Binomial Theorem Test

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Binomial Theorem Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

The term independent of $$x$$ in the expansion of $$\left(\sqrt{\dfrac{x}{3}}+\dfrac{3}{2x^{2}}\right)^{10}$$ will be

A
$$3$$

B
$$5$$

C
$$9$$

D
$$None\ of\ these$$

Solution

As we know that the general term in an expansion $${\left( a + b \right)}^{n}$$ is given as-
$${T}_{r + 1} = {^{n}{C}_{r}} {a}^{r} {b}^{n - r}$$
Given expansion- $${\left( \sqrt{\cfrac{x}{3}} + \cfrac{3}{2 {x}^{2}} \right)}^{10}$$
Here, $$a = \sqrt{\cfrac{}{3}}, \quad b = \cfrac{3}{2 {x}^{2}}, \quad n = 10$$
Therefore,
$${T}_{r + 1} = {^{10}{C}_{r}} {\left( \sqrt{\cfrac{x}{3}} \right)}^{r} {\left( \cfrac{3}{2 {x}^{2}} \right)}^{10 - r}$$
$$\Rightarrow {T}_{r + 1} {^{10}{C}_{r}} \; {x}^{\left( \frac{r}{2} - 20 + 2r \right)} {3}^{\left( 10 - r - \frac{r}{2} \right)} {2}^{\left( r - 10 \right)} = {^{10}{C}_{r}} \; {x}^{\left( \frac{5r}{2} - 20 \right)} {3}^{\left( 10 - \frac{3r}{2} \right)} {2}^{\left( r - 10 \right)}$$
For the term independent of $$x$$-
$$\cfrac{5r}{2} - 20 = 0$$
$$\Rightarrow r = \cfrac{20 \times 2}{5} = 8$$
Hence $${8 + 1}^{th} = {9}^{th}$$ term will be independent of $$x$$.
Question 2
1 / -0

If the fourth term in the expansion of $$\left( p x + \dfrac { 1 } { x } \right) ^ { n }$$ is $$\dfrac { 5 } { 2 } ,$$ then $$n + p$$ is equal to

A
$$\dfrac { 9 } { 2 }$$

B
$$\dfrac { 11 } { 2 }$$

C
$$\dfrac { 13 } { 2 }$$

D
$$\dfrac { 15 } { 2 }$$

Solution
Question 3
1 / -0

The sum of the coefficients of the first three terms in the expansion of $${\left( {x - \frac{3}{{{x^2}}}} \right)^m},\,x \ne 0$$ $$m$$ being a natural number is , $$559$$. Find the term of the expansion containing $$x^3$$

A
-5940

B
1010

C
1001

D
1002

Solution

$$\begin{array}{l} { \left( { x-\frac { 3 }{ { { x^{ 2 } } } } } \right) ^{ m } }{ =^{ m } }{ C_{ o } }{ \left( x \right) ^{ m } }{ \left( { \frac { { -3 } }{ { { x^{ 2 } } } } } \right) ^{ 0 } }{ +^{ m } }{ C_{ 1 } }{ \left( x \right) ^{ m-1 } }{ \left( { \frac { { -3 } }{ { { x^{ 2 } } } } } \right) ^{ 1 } }{ +^{ m } }{ C_{ 2 } }{ x^{ m-2 } }{ \left( { \frac { { -3 } }{ { { x^{ 2 } } } } } \right) ^{ 2 } }+...... \\ { =^{ m } }{ C_{ o } }{ \left( x \right) ^{ m } }+{ \left( { -3 } \right) ^{ m } }{ C_{ 1 } }{ \left( x \right) ^{ m-1-2 } }+{ 9.^{ m } }{ C_{ 2 } }{ \left( x \right) ^{ m-2-4 } }+.... \\ { \Rightarrow ^{ m } }{ C_{ o } }+{ \left( { -3 } \right) ^{ m } }{ C_{ 1 } }+{ 9.^{ m } }{ C_{ 2 } }=559 \\ \Rightarrow 1-3m+9\frac { { m\left( { m-1 } \right) } }{ 2 } =559 \\ \Rightarrow 2-6m+9{ m^{ 2 } }-9m=2\times 559=1118 \\ \Rightarrow 9{ m^{ 2 } }-15m-1116=0 \\ \Rightarrow 3{ m^{ 2 } }-15m-372=0 \\ \Rightarrow 3{ m^{ 2 } }-36m+31m-372=0 \\ \Rightarrow 3m\left( { m-12 } \right) +31\left( { m-12 } \right) =0 \\ \Rightarrow \left( { m-12 } \right) \left( { 3m+31 } \right) =0 \\ \therefore m=12,\, \, and\, m=\frac { { -31 } }{ 3 } \left( { does\, not\, exist } \right) \\ so,\, we\, take\, m=12 \\ { \left( { x-\frac { 3 }{ { { x^{ 2 } } } } } \right) ^{ 12 } } \\ Then \\ { T_{ r+1 } }{ =^{ 12 } }{ C_{ r } }{ x^{ 12r } }{ \left( { \frac { { -3 } }{ { { x^{ 2 } } } } } \right) ^{ r } } \\ ={ \left( { -1 } \right) ^{ r } }{ 3^{ r } }^{ 12 }{ C_{ r } }{ x^{ 12-r-25 } } \\ ={ \left( { -1 } \right) ^{ r } }{ 3^{ r } }^{ 12 }{ C_{ r } }{ x^{ 12-3r } } \\ Containg\, { x^{ 3 } } \\ 12-3r=0 \\ \therefore r=3 \\ Now, \\ { T_{ 4 } }={ T_{ 3+1 } }={ \left( { -1 } \right) ^{ 3 } }{ 3^{ 3 } }\, { \, ^{ 12 } }{ C_{ 3 } }{ x^{ 3 } } \\ =-27\, \, { \, ^{ 12 } }{ C_{ 3 } }{ x^{ 3 } } \\ { T^{ 4 } }=-5940{ x^{ 3 } } \end{array}$$
hence, the option $$A$$ is the correct answer.
Question 4
1 / -0

$$ \binom{n}{0}+2\binom{n}{1}+2^{2}\binom{n}{2}+......++2^{n}\binom{n}{n} $$ is equal to

A
$$ 2^{n} $$

B
$$0$$

C
$$ 3^{n} $$

D
None of these

Solution

Consider $$(1+x)^n=1+^nC_1x+^nC_2x^2+.........+\,^nC_nx^n$$

Put $$x=2$$

$$3^n=1+2^nC_1+4^nC_2+......+2^n \,^nC_n$$

So correct answer is $$3^n$$

option (C)
Question 5
1 / -0

The coefficient of $$x ^ { 8 }$$ in the expansion of $$\left( 1 + x ^ { 4 } \right) ^ { 3 } ( 1 - x ) ^ { 12 }$$ is

A
$$3 + 4 \times 12 C _ { 4 }$$

B
$$3 + ^ { 12 } \mathrm { C } _ { 8 }$$

C
$$^ { 12 } C _ { 8 }$$

D
$$3 - ^ { 12 } \mathrm { C } _ { 8 }$$

Solution
Question 6
1 / -0

$$\dfrac{C_0}{1} + \dfrac{C_2}{3} + \dfrac{C_4}{5} + \dfrac{C_6}{7} ..... = $$

A
$$\dfrac{2^n}{n +1}$$

B
$$\dfrac{2^{n+1} - 1}{n + 1}$$

C
$$\dfrac{2^{n +1}}{n +1}$$

D
None of these

Solution

$$\begin{array}{l} \frac { { { C_{ o } } } }{ 1 } +\frac { { { C_{ 2 } } } }{ 3 } +\frac { { { C_{ 4 } } } }{ 5 } +\frac { { { C_{ 6 } } } }{ 7 } ..... \\ 1+\frac { { n\times \left( { n-1 } \right) } }{ { 3! } } +\frac { { n\times \left( { n-1 } \right) \left( { n-2 } \right) \left( { n-3 } \right) } }{ 5 } +...... \\ \frac { 1 }{ { n+1 } } \left[ { \left( { n+1 } \right) \frac { { \left( { n+1 } \right) \times n\times \left( { n-1 } \right) } }{ { 3! } } +\frac { { n\left( { n+1 } \right) \left( { n-1 } \right) \left( { n-2 } \right) } }{ { 5! } } } \right] \\ \frac { 1 }{ { n+1 } } \left[ { { 2^{ n+1-1 } } } \right] \\ \frac { { { 2^{ n } } } }{ { n+1 } } \end{array}$$
Hence the option $$A$$ is the correct answer.
Question 7
1 / -0

$$C_1 +2C_2 + 3C_3 +4C_4 + ......... + {n+1} nC_n$$

A
$$n2^{n-1}$$

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

C
$$2.2^{n-1}$$

D
Zero

Solution

$$\begin{array}{l} { C_{ 1 } }+2{ C_{ 2 } }+3{ C_{ 3 } }+......+n{ C_{ n } } \\ =n+2\times \frac { { n\left( { n-1 } \right) } }{ { 2! } } +3\times \frac { { n\left( { n-1 } \right) \left( { n-2 } \right) } }{ { 3! } } +....+n\times 1 \\ =n+2\times \frac { { n\left( { n-1 } \right) } }{ 1 } +3\times \frac { { n\left( { n-1 } \right) } }{ 6 } +....+n \\ =n+\frac { { n\left( { n-1 } \right) } }{ 1 } +\frac { { n\left( { n-1 } \right) } }{ 2 } +.....+n \\ =n\left[ { 1+\frac { { n-1 } }{ 1 } +\frac { { n\left( { n-1 } \right) } }{ 2 } +.....+1 } \right] \\ =n\left[ { 1+\frac { { n-1 } }{ { 1! } } +\frac { { n\left( { n-1 } \right) } }{ { 2! } } +.....+1 } \right] \\ put\, \, \, \, \, n-1=N \\ =n\left[ { 1+\frac { N }{ { 1! } } +\frac { { N\left( { N+1 } \right) } }{ { 2! } } +.....+1 } \right] \\ =n\left( { ^{ N }{ C_{ 0 } }{ +^{ N } }{ C_{ 1 } }{ +^{ N } }{ C_{ 2 } }+....{ +^{ N } }{ C_{ N } } } \right) \\ =n{ 2^{ N } } \\ =n{ 2^{ n-1 } } \\ Hence,\, the\, option\, A\, is\, the\, correct\, answer \end{array}$$
Question 8
1 / -0

$$(1+x)^{21}+(1+x)^{22}+..+(1+x)^{30}$$ in the expansion of this what is the coefficient of $$x^{5}$$ is

A
$$^{31} C_5-^{21} C_5$$

B
$$^{31} C_4-^{21} C_4$$

C
$$^{31} C_6-^{21} C_6$$

D
$$None\ of\ these$$

Solution
Question 9
1 / -0

The sum $$^ { 20 } \mathrm { C } _ { 0 } + ^ { 20 } \mathrm { C } _ { 1 } + ^ { 20 } \mathrm { C } _ { 2 } + \ldots \ldots . ^ { 20 } \mathrm { C } _ { 10 }$$ is equal to

A
$$2 ^ { 19 } + \frac { 20 ! } { ( 10 ! ) ^ { 2 } }$$

B
$$2 ^ { 19 } - \frac { 1 } { 2 } \cdot \frac { 20 ! } { ( 10 ! ) ^ { 2 } }$$

C
$$2 ^ { 19 } + ^ { 20 } \mathrm { C } _ { 10 }$$

D
none of these

Solution

As we know that $$(1+x)^{20}=^{20}C_{0}+^{20}C_{1} x+^{20}C_{2} x^2+\cdots+^{20}C_{20} x^{20}$$

Put $$x=1$$

$$\implies 2^{20}=^{20}C_{0}+^{20}C_{1}+^{20}C_2+\cdots+^{20}C_{20}$$
$$\implies 2^{20}=^{20}C_{0}+^{20}C_{1}+\cdots+^{20}C_{10}+^{20}C_{9}+\cdots+^{20}C_{1}+^{20}C_{0}$$

$$2^{20}=2(^{20}C_0+^{20}C_1+^{20}C_2+\cdots+^{20}C_{10})-^{20}C_{10}$$

$$\implies ^{20}C_{0}+^{20}C_1+^{20}C_2+\cdots+^{20}C_{10}=2^{19}+\dfrac{1}{2}^{20}C_{10}=2^{19}+\dfrac{1}{2}\cdot \dfrac{20!}{(10!)^2}$$
Question 10
1 / -0

Coefficient of $$x^ {79}$$ in the expansion of $$\left(x+x^ {2}+x^ {4}\right)$$ is equal to-

A
$$0$$

B
$$150x3^ {19}$$

C
$$1$$

D
$$3^ {20}$$

Solution

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Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 50

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Binomial Theorem Test - 50

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Rank

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SHARING IS CARING

Question 1

$$3$$

$$5$$

$$9$$

$$None\ of\ these$$

Solution

Question 2

$$\dfrac { 9 } { 2 }$$

$$\dfrac { 11 } { 2 }$$

$$\dfrac { 13 } { 2 }$$

$$\dfrac { 15 } { 2 }$$

Solution

Question 3

-5940

1010

1001

1002

Solution

Question 4

$$ 2^{n} $$

$$0$$

$$ 3^{n} $$

None of these

Solution

Question 5

$$3 + 4 \times 12 C _ { 4 }$$

$$3 + ^ { 12 } \mathrm { C } _ { 8 }$$

$$^ { 12 } C _ { 8 }$$

$$3 - ^ { 12 } \mathrm { C } _ { 8 }$$

Solution

Question 6

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

$$\dfrac{2^{n+1} - 1}{n + 1}$$

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

None of these

Solution

Question 7

$$n2^{n-1}$$

$$2^{n+1}$$

$$2.2^{n-1}$$

Zero

Solution

Question 8

$$^{31} C_5-^{21} C_5$$

$$^{31} C_4-^{21} C_4$$

$$^{31} C_6-^{21} C_6$$

$$None\ of\ these$$

Solution

Question 9

$$2 ^ { 19 } + \frac { 20 ! } { ( 10 ! ) ^ { 2 } }$$

$$2 ^ { 19 } - \frac { 1 } { 2 } \cdot \frac { 20 ! } { ( 10 ! ) ^ { 2 } }$$

$$2 ^ { 19 } + ^ { 20 } \mathrm { C } _ { 10 }$$

none of these

Solution

Question 10

$$0$$

$$150x3^ {19}$$

$$1$$

$$3^ {20}$$

Solution

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