Binomial Theorem Test

Result

Binomial Theorem Test - 70

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Weekly Quiz Competition

Question 1
1 / -0

The number of irrational terms in the expansion of$$( \sqrt [ 8 ] { 5 } + \sqrt [ 6 ] { 2 } ) ^ { 100 } ,$$ is

A
$$97$$

B
$$98$$

C
$$96$$

D
$$99$$

Solution
Question 2
1 / -0

The number of irrational terms in the expansion of$$( \sqrt [ 8 ] { 5 } + \sqrt [ 6 ] { 2 } ) ^ { 100 } ,$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$
Question 3
1 / -0

The value of m, for which the coefficients of the $$\left( {2m + 1} \right)$$ and $${\left( {4m + 5} \right)^{th}}$$ terms in the expansion $${\left( {1 + x} \right)^{10}}$$ are equal, is:

A
3

B
1

C
5

D
8

Solution
Question 4
1 / -0

The number of terms in the expansion of $$ ( 2 x + 3 y - 4 z ) ^ { n } $$ ,is

A
$$ n + 1 $$

B
$$ n + 3 $$

C
$$ \frac { ( \mathrm { n } + 1 ) ( \mathrm { n } + 2 ) } { 2 } \quad $$

D
none of these

Solution
Question 5
1 / -0

COEFFICIENT OF $$x^k$$ IN $$(ax + b)^n$$
The coefficient if x in the expansion of $$\displaystyle (1 2x^3 + 3x^5)\left ( 1 + \dfrac{1}{x} \right )^8 $$ is

A
$$56$$

B
$$65$$

C
$$154$$

D
$$62$$

Solution
Question 6
1 / -0

In the expansion of $${ \left( { x }^{ 2 }+1+\frac { 1 }{ { x }^{ 2 } } \right) }^{ n }$$, $$n\in N$$, then

A
number of terms= 2n+1

B
term independent of x=$${ 2 }^{ n-1 }$$

C
coefficient of $${ x }^{ 2n-2 }$$= n

D
coefficient of $${ x }^{ 2 }$$= n

Solution
Question 7
1 / -0

The sum of coefficients of integral powers of $$x$$ in the binomial expansion of $${\left(1-2\sqrt{x}\right)}^{-n}$$ is

A
$$\frac{1}{2}\left({3}^{50}-1\right)$$

B
$$\frac{1}{2}\left({2}^{50}+1\right)$$

C
$$\frac{1}{2}\left({3}^{50}+1\right)$$

D
$$\frac{1}{2}\left({3}^{50}\right)$$

Solution
Question 8
1 / -0

The coefficient of $$x ^ { 4 }$$ in the expansion of $$\left( 1 + x + x ^ { 2 } + x ^ { 3 } \right) ^ { 11 }$$ is

A
$$900$$

B
$$909$$

C
$$990$$

D
$$999$$

Solution
Question 9
1 / -0

In the expression of $$(1+x)^{50}$$. Let S be the sum of coefficient of odd power of x. then S is:

A
0

B
$$2^{49}$$

C
$$2^{50}$$

D
$$2^{51}$$

Solution
Question 10
1 / -0

If the last term in the binomial expansion of $$ (2{ }^{ \frac { 1 }{ 3 } }-\frac { 1 }{ \sqrt { 2 } } ){ }^{ n }\quad is\quad (\frac { 1 }{ { 3 }^{ \frac { 5 }{ 3 } } } ){ }^{ log{ }_{ 3 }8 }$$ , then the 5th term from the beginning is

A
210

B
420

C
105

D
none of these

Solution

Given,

$$\left ( 2^{\frac{1}{3}}-\dfrac{1}{\sqrt 2} \right )^n$$

last term of $$(x-a)^n=(-1)^n(a)^n$$

last term of given equation,

$$(-1)^n\left ( \dfrac{1}{2^{\frac{n}{2}}} \right )=\left ( \dfrac{1}{3\cdot 9^{\frac{1}{3}}} \right )^{\log_3 8}$$

$$=\left ( \dfrac{1}{3\cdot 3^{\frac{2}{3}}} \right )^{\log_3 8}$$

$$=\left ( \dfrac{1}{ 3^{\frac{5}{3}}} \right )^{\log_3 8}$$

$$=\left ( \dfrac{1}{ 3^{\frac{5}{3}\log_3 8}} \right )$$

$$=\left ( \dfrac{1}{ 3^{\log_3 8^{\frac{5}{3}}}} \right )$$

$$=\dfrac{1}{8^{\frac{5}{3}}}$$

$$=\dfrac{1}{2^5}$$

$$\Rightarrow \dfrac{n}{2}=5$$

$$\therefore n=10$$

now,

$$\left ( 2^{\frac{1}{3}}-\dfrac{1}{\sqrt 2} \right )^{10}$$

$$T_5=T_{4+1}=^{10}C_4(2^{\frac{1}{3}})^{10-4}\left ( \dfrac{1}{2^{\frac{1}{2}}} \right )^4(-1)^4$$

$$=^{10}C_4(2^{\frac{6}{3}})\left ( \dfrac{1}{2^{\frac{4}{2}}} \right )$$

$$=^{10}C_4(2^2)\left ( \dfrac{1}{2^2} \right )$$

$$=^{10}C_4$$

$$=210$$