Binomial Theorem Test

Result

Binomial Theorem Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

Coefficient of $$x^ {15}$$ in $$(1+x+x^ {3}+x^ {4})$$^ {n} is

A
$$\displaystyle \sum _{ r=0 }^{ 5 }{ ^{ n } } { C }_{ 5-r }.^{ n }{ C }_{ 3r }$$

B
$$\displaystyle \sum _{ r=0 }^{ 5 }{ ^{ n } } { C }_{ 5r }$$

C
$$\displaystyle \sum _{ r=0 }^{ 5 }{ ^{ n } } { C }_{ 2r }$$

D
$$\displaystyle \sum _{ r=0 }^{ 3 }{ ^{ n } } { C }_{ 3-r }.^{ n }{ C }_{ 5r }$$

Solution
Question 2
1 / -0

The coefficient of $$x^m$$ in the expansion of $$(1+)^{m+n}$$ is -

A
$$\frac { m!n! }{ (m+n)! } $$

B
$$(m+n)!$$

C
$$\frac { (m+n)! }{ m!n! } $$

D
none of these

Solution
Question 3
1 / -0

The 11$$^{th}$$ term from last, in expansion of (2x+$$\frac{1}{^{x2}}$$ is

A
$$^{25}C_{15}$$$$\frac{2^{10}}{x^{20}}$$

B
$$^{-25}C_{15}\frac{2^{10}}{x^{20}}$$

C
$$^{25}C_{14}\frac{2^{11}}{x^{11}}$$

D
$$^{25}C_{15}\frac{2^{10}}{x^{20}}$$

Solution
Question 4
1 / -0

The total number of irrational terms in the binomial expansion of $${ \left( { 7 }^{ 1/5 }-{ 3 }^{ 1/10 } \right) }^{ 60 }$$ is :

A
49

B
48

C
54

D
55

Solution
Question 5
1 / -0

the number of terms in the expansion of $${\left( {x + y - x} \right)^3}$$ is

A
4

B
6

C
9

D
10

Solution
Question 6
1 / -0

The number of distinct terms in the expansion of $${ \left( { x+y }^{ 2 } \right) }^{ 13 }+{ \left( { x }^{ 2 }+y \right) }^{ 14 }$$ is...

A
29

B
28

C
25

D
27

Solution
Question 7
1 / -0

Find the middle terms of the equation of $$\left(x^4 - \dfrac{1}{x^3}\right)^{11}$$.

A
$$-462\ x^9, 462\ x^2$$

B
$$-462\ x^8, 462\ x^4$$

C
$$462\ x^7, -462\ x^3$$

D
None of these

Solution

Given expression is $$\left(x^4-\dfrac{1}{x^3}\right)^{11}$$
Here index, $$n=11$$ (odd)
so, there are two middle terms, which are $$\left(\dfrac{11+1}{2}\right)^{th}$$ i.e, $$6^{th}$$ term and $$\left(\dfrac{11+3}{2}\right )^{th}$$ i.e $$7^{th}$$ term.

$$T_6=T_{5+1}=^{11}C_5(x^4)^{11-5}\left(\dfrac{-1}{x^3}\right)^5$$

$$=\dfrac{11!}{6!5!}(x^4)^6 \left(\dfrac{-1}{x^{15}}\right)$$

$$=\dfrac{-11\times 10\times 9\times 8\times 7\times 6!}{6!\times 5\times 4\times 3\times 2} \dfrac{x^{24}}{x^{15}}$$

$$=-462 x^9$$

$$T_7=T_{6+1}=^{11}C_6(x^4)^{11-6}\left(\dfrac{-1}{x^3}\right)^6$$

$$=\dfrac{11!}{5!\times 6!}(x^4)^5 \left(\dfrac{1}{x^{18}}\right)$$

$$=462 x^2$$
Question 8
1 / -0

The coefficient of $$ x^{2 n+1} $$ in the expansion of $$ E=\frac{1}{(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right) \ldots\left(1+x^{2 m}\right)}|x|<1 $$ is

A
$$3$$

B
$$2$$

C
$$1$$

D
$$0$$

Solution
Question 9
1 / -0

The coefficient of $$ x^{5} $$ in the expansion of $$ \left(x^{2}-x-2\right)^{5} $$ is

A
$$-83$$

B
$$-82$$

C
$$-81$$

D
$$0$$

Solution

$${\textbf{Step -1: Identify binomial coefficients and number of terms in a binomial expansion.}}$$

$$(x^2-x-2)^5=(x^2-2x+x-2)^5$$

$$=(x(x-2)+1(x-2))^5$$

$$=((x-2)(x+1))^5$$

$$=(x-2)^5(x+1)^5$$

$$=[^{5}C_{0}x^5+^{5}C_{1}x^4.(-2)+^{5}C_{2}x^3.(-2)^2+^{5}C_{3}x^2.(-2)^3+^{5}C_{4}x.(-2)^4+(-2)^5]$$

$$\times[^{5}C_{0}x^5+^{5}C_{1}x^4+^{5}C_{2}x^3+^{5}C_{3}x^2+^{5}C_{4}x+1]$$

$$\therefore\text{coefficient of }x^5\text{ in the expansion of the product }(x-2)^5(x+1)^5$$

$$=-2^5+1+^5C_2 .^5C_3(-2)^3+^5C_3 .^5C_2(-2)^2+^5C_4 .^5C_1(-2)^1+^5C_1 .^5C_4(-2)^4$$

$$=-32+1-800+400-50+400$$

$$=-81$$

$${\textbf{Hence, option C is correct.}}$$
Question 10
1 / -0

Coefficient of $$ x^{5} $$ in the expansion of $$ \left(x^{2}-x-2\right)^{5} $$ is

A
$$-82$$

B
$$-83$$

C
$$-81$$

D
$$0$$

Solution

$${\textbf{Step -1: Identify binomial coefficients and number of terms in a binomial expansion.}}$$
$$(x^2-x-2)^5=(x^2-2x+x-2)^5$$
$$=(x(x-2)+1(x-2))^5$$
$$=((x-2)(x+1))^5$$
$$=(x-2)^5(x+1)^5$$
$$=[^{5}C_{0}x^5+^{5}C_{1}x^4.(-2)+^{5}C_{2}x^3.(-2)^2+^{5}C_{3}x^2.(-2)^3+^{5}C_{4}x.(-2)^4+(-2)^5]$$
$$\times[^{5}C_{0}x^5+^{5}C_{1}x^4+^{5}C_{2}x^3+^{5}C_{3}x^2+^{5}C_{4}x+1]$$
$$\therefore\text{coefficient of }x^5\text{ in the expansion of the product }(x-2)^5(x+1)^5$$
$$=-2^5+1+^5C_2 .^5C_3(-2)^3+^5C_3 .^5C_2(-2)^2+^5C_4 .^5C_1(-2)^1+^5C_1 .^5C_4(-2)^4$$
$$=-32+1-800+400-50+400$$
$$=-81$$
$${\textbf{Hence, option C is correct.}}$$