By the properties of binomial coefficients.
$$\:^{2n-1}C_{n-1}+\:^{2n-1}C_{n}$$
$$=\:^{2n}C_{n}$$
Hence $$B+C=A$$
Question 4
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The middle term in the expansion of$$\displaystyle (1+x)^{2n}$$is

A
$$\displaystyle \frac{1\cdot 3\cdot5...(2n-1)}{n!}2^{n}\cdot x^{n}$$

B
$$\displaystyle \frac{1\cdot 3\cdot5...(2n-1)}{n!}2^{n}\cdot x^{n}x^{n}$$

C
$$\displaystyle ^{2n}C_{n}$$

D
$$\displaystyle ^{2n}C_{n}-1x^{n-1}$$

Solution

The middle term in the above binomial expansion will be $$(n+1)^{th}$$ term.
Therefore coefficient will be
$$\:^{2n}C_{n}$$
$$=\dfrac{2n!}{(n!)(n!)}$$
$$=\dfrac{2^{n}x^{n}(1.3.5...(2n-1)}{n!}$$
Question 5
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If the coeffecient of the middle term in the expansion of $${ (1+x) }^{ 2n+2 }$$ is $$\alpha $$ and coeffecient of middle terms in the expansion of $${ (1+x) }^{ 2n+1 }$$ are $$\beta $$ and $$\gamma $$, then relate $$\alpha, \beta$$ and $$ \gamma $$

A
$$\beta -\gamma =\alpha $$

B
$$\gamma -\beta =\alpha $$

C
$$\beta +\gamma =\alpha $$

D
none of these

Solution

$$(1+x)^{2n+2}$$.
The middle term will be $$(\frac{N}{2}+1)$$ th term.
$$=\frac{2n+2}{2}+1$$ th term
$$=(n+2)^{th}$$ term
Hence coefficient of the middle term will be
$$\:^{2n+2}C_{n+1}$$
$$=\alpha$$.
For
$$(1+x)^{2n+1}$$.
The middle terms will be $$(\frac{N+1}{2}+1)$$ and $$(\frac{N+1}{2})$$ terms
$$=\frac{2n+2}{2}+1$$ th term and $$\frac{2n+2}{2}^{th}$$ term.
$$=(n+2)^{th}$$ term and $$(n+1)^{th}$$ term.
Hence coefficient of the middle terms will be
$$\:^{2n+1}C_{n+1}=\beta$$ and $$\:^{2n+1}C_{n}=\gamma$$
By the properties of binomial coefficients.
$$\:^{2n+1}C_{n+1}+\:^{2n+1}C_{n}$$
$$=\:^{2n+2}C_{n+1}$$
Hence $$\beta+\gamma=\alpha$$
Question 6
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The middle term in the expansion of $$\displaystyle \left ( x+\frac{1}{x} \right )^{10}$$,is

A
$$\displaystyle ^{10}C_{1}\frac{1}{x}$$

B
$$\displaystyle ^{10}C_{5}$$

C
$$\displaystyle ^{10}C_{6}$$

D
$$\displaystyle ^{10}C_{7}x$$

Solution

The middle term would be the 6th term.
Hence
$$T_{5+1}=\:^{10}C_{5}.$$
Question 7
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The middle term in the expansion of $${ (1+x) }^{ 2n }$$ is

A
$$\displaystyle\frac { { 2 }^{ n }{ x }^{ n }(1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)) }{ n! } $$

B
$$\displaystyle\frac { { 2n }{ x }^{ n }(1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)) }{ n! } $$

C
$$\displaystyle\frac { { 2n }{ x }^{ n }(1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)) }{ n-1 }$$

D
$$\displaystyle\frac { { 2 }^{ n }{ x }^{ n }(1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)) }{ n-1 } $$

Solution

The middle term will be

$$(\dfrac{N}{2}+1)th$$ term.

Hence the middle term here will be

$$(\dfrac{2n}{2}+1)th$$ term.

$$=(n+1)th$$ term.

Hence $$r=n$$.

Therefore the coefficient will be

$$\:^{2n}C_{n}$$

$$=\dfrac{2n!}{{n!}{n!}}$$

$$=\dfrac{2^nx^n(1.3.5...(2n-1))(n!)}{n!n!}$$

$$=\dfrac{2^nx^n(1.3.5...(2n-1)}{n!}$$

Hence, option A is correct.
Question 8
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In the second term in the expansion $$\displaystyle { \left( \sqrt [ 13 ]{ a } +\frac { a }{ \sqrt { { a }^{ -1 } } } \right) }^{ n }$$ is $$\displaystyle 14{ a }^{ \frac { 5 }{ 2 } }$$ , then the value of $$\displaystyle \frac { _{ }^{ n }{ { C }_{ 3 }^{ } } }{ ^{ n }{ { C }_{ 2 }^{ } } } $$ is

A
$$8$$

B
$$12$$

C
$$4$$

D
None of these

Solution

Given: $${ T }_{ 2 }=14{ a }^{\tfrac 52 }$$

$$\displaystyle \Rightarrow { _{ }^{ n }{ C } }_{ 1 }^{ }{ \left( { a }^{\tfrac 1{13} } \right) }^{ n-1 }.{ \left( \dfrac { a }{ { a }^{ -\tfrac 12 } } \right) }^{ 1 }=14{ a }^{\tfrac 52 }$$

$$\displaystyle \Rightarrow n.{ a }^{ \dfrac { n-1 }{ 13 } }.{ a }^{ \dfrac { 3 }{ 2 } }=14{ a }^{ \dfrac { 5 }{ 2 } }$$

$$\displaystyle \Rightarrow n.{ a }^{ \dfrac { n-1 }{ 13 } }=14a\Rightarrow n.{ a }^{ \dfrac { n-14 }{ 13 } }=14\Rightarrow n=14$$

$$\displaystyle \therefore \dfrac { { _{ }^{ n }{ C } }_{ 3 }^{ } }{ { _{ }^{ n }{ C } }_{ 2 }^{ } } =\dfrac { { _{ }^{ 14 }{ C } }_{ 3 }^{ } }{ { _{ }^{ 14 }{ C } }_{ 2 }^{ } } =4$$
Question 9
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The coeffecients of the middle term in the binomial expansion in powers of $$x$$ of $${ (1+\alpha x) }^{ 4 }$$ and $${ (1+\alpha x) }^{ 6 }$$ is the same if $$\alpha $$ equals

A
$$-\cfrac { 5 }{ 3 } $$

B
$$\cfrac { 10 }{ 3 } $$

C
$$\cfrac { 3 }{ 10 } $$

D
$$\cfrac { 3 }{ 5 } $$

Solution

The middle terms of the above given binomial expressions will be
$$\:^{4}C_{2}(\alpha)^{2}x^2$$ and $$\:^{6}C_{3}(\alpha)^{3}x^3$$
Since the coefficients are equal,
$$\:^{4}C_{2}(\alpha)^{2}=\:^{6}C_{3}(\alpha)^{3}$$
$$6(\alpha)^{2}=20(\alpha)^{3}$$
$$\dfrac{3}{10}=\alpha$$
Question 10
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The coefficient of 1/x in the expansion of $$(1 + x)^n (1 + 1/x)^n$$ is

A
$$\displaystyle \frac{n!}{(n-1)!(n+1)!}$$

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

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

D
none of these

Solution

$$(1+x)^{n}(1+\dfrac{1}{x})^{n}$$
$$=\dfrac{(1+x)^{2n}}{x^n}$$
$$=\dfrac{1}{x^n}[(1+x)^{2n}]$$
$$T_{r+1}=\:^{2n}C_{r}x^{r-n}$$
Now for $$\dfrac{1}{x}$$
$$r-n=-1$$
$$r=n-1$$
Substituting, we get
$$T_{n}=\:^{2n}C_{n-1}x^{-1}$$
Coefficient implies
$$\:^{2n}C_{n-1}$$
$$=\dfrac{(2n)!}{(n+1)!(n-1)!}$$

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Binomial Theorem Test - 19

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Binomial Theorem Test - 19

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Question 1

$$\displaystyle\cfrac { n(n+1) }{ 3! } $$

$$\displaystyle\cfrac { n(n-1) }{ 2 } $$

$$\displaystyle\cfrac { n(n+1) }{ 2 } $$

None of these

Solution

Question 2

$$\displaystyle ^{2n}C_{n}$$

$$\displaystyle ^{-2n}C_{n}$$

$$\displaystyle ^{-2n}C_{n-1}$$

none of these

Solution

Question 3

$$\displaystyle A+B=C$$

$$\displaystyle B+C=A$$

$$\displaystyle C+A=B$$

$$\displaystyle A+B+C=0$$

Solution

Question 4

$$\displaystyle \frac{1\cdot 3\cdot5...(2n-1)}{n!}2^{n}\cdot x^{n}$$

$$\displaystyle \frac{1\cdot 3\cdot5...(2n-1)}{n!}2^{n}\cdot x^{n}x^{n}$$

$$\displaystyle ^{2n}C_{n}$$

$$\displaystyle ^{2n}C_{n}-1x^{n-1}$$

Solution

Question 5

$$\beta -\gamma =\alpha $$

$$\gamma -\beta =\alpha $$

$$\beta +\gamma =\alpha $$

none of these

Solution

Question 6

$$\displaystyle ^{10}C_{1}\frac{1}{x}$$

$$\displaystyle ^{10}C_{5}$$

$$\displaystyle ^{10}C_{6}$$

$$\displaystyle ^{10}C_{7}x$$

Solution

Question 7

$$\displaystyle\frac { { 2 }^{ n }{ x }^{ n }(1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)) }{ n! } $$

$$\displaystyle\frac { { 2n }{ x }^{ n }(1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)) }{ n! } $$

$$\displaystyle\frac { { 2n }{ x }^{ n }(1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)) }{ n-1 }$$

$$\displaystyle\frac { { 2 }^{ n }{ x }^{ n }(1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)) }{ n-1 } $$

Solution

Question 8

$$8$$

$$12$$

$$4$$

None of these

Solution

Question 9

$$-\cfrac { 5 }{ 3 } $$

$$\cfrac { 10 }{ 3 } $$

$$\cfrac { 3 }{ 10 } $$

$$\cfrac { 3 }{ 5 } $$

Solution

Question 10

$$\displaystyle \frac{n!}{(n-1)!(n+1)!}$$

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

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

none of these

Solution

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