Binomial Theorem Test

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Binomial Theorem Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

The $$4th$$ term from the end in the expansion of $${ \left( \cfrac { { x }^{ 3 } }{ 2 } -\cfrac { 2 }{ { x }^{ 2 } } \right) }^{ 7 }$$ is

A
$$35x$$

B
$$70x^{2}$$

C
$$35x^{2}$$

D
$$70x$$

Solution

For the above question
$$T_{r+1}=\:^{7}C_{r}x^{21-5r}2^{2r-7}$$
For the fourth term, from the end $$r=4$$
$$T_{5+1}=\:^{7}C_{4}x^{1}2^{}$$
$$=(35)(2)x$$
$$=70x$$
Question 2
1 / -0

The middle term in the expansion of $${ \left( \cfrac { a }{ x } +bx \right) }^{ 12 }$$ is

A
$$924{ a }^{ 6 }{ b }^{ 6 }$$

B
$$924{ a }^{ 6 }{ b }^{ 5 }$$

C
$$924{ a }^{ 5 }{ b }^{ 5 }$$

D
$$924{ a }^{ 5 }{ b }^{ 6 }$$

Solution

The middle term will be the 7th term.
Hence
$$T_{6+1}=\:^{12}C_{6}(\frac{a}{x})^{6}(bx)^{6}=924a^{6}b^{6}$$
Question 3
1 / -0

Find the middle term in the expansion of $${ \left( 3x-\cfrac { { x }^{ 3 } }{ 6 } \right) }^{ 9 }$$.

A
$${_{ }^{ 9 }{ C } }_{ 6 }{ \left( 3x \right) }^{ 5 }$$

B
$${ _{ }^{ 9 }{ C } }_{ 5 }{ \left( 3x \right) }^{ 4}$$

C
Both A & B

D
none of the above

Solution
Question 4
1 / -0

There will be no term containing $${ x }^{ 2r }$$ in the expansion of $${ \left( x+{ x }^{ -2 } \right) }^{ n-3 }$$ if $$(n-2r)$$ is positive but not a multiple of

A
$$2$$

B
$$3$$

C
$$5$$

D
$$11$$

Solution

$$(x+\dfrac{1}{x^2})^{n-3}$$
$$T_{r+1}=\:^{n-3}C_{r}x^{n-3-3r}$$
For $$x^{2r}$$
$$n-3-3r=2r$$
$$n-3=5r$$
or
$$n-2r=3+3r$$
$$=3(1+r)$$
$$=3k$$
Hence $$n-2r$$ has to be a multiple of 3 for the existence of $$x^{2r}$$
Question 5
1 / -0

If $${ \left( 8+3\sqrt { 7 } \right) }^{ n }=\alpha +\beta $$ where $$n$$ and $$\alpha$$ are positive integers and $$\beta$$ is a positive proper fraction,then

A
$$ (1-\beta)(\alpha+\beta)=1$$

B
$$ (1+\beta)(\alpha+\beta)=1$$

C
$$ (1-\beta)(\alpha-\beta)=1$$

D
$$ (1+\beta)(\alpha-\beta)=1$$

Solution

Given, $$\alpha+\beta=(8+3\sqrt{7})^{n}$$

And

$$1-\beta$$ $$=\dfrac{1}{\alpha+\beta}$$ $$=\dfrac{1}{(8+3\sqrt{7})^{n}}$$

Rationalizing, we get

$$1-\beta$$ $$=\dfrac{1}{(8+3\sqrt{7})^{n}}\dfrac{(8-3\sqrt{7})^{n}}{(8-3\sqrt{7})^{n}}$$

$$=\dfrac{(8-3\sqrt{7})^{n}}{(64-63)^n}$$ ........ $$a^2-b^2=(a-b)(a+b)$$

$$=(8-3\sqrt{7})^{n}$$

Now
$$(1-\beta)(\alpha+\beta)$$

$$=(8-3\sqrt{7})^{n}(8+3\sqrt{7})^{n}$$

$$=(64-63)^{n}$$ ........ $$a^2-b^2=(a-b)(a+b)$$

$$=1$$
Option A is correct
Question 6
1 / -0

If $${ \left( 1+x \right) }^{ n }={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }+...+{ C }_{ n }{ x }^{ n }$$, then $$\displaystyle \sum _{ 0\le i\le }^{ }{ \sum _{ j\le n }^{ }{ { \left( { C }_{ i }+{ C }_{ j } \right) }^{ 2 } } = } $$

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

B
$$n._{ }^{ 2n }{ { C }_{ n }^{ } }+{ 2 }^{ 2n }$$

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

D
None of these

Solution

$$\sum _{ 0\le i\le }^{ }{ \sum _{ j\le n }^{ }{ { \left( { C }_{ i }+{ C }_{ j } \right) }^{ 2 } } } \quad i=0,1,2,...,\left( n-1 \right) ;j=1,2,3,...,n$$ and $$i<j$$
$$=n\left( { C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+...+{ C }_{ n }^{ 2 } \right) +2\sum { \sum { { C }_{ i }{ C }_{ j } } } 0\le i\le j\le n\\ =n.^{ 2n }{ { C }_{ n } }+\left[ { \left( { C }_{ 0 }+{ C }_{ 1 }+...+{ C }_{ n } \right) }^{ 2 }-\left( { C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+...+{ C }_{ n } \right) \right] \\ =n.^{ 2n }{ { C }_{ n } }+{ { \left( { 2 }^{ n } \right) } }^{ 2 }-^{ 2n }{ { C }_{ n } }=\left( n-1 \right) .^{ 2n }{ { C }_{ n } }+{ 2 }^{ 2n }$$
Question 7
1 / -0

If the second term in the expansion $${ \left[ a^{\dfrac {1}{13}} +\dfrac { a }{ \sqrt { { a }^{ -1 } } } \right] }^{ n }$$ is $$14\ { a }^{ 5/2 }$$, then the value of $$\dfrac {^{n}C_{3}}{^{n}C_{2}}$$ is

A
$$4$$

B
$$3$$

C
$$12$$

D
$$6$$

Solution

The first term will have a coefficient of $$\:^{n}C_{1}=n$$
Now, this coefficient has to be equal to $$14.$$
Therefore $$n=14$$
Now
$$\dfrac{\:^{14}C_{3}}{\:^{14}C_{2}}=\dfrac{14!(12!)(2!)}{11!(3!)(14!)}=\dfrac{12}{3}=4$$
Question 8
1 / -0

If the number of terms in $${ \left( x+1+\cfrac { 1 }{ x } \right) }^{ n }\quad (n\in { I }^{ + })$$ is 401, then $$n$$ is greater than

A
201

B
200

C
199

D
none of these

Solution

If we substitute 2 in place of 1 in the above expression, we get
$$(x+2+\frac{1}{x})^{n}$$
$$((\sqrt{x}+\frac{1}{\sqrt{x}}))^{2n}$$
Since the total number of terms is 401
$$2n+1=401$$
$$2n=400$$
$$n=200$$
However the actual question is
$$(x+1+\frac{1}{x})^{n}$$
Hence n is greater than 199.
Question 9
1 / -0

If $$\displaystyle{ a }_{ n }=\sum _{ r=0 }^{ n }{ \cfrac { 1 }{ { _{ }^{ n }{ C } }_{ r } } } $$then $$\displaystyle\sum _{ r=0 }^{ n }{ \cfrac { r }{ { _{ }^{ n }{ C } }_{ r } } }$$ equals

A
$$(n-1){ a }_{ n }$$

B
$$n{ a }_{ n }$$

C
$$\cfrac { 1 }{ 2 } n{ a }_{ n }$$

D
$$\cfrac { (n-1) }{ 2 } { a }_{ n }$$

Solution

We have $$\displaystyle { a }_{ n }=\sum _{ r=0 }^{ n }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } =\sum _{ r=0 }^{ \frac { n }{ 2+1 } }{ \frac { 1 }{ ^{ n }{ C }_{ r } } +\frac { 1 }{ ^{ n }{ { C }_{ n-r } } } } $$
$$\displaystyle \Rightarrow { a }_{ n }=\sum _{ r=0 }^{ \frac { n+1 }{ 2 } }{ \frac { 2 }{ ^{ n }{ C_{ r } } } =2 } \sum _{ r=0 }^{ \frac { n+1 }{ 2 } }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } $$
$$\displaystyle \therefore \sum _{ r=0 }^{ n }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } =\sum _{ r=0 }^{ \frac { n+1 }{ 2 } }{ \frac { r+\left( n-r \right) }{ ^{ n }{ { C }_{ r } } } } $$
$$\displaystyle \Rightarrow n\sum _{ r=0 }^{ n }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } =n\left( \frac { { a }_{ n } }{ 2 } \right) =\frac { n }{ 2 } { a }_{ n }$$
Question 10
1 / -0

The total number of terms in the expansion of $${ \left( x+a \right) }^{ 100 }+{ \left( x-a \right) }^{ 100 }$$ after simplification is

A
202

B
51

C
50

D
49

Solution

In the above binomial expansion, the terms at the even places will get eliminated, and we would be left with twice the sum of the terms at odd places.
Hence there will be
$$\frac{n}{2}+1$$
$$=\frac{100}{2}+1$$
$$=51$$ terms

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Binomial Theorem Test - 22

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Binomial Theorem Test - 22

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Question 1

$$35x$$

$$70x^{2}$$

$$35x^{2}$$

$$70x$$

Solution

Question 2

$$924{ a }^{ 6 }{ b }^{ 6 }$$

$$924{ a }^{ 6 }{ b }^{ 5 }$$

$$924{ a }^{ 5 }{ b }^{ 5 }$$

$$924{ a }^{ 5 }{ b }^{ 6 }$$

Solution

Question 3

$${_{ }^{ 9 }{ C } }_{ 6 }{ \left( 3x \right) }^{ 5 }$$

$${ _{ }^{ 9 }{ C } }_{ 5 }{ \left( 3x \right) }^{ 4}$$

Both A & B

none of the above

Solution

Question 4

$$2$$

$$3$$

$$5$$

$$11$$

Solution

Question 5

$$ (1-\beta)(\alpha+\beta)=1$$

$$ (1+\beta)(\alpha+\beta)=1$$

$$ (1-\beta)(\alpha-\beta)=1$$

$$ (1+\beta)(\alpha-\beta)=1$$

Solution

Question 6

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

$$n._{ }^{ 2n }{ { C }_{ n }^{ } }+{ 2 }^{ 2n }$$

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

None of these

Solution

Question 7

$$4$$

$$3$$

$$12$$

$$6$$

Solution

Question 8

201

200

199

none of these

Solution

Question 9

$$(n-1){ a }_{ n }$$

$$n{ a }_{ n }$$

$$\cfrac { 1 }{ 2 } n{ a }_{ n }$$

$$\cfrac { (n-1) }{ 2 } { a }_{ n }$$

Solution

Question 10

202

51

50

49

Solution

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