Binomial Theorem Test

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Binomial Theorem Test - 39

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Question 1
1 / -0

The coefficient of $$x^{2012}$$ in $$\dfrac{1+x}{(1+x^2)(1-x)}$$ is.

A
1

B
2

C
3

D
4

Solution

$$\displaystyle{\frac { (1+x) }{ (1+{ x }^{ 2 })(x-1) } =\frac { x }{ (1+{ x }^{ 2 }) } +\frac { 1 }{ (1-x) } }$$
The general term for $$\displaystyle{{ \left( 1-x \right) }^{ -1 }=1+x+{ x }^{ 2 }+{ x }^{ 3 }+.........}$$
$$\displaystyle{{ \left( 1-x \right) }^{ -1 }=\sum _{ 0 }^{ \infty }{ { x }^{ n } } .........(i)}$$
Replacing $$x$$ by $$-{ x }^{ 2 }$$
$$\displaystyle{\frac { 1 }{ 1+{ x }^{ 2 } } =1-{ x }^{ 2 }+{ x }^{ 4 }-{ x }^{ 6 }.......\\ \frac { 1 }{ 1+{ x }^{ 2 } } =\sum _{ 0 }^{ \infty }{ { { (-1) }^{ n }x }^{ 2n } }} $$
Multiplying both sides by $$x$$
$$\displaystyle{\frac { x }{ 1+{ x }^{ 2 } } =\sum _{ 0 }^{ \infty }{ { { (-1) }^{ n }x }^{ 2n+1 } } ....(ii)}$$
From (i) clearly the cofficient of $${ x }^{ 2012 }$$ is $$1$$ and there is no term with even cofficient in $$(ii)$$
So the cofficient of $$\displaystyle{{ x }^{ 2012 }}$$ in the given expression is $$1$$
Question 2
1 / -0

Let $$n$$ be a positive integer and, $${ \left( 1+x \right) }^{ n }={ a }_{ 0 }+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+\dots +{ a }_{ n }{ x }^{ n }$$. What is $${ a }_{ 0 }+{ a }_{ 1 }+{ a }_{ 2 }+\dots +{ a }_{ n }$$ equal to?

A
$$1$$

B
$${ 2 }^{ n }$$

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

D
$${ 2 }^{ n+1 }$$

Solution

Given expansion is $${(1+x)}^n=a_0+a_1 x+a_2 x^2+....a_n x^n$$
Substituting $$x=1$$ on both the sides we get,
$${(1+1)}^n=a_0+a_1\times 1+a_2\times 1^2+....a_n\times 1^n$$
$$\implies 2^n=a_0+a_1+a_2+...a_n$$
Thus, the value of $$a_0+a_1+a_2+...a_n$$ is $$2^n$$.
Question 3
1 / -0

$$5^{th}$$ term from the end in the expansion of $$\left( \dfrac{x^2}{2} - \dfrac{2}{x^2} \right )^{12}$$ is

A
$$-7920x^{-4}$$

B
$$7920x^{4}$$

C
$$7920x^{-4}$$

D
$$-7920x^{4}$$

Solution

$$ { 5 }^{ th }$$ term in the expansion of$${ \left( \cfrac { { x }^{ 2 } }{ 2 } -\cfrac { 2 }{ { x }^{ 2 } } \right) }^{ 12 }$$ is
$$ { T }_{ r+1 }= { }^nC_r { x }^{ r }.{ y }^{n- r }$$
$$ { T }_{ 5 }= { }^{12}C_4 \left[{ \left( { \cfrac { { x }^{ 2 } }{ 2 } } \right) ^{ 4 }{ \left( \cfrac { -2 }{ { x }^{ 2 } } \right) }^{ 8 } }\right]$$
$$ \quad \quad = \cfrac { 12! }{ 4!8! } .\cfrac { { x }^{ 8 } }{ { 2 }^{ 4 } } .\cfrac { { 2 }^{ 8 } }{ { x }^{ 16 } }$$
$$ \quad \quad = { 7920x }^{ -4 } $$
Question 4
1 / -0

The value of $${ }^nC_0 - { }^nC_1 + { }^n C_2 - .... + (-1)^{n^n}C_n$$ is:

A
$$1$$

B
$$0$$

C
$$2^n$$

D
$$n$$

Solution

By Binomial Expansion we know,
$${ \left( a+b \right) }^{ n }=^{ n }{ C }_{ 0 }{ a }^{ 0 }{ b }^{ n }+^{ n }{ C }_{ 1 }{ a }^{ 1 }{ b }^{ n-1 }+....+^{ n }{ C }_{ n-1 }{ a }^{ n-1 }{ b }^{ 1 }+^{ n }{ C }_{ n }{ a }^{ n }{ b }^{ 0 }$$
Now putting $$a=-1$$ & $$b=+1$$
we have,
$${ \left( -1+1 \right) }^{ n }=^{ n }{ C }_{ 0 }{ \left( -1 \right) }^{ 0 }{ \left( +1 \right) }^{ n-1 }+^{ n }{ C }_{ 0 }{ \left( -1 \right) }^{ 1 }{ \left( +1 \right) }^{ n-1 }+.....+^{ n }{ C }_{ n-1 }{ \left( -1 \right) }^{ n-1 }\left( +1 \right) +^{ n }{ C }_{ n }{ \left( -1 \right) }^{ n }{ \left( +1 \right) }^{ 0 }$$
$$\Rightarrow 0 =^{ n }{ C }_{ 0 }{ \left( -1 \right) }^{ 0 }{ \left( +1 \right) }^{ n-1 }+^{ n }{ C }_{ 0 }{ \left( -1 \right) }^{ 1 }{ \left( +1 \right) }^{ n-1 }+.....+^{ n }{ C }_{ n-1 }{ \left( -1 \right) }^{ n-1 }\left( +1 \right) +^{ n }{ C }_{ n }{ \left( -1 \right) }^{ n }$$
Question 5
1 / -0

Consider the expansion of $$(1+x)^{2n+1}$$
The average of the coefficients of the two middle terms in the expansion is

A
$$^{2n+1}C_{n+2}$$

B
$$^{2n+1}C_{n}$$

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

D
$$^{2n}C_{n+1}$$

Solution

Since, $$2n+1$$ is odd.
Hence, $$\cfrac{2n+1+1}{2}$$ and $$\cfrac{2n+1+3}{2}$$ are two middle terms.
i.e. $$(n+1)th$$ and $$(n+2)th$$ terms are two middle terms.
$$\therefore\cfrac{C^{2n+1}_n+ C^{2n+1}_{n+1}}{2}=\cfrac{C^{2n+1+1}_{n+1}}{2}$$
$$=\cfrac12C^{2n+2}_{n+1}=\cfrac12.\cfrac{2n+2}{n+1}C^{2n+1}_{n}=C^{2n+1}_{n}$$
Hence, B is the correct option.
Question 6
1 / -0

In the expansion of $$\left( x^3 - \dfrac{1}{x^2} \right )^n , n \in N$$, if the sum of the coefficient of $$x^5$$ and $$x^{10}$$ is $$0$$, then $$n$$ is :

A
$$25$$

B
$$20$$

C
$$15$$

D
None of these

Solution

Term of $${ x }^{ 5 }=^{ n }{ C }_{ r }\quad { x }^{ 3r }{ \left( \cfrac { -1 }{ { x }^{ 2 } } \right) }^{ n-r }=^{ n }{ C }_{ r }\quad { x }^{ 3r }. { x }^{ 2r-2n }. { \left( -1 \right) }^{ n-r }$$
So $$3r+2r-2n=5$$
$$r=\cfrac { 5+2n }{ 5 } \longrightarrow \left( 1 \right) $$
Also for $${ x }^{ 10 }$$ be $$\left( { r }^{ 1 }+1 \right) ^{ th }$$ term
So term with $${ x }^{ 10 }=^{ n }C_{ { r }^{ 1 } }{ x }^{ { 3r }^{ 1 } }{ \left( \cfrac { -1 }{ { x }^{ 2 } } \right) }^{ n-r }$$
$$\Rightarrow 3{ r }^{ 1 }+2{ r }^{ 1 }-2n=10$$
$${ r }^{ 1 }=\cfrac { 2n+10 }{ 5 } \longrightarrow \left( 2 \right) $$
Now we know if $$\left| ^{ n }{ C }_{ r } \right| =\left| ^{ n }{ C }_{ { r }^{ 1 } } \right| \Rightarrow n=r+{ r }^{ 1 }$$
Now we add co oefficient of $${ x }^{ 5 }\& { x }^{ 10 }$$
$$^{ n }{ C }_{ r }{ \left( -1 \right) }^{ n-r }+^{ n }{ C }_{ { r }^{ 1 } }{ \left( -1 \right) }^{ { n-r }^{ 1 } }=0$$
Co oefficient of $${ x }^{ 5 }\quad is\quad ^{ n }{ C }_{ r }{ \left( -1 \right) }^{ n-r }$$; co oeffiecient of $${ x }^{ 10 }\quad is\quad ^{ n }{ C }_{ { r }^{ 1 } }{ \left( -1 \right) }^{ { n-r }^{ 1 } }$$
$$\Rightarrow ^{ n }{ C }_{ r }{ \left( -1 \right) }^{ n-r }=-\left( ^{ n }{ C }_{ { r }^{ 1 } }\quad { \left( -1 \right) }^{ { n-r }^{ 1 } } \right) $$
$$\Rightarrow \left| ^{ n }{ C }_{ r } \right| =\left| ^{ n }{ C }_{ { r }^{ 1 } } \right| $$
So, $$n=r+{ r }^{ 1 }$$
$$n=\cfrac { 2n+10 }{ 5 } +\cfrac { 2n+5 }{ 5 } $$
$$5n=4n+15$$
$$n=15$$
Question 7
1 / -0

Consider the expansion of $$(1+x)^{2n+1}$$
The sum of the coefficients of all the terms in the expansion is

A
$$2^{2n-1}$$

B
$$4^{n-1}$$

C
$$2\times 4^n$$

D
None of the above

Solution

We can get the sum of all coefficients by putting $$x=1$$ in the expansion, because for calculating the coefficients we need the terms independent of $$x$$.
To find the sum of coefficients of all terms, put $$x=1$$ in the given expression$$(1+x)^{2n+1}$$ we get
$$2^{2n+1}=2.2^{2n}=2.4^n$$
Hence, C is the correct option.
Question 8
1 / -0

How many terms are there in the expansion of $${ \left( 1+2x+{ x }^{ 2 } \right) }^{ 10 }$$?

A
$$11$$

B
$$20$$

C
$$21$$

D
$$30$$

Solution

Now,
$$(1+2x+x^{ 2 })^{ 10 }=((1+x)^{ 2 })^{ 10 }=(1+x)^{ 20 }$$
Now, the number of terms in the expansion of $$(1+x)^{ n }$$ are $$n+1$$.
Thus, the number of terms in the expansion of $$(1+x)^{20}$$ will be $$20+1=21$$.
Hence, option C is correct.
Question 9
1 / -0

The value of the expression $${ _{ }^{ k-1 }{ C } }_{ k-1 }+{ _{ }^{ k }{ C } }_{ k-1 }+....{ _{ }^{ n+k-2 }{ C } }_{ k-1 }$$ is given by :

A
$${ _{ }^{ n+k-1 }{ C } }_{ k-1 }$$

B
$${ _{ }^{ n+k-1 }{ C } }_{ k }$$

C
$${ _{ }^{ n+k }{ C } }_{ k }$$

D
None of these

Solution

We know,
$$\displaystyle \sum _{ m=k }^{ n }{ { { ^{ m }{ C }_{ r}=\ ^{ n+1 }{ C }_{ r+1\\ } } } } $$
According to the question,
$$ m= n+k-2$$
$$ r= k-1$$
Sum of series is given by
$$= \ ^{ m+1 }{ C }_{ k+1\\ }$$ $$ =\ ^{ n+k-1 }{ C }_{ k }$$
Question 10
1 / -0

Directions For Questions

For the next three (03) items that follow :
In the expansion of $$\left ( x^3-\frac{1}{x^2} \right )^n$$ where n is a positive integer, the sum of the coefficients of $$x^5$$ and $$x^{10}$$ is 0.

...view full instructions

What is n equal to ?

A
$$5$$

B
$$10$$

C
$$15$$

D
None of the above

Solution

Using binomial theorem, we can write
$$\displaystyle \left({ x }^{ 3 }-\cfrac { 1 }{x^{2}} \right)^{ n }=\sum_{ r=0 }^{ n }\ { _{ r }^{ n }{ C{ \left(x ^{ 3 }\right) }^{ n-r } } } { \left(\cfrac { -1 }{ { x }^{ 2 } } \right) }^{ r }\\ =\displaystyle \sum_{ r=0 }^{ n }{ { \left(-1\right) }^{ r }\ {}_{ r }^{ n }{ C }{ x }^{ \left(3n-5r\right) } } $$
For $${x}^{3}$$ and $${x}^{10}$$ let value of r be $${ r }_{ 5 }$$ and $${ r }_{ 10 }$$ respectively
$$3n-5{r}_{10}=5$$ ......... $$(i)$$
$$3n-5{r}_{10}=10$$ .......... $$(ii)$$
$${ \left(-1\right) }^{ { r }_{ 5 } }\quad _{ { r }_{ 5 } }^{ n }{ C\quad }{ x }^{ \left(3n-5{ r }_{ 5 }\right) }$$$$=-{ \left(-1\right) }^{ { r }_{ 10} }\quad _{ { r }_{ 10 } }^{ n }{ C\quad }{ x }^{ \left(3n-5{ r }_{ 10}\right) }$$
Given sum of coefficient of $${x}^{5}$$ and $${x}^{10}$$ is 0
All terms are positive, but still there is a negative sign, means one of $${r}_{5}$$ and $${r}_{10}$$ is even and other is odd
By properties of combinations
$$_{ { r }_{ 5 } }^{ n }{ C}$$$$={}_{ { r }_{ 10 } }^{ n }{ C\quad }$$
$${r}_{5}+{r}_{10}=n$$ .......... $$(iii)$$
Adding $$(i)$$ and $$(ii)$$
$$6n-5\left({r}_{5}+{r}_{10}\right)=15$$
From $$(iii),$$
$${r}_{5}+{r}_{10}=n$$
$$\Rightarrow n=15$$

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Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 39

Result

Binomial Theorem Test - 39

Score

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Rank

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SHARING IS CARING

Question 1

1

2

3

4

Solution

Question 2

$$1$$

$${ 2 }^{ n }$$

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

$${ 2 }^{ n+1 }$$

Solution

Question 3

$$-7920x^{-4}$$

$$7920x^{4}$$

$$7920x^{-4}$$

$$-7920x^{4}$$

Solution

Question 4

$$1$$

$$0$$

$$2^n$$

$$n$$

Solution

Question 5

$$^{2n+1}C_{n+2}$$

$$^{2n+1}C_{n}$$

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

$$^{2n}C_{n+1}$$

Solution

Question 6

$$25$$

$$20$$

$$15$$

None of these

Solution

Question 7

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

$$4^{n-1}$$

$$2\times 4^n$$

None of the above

Solution

Question 8

$$11$$

$$20$$

$$21$$

$$30$$

Solution

Question 9

$${ _{ }^{ n+k-1 }{ C } }_{ k-1 }$$

$${ _{ }^{ n+k-1 }{ C } }_{ k }$$

$${ _{ }^{ n+k }{ C } }_{ k }$$

None of these

Solution

Question 10

$$5$$

$$10$$

$$15$$

None of the above

Solution

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