Binomial Theorem Test

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Binomial Theorem Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

The number of real negative terms in the bionomial expansion of $$ (1+ix)^{4n-2} ,n \in N,x>0 $$ is

A
$$n$$

B
$$n+1$$

C
$$n-1$$

D
$$2n$$

Solution

$${ \left( 1+ix \right) }^{ 4x-2 }=\sum _{ r=0 }^{ ^{ 4x-2 } }{ ^{ ^{ 4x-2 } }{ { C }_{ r } } } { \left( ix \right) }^{ r }$$
For real negative terms, we have to get $$i^2;$$ Since $$(i^2=-1).$$ For that, $$r$$ must be equal to $$4k-2,$$ where $$k\in I$$
Hence, the total real negative terms in the expansion$$ =n$$
$$p.s.$$     $$i=\sqrt { -1 } $$
$$\Rightarrow$$       $$i^2=-1$$
$$\Rightarrow$$       $$i^3=-\sqrt { -1 } =-i$$
$$\Rightarrow$$       $$i^4={ \left( { i }^{ 2 } \right) }^{ 2 }=(-1)^2=1$$
Hence, the answer is $$n.$$
Question 2
1 / -0

Find $$7^{th}$$ term of $$\displaystyle \left ( \frac{4x}{5}-\frac{5}{2x} \right )^{9}$$

A
$$\dfrac{10050}{x^{3}}$$

B
$$\dfrac{10500}{x^{3}}$$

C
$$\dfrac{1050}{x^{3}}$$

D
$$\dfrac{1000}{x^{3}}$$

Solution

$$7^{th}$$ term of $$\displaystyle \left ( \frac{4x}{5}-\frac{5}{2x} \right )^{9}$$
$$\displaystyle T_{6+1}=^{9}\textrm{C}_{6}\left ( \frac{4x}{5} \right )^{9-6}\left ( -\frac{5}{2x} \right )^{6}$$
$$\displaystyle =\frac{9!}{3!6!}\left ( \frac{4x}{5} \right )^{3}\left ( \frac{5}{2x} \right )^{6}=\frac{10500}{x^{3}}$$
Question 3
1 / -0

Find $$28^{th}$$ term of $$\left ( 5x+8y \right )^{30}$$

A
$$^{30}\textrm{C}_{27}\left ( 5x \right )^{27}\left ( 8y \right )^{3}$$

B
$$^{30}\textrm{C}_{28}\left ( 5x \right )^{2}\left ( 8y \right )^{28}$$

C
$$^{30}\textrm{C}_{27}\left ( 5x \right )^{3}\left ( 8y \right )^{27}$$

D
$$^{30}\textrm{C}_{28}\left ( 5x \right )^{28}\left ( 8y \right )^{2}$$

Solution

$$\displaystyle T_{27+1}=^{30}\textrm{C}_{27}\left ( 5x \right )^{30-27}\left ( 8y \right )^{27}=\frac{30!}{3!27!}\left ( 5x \right )^{3}.\left ( 8y \right )^{27}$$
Question 4
1 / -0

The number of integral terms in $${(\sqrt {3}+\sqrt [ 8 ]{ 2 } )}^{64}$$ is-

A
$$8$$

B
$$7$$

C
$$9$$

D
$$6$$

Solution

The general term of expansion $$(x+y)^{n}$$ is $${ ^{ n }{ C } }_{ r }{ x }^{ n-r }{ y }^{ r }$$
So the general term of $${ (\sqrt { 3 } +\sqrt [ 8 ]{ 2 } ) }^{ 64 }$$ is $${ ^{ 64 }{ C } }_{ r }{ 3 }^{ \frac { 64-r }{ 2 } }{ 2 }^{ \frac { r }{ 8 } }$$
For the term to be integer , $$r$$ must be divided by $$8$$ and $$64-r$$ must be divided by $$2$$
The possible values of $$r$$ are $$0,8,16,24,32,40,48,56,64$$
the number of integral values is $$9$$
Question 5
1 / -0

Value of $$\displaystyle \sum_{k=1}^{\infty} \sum _{r=0}^{k} \frac{1}{3^{k}} (^{k}C_{r})$$ is

A
$$\displaystyle \frac{2}{3}$$

B
$$\displaystyle \frac{4}{3}$$

C
$$2$$

D
$$1$$

Solution

$$\sum _{ r=0 }^{ k }{ \dfrac { 1 }{ { 3 }^{ k } } } \left( { ^{ k }{ C } }_{ r } \right) $$
$$\Rightarrow \dfrac { 1 }{ { 3 }^{ k } } \sum _{ r=0 }^{ k }{ \left( { ^{ k }{ C } }_{ r } \right) } =\dfrac { 1 }{ { 3 }^{ k } }.2^k$$
$$={ \left( \dfrac { 2 }{ 3 } \right) }^{ k }$$
$$\sum _{ k=1 }^{ \infty }{ { \left( \dfrac { 2 }{ 3 } \right) }^{ k } } $$
$$=\dfrac{2}{3}+{ \left( \dfrac { 2 }{ 3 } \right) }^{ 2 }+{ \left( \dfrac { 2 }{ 3 } \right) }^{ 3 }........$$
Sum of an infinite GP
$$\Rightarrow S=\sum _{ k=1 }^{ \infty }{ { \left( \dfrac { 2 }{ 3 } \right) }^{ k } } =\dfrac { \dfrac { 2 }{ 3 } }{ 1-\dfrac { 2 }{ 3 } }=2 $$
Hence, the answer is $$2.$$
Question 6
1 / -0

If the sum of the coefficients in the expansion $$(1-3x+10x^{2} )^{n} $$ is $$a$$ and sum of coefficients in the expansion of $$ (1+x^{2})^{n}$$is b,then

A
$$a=3b$$

B
$$a=b^{3}$$

C
$$b=a^{3}$$

D
none of these

Solution

$${ \left( 1-3x+10{ x }^{ 2 } \right) }^{ n }=\sum { \dfrac { (n)! }{ ({ r }_{ 1 })!({ r }_{ 2 })!({ r }_{ 3 })! } } \left\{ { (1) }^{ { r }_{ 1 } }.({ -3x) }^{ { r }_{ 2 } }.({ 10x) }^{ { r }_{ 3 } } \right\} $$
We can get the sum of the coefficient by taking $$x=1$$
Hence, $$a={(1-3+10)}^n=8^n=2^{3n}$$
Similarly $${(1+x^2)}^n=\sum _{ r=0 }^{ n }{ ^{ n }{ { C }_{ r } } } ({ x }^{ 2 }{ ) }^{ r }$$
Taking $$x=1;$$ $$b=2^n$$
Hence, $$a=b^3$$
Hence, the answer is $$a=b^3.$$
Question 7
1 / -0

Which term is the constant term in the expansion of $$\left ( 2x\, -\, \frac{7}{3x} \right )^6$$ ?

A
2nd term

B
3rd term

C
4th term

D
5th term

Solution
Question 8
1 / -0

The tenth term in the expansion of $$\left (2x^2 + \dfrac{1}{x}\right )^{12}$$ is:

A
$$\dfrac{1760}{x^3}$$

B
$$-\dfrac{1760}{x^3}$$

C
$$\dfrac{1760}{x^2}$$

D
none of the above

Solution

Given
$$\left ( 2x^2+\dfrac{1}{x} \right )^{12}$$
General term of given expansion is
$$T_{r+1}=\sum_{r=0}^{12}\displaystyle^{12}C_{r}(2x^2)^{12-r}\left( \dfrac{1}{x} \right )^{r} $$
We need $$10^{th}$$ term hence put $$r=9$$
$$T_{9+1}=\displaystyle^{12}C_{9}(2x^2)^{12-9}\left( \dfrac{1}{x} \right )^{9} $$

$$T_{10}=\displaystyle^{12}C_{9}(2x^2)^{3}\left( \dfrac{1}{x} \right )^{9} $$

$$T_{10}=\displaystyle^{12}C_{9}\cdot 2^3 (x)^{6-9} $$

$$T_{10}=\dfrac{12!}{9!3!}\cdot 8 (x)^{-3} $$

$$T_{10}=\dfrac{12\times 11 \times 10 }{3\times 2\times x^3}\cdot 8 $$

$$T_{10}=\dfrac{1760}{x^3} $$
Question 9
1 / -0

The $$7^{th}$$ term from the end in the expansion of $$\left (x-\dfrac {2}{x^2}\right )^{10}$$ is equal to:

A
$$^{10}C_4\cdot 2^4\left (\dfrac {1}{x^2}\right )$$

B
$$^{10}C_4\cdot 2^4$$

C
$$-^{10}C_4\cdot 2^3x$$

D
none of the above

Solution

Given
$$\left ( x-\dfrac{2}{x^2} \right )^{10}$$
General term of given expansion is
$$T_{r+1}=\sum_{r=0}^{10}\displaystyle^{10}C_{r}(x)^{10-r}\left( \dfrac{2}{x^2} \right )^{r} (-1)^r$$
We know that
The $$r^{th}$$ term from end is equal to $$(n-r+2)^{th}$$ term from begining
Hence
We need $$7^{th}$$ term from end i.e $$(10-7+2)^{th}=5^{th}$$ term from end So put $$r=4$$
$$T_{4+1}=\displaystyle^{10}C_{4}(x)^{10-4}\left( \dfrac{2}{x^2} \right )^{4} (-1)^4$$

$$T_{5}=\displaystyle^{10}C_{4}(x)^{6}\left( \dfrac{2^4}{x^8} \right ) $$

$$T_{5}=\displaystyle^{10}C_{4}\cdot 2^4 \left(\dfrac{1}{x^2}\right) $$
Question 10
1 / -0

If the middle term in the expansion of $$\left (x^2+\dfrac {1}{x}\right )^n$$ is $$924x^6$$, then $$n=$$

A
$$10$$

B
$$12$$

C
$$14$$

D
$$none\ of\ these$$

Solution

For $$ \left ( x^{2}+\dfrac{1}{x} \right )^{n} $$ have middle term.
n is even
$$ \Rightarrow n = 2a; aEN $$
middle term $$ ^{2a}C_{a} (x^{2})^{a}.\dfrac{1}{x^{a}} = ^{2a}C_{a}.x^{a} $$
$$ = 924x^{6} $$
$$ \Rightarrow a = 6 $$
Also for $$ a = 6, ^{2a}C_{a} = ^{12}C_{6} = 924 $$
So, $$ n = 2a = 12 $$

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Binomial Theorem Test - 35

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Binomial Theorem Test - 35

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Question 1

$$n$$

$$n+1$$

$$n-1$$

$$2n$$

Solution

Question 2

$$\dfrac{10050}{x^{3}}$$

$$\dfrac{10500}{x^{3}}$$

$$\dfrac{1050}{x^{3}}$$

$$\dfrac{1000}{x^{3}}$$

Solution

Question 3

$$^{30}\textrm{C}_{27}\left ( 5x \right )^{27}\left ( 8y \right )^{3}$$

$$^{30}\textrm{C}_{28}\left ( 5x \right )^{2}\left ( 8y \right )^{28}$$

$$^{30}\textrm{C}_{27}\left ( 5x \right )^{3}\left ( 8y \right )^{27}$$

$$^{30}\textrm{C}_{28}\left ( 5x \right )^{28}\left ( 8y \right )^{2}$$

Solution

Question 4

$$8$$

$$7$$

$$9$$

$$6$$

Solution

Question 5

$$\displaystyle \frac{2}{3}$$

$$\displaystyle \frac{4}{3}$$

$$2$$

$$1$$

Solution

Question 6

$$a=3b$$

$$a=b^{3}$$

$$b=a^{3}$$

none of these

Solution

Question 7

2nd term

3rd term

4th term

5th term

Solution

Question 8

$$\dfrac{1760}{x^3}$$

$$-\dfrac{1760}{x^3}$$

$$\dfrac{1760}{x^2}$$

none of the above

Solution

Question 9

$$^{10}C_4\cdot 2^4\left (\dfrac {1}{x^2}\right )$$

$$^{10}C_4\cdot 2^4$$

$$-^{10}C_4\cdot 2^3x$$

none of the above

Solution

Question 10

$$10$$

$$12$$

$$14$$

$$none\ of\ these$$

Solution

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