Binomial Theorem Test

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Binomial Theorem Test - 13

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Question 1
1 / -0

If the coefficients of $$ (r+2)^{th} $$ and $$ (2r+1) ^{th}$$ terms $$ (r \neq 1) $$ are equal in the expansion of $$ (1+x)^{43} $$, then $$ r = $$

A
$$12$$

B
$$13$$

C
$$14$$

D
$$15$$

Solution

Since coefficient of $$(r+2)^{th}$$ term is equal to coefficient of $$(2r+1)^{th}$$ term in the binomial expansion of $$(1+x)^{43}$$
Therefore $$\:^{43}C_{r+1}=\:^{43}C_{2r}$$
Now $$\:^{n}C_{r}=\:^{n}C_{n-r}$$
$$\therefore 43-(r+1)=2r$$
$$\Rightarrow 42=3r$$
$$\Rightarrow r=14$$
Question 2
1 / -0

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

A
$$ ^{2n+1}C_n.x $$

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

C
$$ (-1)^{n}.^{2n+1}C_{n} $$

D
$$ (-1)^{n}.^{2n+1}C_{n}.x $$

Solution

Let the middle term of $$\left (x-\dfrac{1}{x}\right)^{2n+1}$$ be $$M$$.
The middle term is $$\dfrac{1}{2}(2n+1+1) $$ i.e $$(n+1)^{th}$$ term.
$$M=T_{n+1}$$
We k.n.t $$T_r$$ of expansion $$(a+b)^n$$ is $$T_r=^nC_{r-1}a^{n-r+1}b^{r-1} $$
$$\therefore M=T_{n+1}=^{2n+1}C_n x^{2n+1-n} \left (-\dfrac{1}{x}\right)^n = (-1)^n .^{2n+1}C_n.x^{n+1-n}=(-1)^n.^{2n+1}C_n.x $$
Option (D) is correct.
Question 3
1 / -0

The ratio of the coefficient of $$ x^{10} $$ in $$ (1-x^2)^{10} $$ and the term independent of $$x$$ in $$ \left ( x - \dfrac{2}{x} \right )^{10} $$ is

A
$$32 : 1$$

B
$$-32 : 1$$

C
$$-1 : 32$$

D
$$1 : 32$$

Solution

Given $$(1-x^2)^{10}$$
$$T_{r+1}=(-1)^{r}\:^{10}C_{r}x^{2r}$$...(i)
For the coefficient of $$x^{10}$$
$$2r=10$$
$$r=5$$
$$T_{6}=-\:^{10}C_{5}x^{10}$$ ...(a)
$$T_{r+1}=(-1)^{r}\:^{10}C_{r}x^{10-2r}2^{r}$$...(ii)
For the term independent of $$x$$
$$2r=10$$
$$r=5$$
Substituting in (ii), we get
$$T_{6}=-\:^{10}C_{5}2^{5}$$...(b)
Hence $$\dfrac{a}{b}=1:32$$
Question 4
1 / -0

The $$3$$rd, $$4$$th and $$5$$th terms in the expansion of $$ (1+x)^n $$ are $$60, 160$$ and $$240$$ respectively, then $$ x = $$

A
$$2$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

It is given that in $$(1+x)^{n}$$
$$T_{3}=\:^nC_{2}x^2=60$$
$$T_{4}=\:^nC_{3}x^3=160$$
$$T_{5}=\:^nC_{4}x^4=240$$
Therefore taking ratios we get
$$\displaystyle \frac{\:^nC_{2}x^2}{\:^nC_{3}x^3}=\frac{3}{8}$$

$$\displaystyle \frac{3}{x(n-2)}=\frac{3}{8}$$
$$8=nx-2x$$
$$nx=2x+8$$ $$...(i)$$

$$\displaystyle \frac{\:^nC_{3}x^3}{\:^nC_{4}x^4}=\frac{160}{240}$$

$$\displaystyle \frac{4}{x(n-3)}=\frac{160}{240}$$
$$\displaystyle \frac{1}{nx-3x}=\frac{40}{240}$$
$$6=nx-3x$$
$$nx=3x+6$$ $$...(ii)$$
Substituting the value of $$nx$$ in equation $$(i)$$ we get
$$3x+6=2x+8$$
$$x=2$$
Question 5
1 / -0

Coefficient of $$ x^{16} $$ in $$ (1+x+x^2)(1-x)^{15} $$

A
16

B
-14

C
14

D
-16

Solution

$$(1+x+x^2)(1-x)^{15}$$
$$=(1+x+x^2)(1-15x+\:^{15}C_{2}x^2-\:^{15}C_{3}x^3...-\:^{15}C_{15}x^{15})$$
Hence coefficient of $$x^{16}$$ will be.
$$-\:^{15}C_{15}x.x^{15}+\:^{15}C_{14}x^2.x^{14}$$
Hence coefficient will be.
$$=-1+15$$
$$=14$$
Question 6
1 / -0

The number of terms in the expansion of $$ (1+5\sqrt{2}x)^9 + (1-5\sqrt{2}x)^9 $$ is :

A
$$5$$

B
$$10$$

C
$$18$$

D
$$20$$

Solution

$$(1+5\sqrt{2}x)\:^{9}+(1-5\sqrt{2}x)\:^{9}$$
Applying binomial theorem for expansion of the following
$$=1+(5\sqrt{2}x)\:(^{9}C_{1})+(5\sqrt{2}x)^{2}\:(^{9}C_{2})...(\:^{9}C_{9}(5\sqrt{2}x)^{9})$$
$$+1-(5\sqrt{2}x)\:(^{9}C_{1})+(5\sqrt{2}x)^{2}\:(^{9}C_{2})...-(\:^{9}C_{9}(5\sqrt{2}x)^{9})$$
$$=2[1+(5\sqrt{2}x)\:^2\:(^{9}C_{2})+(5\sqrt{2}x)\:^4\:(^{9}C_{4})...(5\sqrt{2}x)\:^8(\:^{9}C_{8})]$$
The powers of $$5\sqrt{2}x$$ are in $$A.P$$
$$0,2,4,6,8$$
Hence including the term $$1\:or\: x^0$$ there are $$\dfrac{9+1}{2} = 5$$ terms.
Hence, answer is $$A.$$
Question 7
1 / -0

Coefficient of $$ x^5 $$ in $$ (1+x^2)^5(1+x)^4 $$ is

A
60

B
80

C
90

D
100

Solution

$$(1+x)^{4}(1+x^2)^{5}$$
$$(1+4x+6x^2+4x^3+x^4)(1+5x^2+10x^{4}+10x^6+5x^8+x^{10})$$
Hence coefficient of $$x^5$$ will be
$$4(10)+4(5)$$
$$=40+20$$
$$=60$$
Question 8
1 / -0

A. $$ ^{2n}C_n = C_0^2 + C_1^2 + C_2^2 + C_3^2 + \dots \dots + C_n^2 $$

B. $$ ^{2n}C_n = $$ term independent of $$x$$ in $$ (1+x)^n \left(1+\frac{1}{x} \right)^n $$

C. $$ ^{2n}C_n = \dfrac{1.3.5.7 \ldots \ldots (2n-1)}{n!} $$ then

A
A, B are false, C is true

B
A is false, B and C are true

C
A and B are true; C is false

D
A, B, C are true

Solution

$$\:^{2n}C_{n}$$ $$=\dfrac{2n!}{(n!)^{2}}$$

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

Hence option C is false.

Consider Option B
$$(1+x)^{n}\cdot (1+\dfrac{1}{x})^{n}$$
$$=\dfrac{(1+x)^{n}(1+x)^{n}}{x^{n}}$$
$$=\dfrac{(1+x)^{2n}}{x^{n}}$$.

The general term in this case is
$$T_{r+1}=\:^{2n}C_{r}x^{r}.x^{-n}$$
$$=\:^{2n}C_{r}x^{r-n}$$
For term independent of $$x$$
$$r-n=0$$
$$n=r$$.

Hence the term independent of $$x$$ will be
$$\:^{2n}C_{n}$$
Thus B is true.

Consider option A.
Consider the following series.
$$\:^{m}C_{0}\:^{n}C_{r}+\:^{m}C_{1}\:^{n}C_{r-1}+...\:^{m}C_{r}\:^{n}C_{0}$$
$$=$$ Coefficient of $$x^{r}$$ in $$(1+x)^{m}.(x+1)^{n}$$
$$=$$ Coefficient of $$x^{r}$$ in $$(1+x)^{m+n}$$
$$=\:^{m+n}C_{r}$$

In the above case, $$r=n$$ and $$m=n$$
Hence, $$\:^{n}C_{0} ^{2}+\:^{n}C_{1} ^{2}+\:^{n}C_{2} ^{2}+\:^{n}C_{3} ^{2}+..\:^{n}C_{n} ^{2}$$
$$=\:^{n}C_{0}\:^{n}C_{n}+\:^{n}C_{1}\:^{n}C_{n-1}+...\:^{n}C_{2}\:^{n}C_{n-2}$$
$$=\:^{2n}C_{n}$$

Hence A is also true.
Question 9
1 / -0

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

A
$$202$$

B
$$51$$

C
$$101$$

D
$$50$$

Solution

$$(x+a)\:^{100}+(x-a)\:^{100}$$
Applying binomial theorem for expansion of the following
$$(x+a)\:^{100}+(x-a)\:^{100}$$
$$=x^{100}+x^{99}\:(^{100}C_{1}a)+x^{98}\:(^{100}C_{2}a^2)...(\:^{100}C_{100}a^{100})$$
$$+x^{100}-x^{99}\:(^{100}C_{1}a)+x^{98}\:(^{100}C_{2}a^2)...(\:^{100}C_{100}a^{100})$$
$$=2[x^{100}+x^{98}\:(^{100}C_{2}a^2)+x^{96}\:(^{100}C_{4}a^4)...(\:^{100}C_{100}a^{100})]$$
The powers of $$a$$ are in A.P
$$a^0,a^2,a^4,a^6...a^{100}$$
$$0,2,4,6...100$$
Hence, including the term $$a^0\:or\:x^{100}$$ there are $$51$$ terms.
Hence, answer is $$B.$$
Question 10
1 / -0

Coefficient of $$ x^{10} $$ in $$ (1+2x^4)(1-x)^8 $$ is

A
-56

B
56

C
112

D
-112

Solution

$$(1+2x^4)(1-x)^{8}$$
$$=(1+2x^4)(1-8x+\:^{8}C_{2}x^2-\:^{8}C_{3}x^3...+\:^{8}C_{2}x^8)$$
Hence coefficient of $$x^{10}$$ will be.
$$2\times\:^{8}C_{6}x^4(x^6)$$
Hence coefficient of $$x^{10}$$
$$=2\:^{8}C_{2}$$
$$=2(\displaystyle \frac{8\times7}{2})$$
$$=56$$

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Binomial Theorem Test - 13

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Binomial Theorem Test - 13

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Question 1

$$12$$

$$13$$

$$14$$

$$15$$

Solution

Question 2

$$ ^{2n+1}C_n.x $$

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

$$ (-1)^{n}.^{2n+1}C_{n} $$

$$ (-1)^{n}.^{2n+1}C_{n}.x $$

Solution

Question 3

$$32 : 1$$

$$-32 : 1$$

$$-1 : 32$$

$$1 : 32$$

Solution

Question 4

$$2$$

$$4$$

$$5$$

$$6$$

Solution

Question 5

16

-14

14

-16

Solution

Question 6

$$5$$

$$10$$

$$18$$

$$20$$

Solution

Question 7

60

80

90

100

Solution

Question 8

A, B are false, C is true

A is false, B and C are true

A and B are true; C is false

A, B, C are true

Solution

Question 9

$$202$$

$$51$$

$$101$$

$$50$$

Solution

Question 10

-56

56

112

-112

Solution

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