Binomial Theorem Test

Result

Binomial Theorem Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

If the coefficients of $$2^{nd}, 3^{rd},$$ and $$4^{th}$$ terms in the expansion of $${(1 + x)^n},n \in N$$ are in A.P.,then $$n=$$

A
$$7$$

B
$$14$$

C
$$2$$

D
None of these

Solution
Question 2
1 / -0

The coefficient of $${ x }^{ n }$$ in $$(1-x+\frac { { x }^{ 2 } }{ 2! } -\frac { { x }^{ 3 } }{ 3! } +.....+\frac { { (-1) }^{ n }{ x }^{ n } }{ n! } )^{ 2 }$$ is

A
$$\frac { { (-n) }^{ n } }{ n! } $$

B
$$\frac { { (-2) }^{ n } }{ n! } $$

C
$$\frac { 1 }{ (n!)^{ 2 } } $$

D
$$-\frac { 1 }{ (n!)^{ 2 } } $$

Solution
Question 3
1 / -0

The middle term in the expansion of $$\left(\dfrac{2}{3}x-\dfrac{3}{2}y\right)^{20}$$ is

A
$$_{ }^{ 20 }{ { C }_{ 10 } }{ x }^{ 10 }{ y }^{ 10 }$$

B
$${ x }^{ 10 }{ y }^{ 10 }$$

C
$$_{ }^{ 20 }{ { C }_{ 10 } }{ \left( 2/3 \right) }^{ 10 }{ \left( xy \right) }^{ 10 }$$

D
$$-x^{ 10 }y^{ 10 }$$

Solution

$$\left(\frac{2 }{3} x-\frac{3}{2} y\right)^{20}\\$$
$$\text { There will be total } 21 \text { terms in this exapnsion } \\$$
$$\text { : : } 11 \text { th term will be middle tem. }$$

$$T_{r+1=}{}^n C_{r} \cdot x^{n-r} \cdot y^{r}$$

$$\Rightarrow T_{11}={ }^{20} c_{10} \cdot\left(\frac{2}{3} x\right)^{20-10}\left(\frac{-3}{2} y\right)^{10}$$

$$={}^{20} {c_{10}} x^{10} \cdot y^{10}$$

option (a)
Question 4
1 / -0

The coefficient of $$x^8$$ in the expansion of $$(1+x+x^3+x^5+x^9)(1+x^2)^5(1_x^4)^6$$ is equal to

A
180

B
100

C
80

D
None of these
Question 5
1 / -0

If coefficient of $$2^{nd}, 3^{rd}$$ and $$4^{th}$$ term in the expansion of $$\left(1+x\right)^{2n}$$ are in A.P. then :

A
$$2n^{2}+9n+7=0$$

B
$$2n^{2}-9n+7=0$$

C
$$2n^{2}-9n-7=0$$

D
$$2n^{2}+9n-7=0$$

Solution

$$^{2n}C_{1},^{2n}C_{2}, ^{2n}C_{3}$$ are in A.P
$$\therefore 2\times ^{2n}C_{2}= ^{2n}C_{1}+^{2n}C_{3}$$
$$=2\times (\frac{2n(2n-1)}{1\times 2})=\frac{2n}{1}+\frac{2n(2n-1)(2n-2)}{1\times 2\times 3}$$
$$=2n-1=1+\frac{(2n-1)(n-1)}{3}$$
$$= 6n-3=3n+2n^{2}-3n+1$$
$$\Rightarrow 2n^{2}-9n+7=0$$
Question 6
1 / -0

The coefficient of $$a^3b^4c$$ in the expansion of $$(1+a+b-c)^9$$ is

A
$$2.^9C_7.^7C_4$$

B
$$-2.^9C_2.^7C_3$$

C
$$^9C_7.^7C_4$$

D
none of these

Solution
Question 7
1 / -0

If the fourth term in the expansion of $$\left(px+\dfrac{1}{x}\right)^{n}$$ is independent of $$x$$, then the value of term is :

A
$$5p^{3}$$

B
$$10p^{3}$$

C
$$20p^{3}$$

D
$$None\ of\ these$$
Question 8
1 / -0

If in the expansion of $$\left(2^{1/3}+\dfrac {1}{3^{1/3}}\right)^{n}$$, the ratio of $$6^{th}$$ term from beginning and from the end is $$1/6$$, then the value of $$n$$ is

A
$$5$$

B
$$7$$

C
$$13$$

D
$$None\ of\ these$$

Solution

As we know that in an expansion $${\left( a + b \right)}^{n}$$, the general term is given as-
$${T}_{r + 1} = {^{n}{C}_{r}} {a}^{n} {b}^{n-r}$$
In the given expansion-
$${\left( {2}^{{1}/{3}} + \cfrac{1}{{3}^{{1}/{3}}} \right)}^{n}$$
$$a = {2}^{{1}/{3}}, \quad b = \cfrac{1}{{3}^{{1}/{3}}}, \quad n = ?$$
Therefore,
$$\Rightarrow \cfrac{{^{n}{C}_{5}} {\left( {2}^{{1}/{3}} \right)}^{5} {\left( \cfrac{1}{{3}^{{1}/{3}}} \right)}^{n-5}}{{^{n}{C}_{n-5}} {\left( {2}^{{1}/{3}} \right)}^{n-5} {\left( \cfrac{1}{{3}^{{1}/{3}}} \right)}^{5}} = \cfrac{1}{6} \quad \left( \text{Given} \right)$$
$$\Rightarrow \cfrac{1}{{\left( {2}^{{1}/{3}} \right)}^{n-10} {\left( {3}^{{1}/{3}} \right)}^{n-10}} = \cfrac{1}{6}$$
$$\Rightarrow \cfrac{1}{{\left( 6 \right)}^{\frac{n-10}{3}}} = \cfrac{1}{{\left( 6 \right)}^{1}}$$
$$\Rightarrow \cfrac{n - 10}{3} = 1$$
$$\Rightarrow n - 10 = 3$$
$$\Rightarrow n = 10 + 3 = 13$$
Question 9
1 / -0

Find the number of terms in expansion of $$(1+x)^{2}+(1-x)^{8}$$

A
$$17$$

B
$$18$$

C
$$5$$

D
$$9$$

Solution
Question 10
1 / -0

Middle term in the expansion of $$(1+3x+3x^2+x^3)^6$$ is

A
$$4^{th}$$

B
$$3^{rd}$$

C
$$10^{th}$$

D
None of these

Solution