Binomial Theorem Test

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Binomial Theorem Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

If the expansion of $$\displaystyle\left(x^3+\frac{1}{x^2}\right)^n$$ contains a term independent of x, then the value of n can be

A
18

B
20

C
24

D
22

Solution

Given, $$\left(x^3+\dfrac{1}{x^2}\right)^n$$
this contains a term independent of x.
so general term $$T_{r+1}$$

$$T_{r+1}={^{n}C_r}x^{n-r}y^r\\$$
$$T_{r+1}={^{n}C_r}(x^3)^{n-r}\left(\dfrac{1}{x^2}\right)^r\\$$
$$T_{r+1}={^{n}C_r}\dfrac{(x)^{3n-3r}}{x^{2r}}\\$$
$$T_{r+1}={^{n}C_r}(x)^{3n-5r}$$

Since, given term is indepedent of x.
so, $$3n-5r=0$$
$$3n=5r$$
$$n=\dfrac{5r}{3}$$

Since n should be an integer, using options
if $$n=9$$ $$n=\dfrac{5\times 9}{3}$$ $$n=15$$ (not in options)

so, $$r=15$$, $$n=\dfrac{5\times 15}{3}$$, $$n=25$$ (not in options)

so, $$r=12$$, $$n=\dfrac{5\times 12}{3}$$, $$n=20$$(in options).
Question 2
1 / -0

In the expansion of $${ \left( \dfrac { 3{ x }^{ 2 } }{ 5 } +\dfrac { 5 }{ 3{ x }^{ 2 } } \right) }^{ 10 }$$ mid term is

A
$$291$$

B
$$242$$

C
$$252$$

D
$$284$$

Solution

Given expansion $$={ \left( \dfrac { 3{ x }^{ 2 } }{ 5 } +\dfrac { 5 }{ 3{ x }^{ 2 } } \right) }^{ 10 }$$
On comparing with $${ \left( x+a \right) }^{ n }$$, we get
$$x=\dfrac { 3{ x }^{ 2 } }{ 5 } $$, $$a=\dfrac { 5 }{ 3{ x }^{ 2 } }$$ and $$n=10$$
$$\because $$ Number of terms $$=10+1=11$$
$$\therefore$$ Mid term $$ =\left( \dfrac { 11+1 }{ 2 } \right)$$th term $$ =6$$th term
Now, $${ T }_{ 6 }={ T }_{ 5+1 }=_{ }^{ 10 }{ { C }_{ 5 } }{ \left[ \dfrac { 3{ x }^{ 2 } }{ 5 } \right] }^{ 10-5 }{ \left[ \dfrac { 5 }{ 3{ x }^{ 2 } } \right] }^{ 5 }$$ $$\left[ \because { T }_{ r+1 }=_{ }^{ n }{ { C }_{ r } }{ x }^{ n-r }{ a }^{ r } \right] $$
$$=_{ }^{ 10 }{ { C }_{ 5 } }{ \left[ \dfrac { 3{ x }^{ 2 } }{ 5 } \right] }^{ 5 }{ \left[ \dfrac { 5 }{ 3{ x }^{ 2 } } \right] }^{ 5 }=_{ }^{ 10 }{ { C }_{ 5 } }=252$$.
Question 3
1 / -0

If the sum of the coefficients in the expansion of $$(l^2x^2-2lx+1)^{50}$$ vanishes then $$l$$ is equal to:

A
$$-1$$

B
$$-2$$

C
$$1$$

D
$$2$$

Solution

$${(l^2x^2-2lx+1)}^{50}$$=$${(lx-1})^{100}$$
If the sum of the coefficcents of the expansion vanishes mens equal to 0
means on putting x=1
$$\Rightarrow$$$$lx-1=0$$
$$\Rightarrow$$$$l-1=0$$
$$\Rightarrow$$$$l=1$$
Question 4
1 / -0

The term independent of $$x$$ in the expansion of $$\left [\sqrt {\dfrac {x}{3}} + \sqrt {\dfrac {3}{2x^{2}}} \right ]^{10}$$ is

A
$$1$$

B
$$^{10}C_{1}$$

C
$$\dfrac {5}{12}$$

D
None of these

Solution

$$T_{r + 1} = ^{10}C_{r} \left (\dfrac {x}{3}\right )^{\tfrac {10 - r}{2}} \left (\dfrac {3}{2x^{2}} \right )^{\tfrac {r}{2}}$$

$$= ^{10}C_{r} x^{\tfrac {10 - r}{2} - r} \left (\tfrac {1}{2}\right )^{\tfrac {r}{2}} 3^{r - 5}$$ Put $$\dfrac {10 - r}{2} - r = 0$$

$$\Rightarrow r = \dfrac {10}{3} \not {\epsilon}$$ whole number

$$\therefore$$ No choice match

$$\therefore$$ (d) is correct answer.
Question 5
1 / -0

If $$\sum _{ r=0 }^{ n-1 }{ { \left( \cfrac { { _{ }^{ n }{ C } }_{ r } }{ { _{ }^{ n }{ C } }_{ r }+{ _{ }^{ n }{ C } }_{ r+1 } } \right) }^{ 3 } } =\cfrac { 4 }{ 5 } $$ then $$n=$$

A
$$4$$

B
$$6$$

C
$$8$$

D
None of these

Solution

We know that $$\left( \begin{matrix} n \\ r+1 \end{matrix} \right) +\left( \begin{matrix} n \\ r \end{matrix} \right) =\left( \begin{matrix} n+1 \\ r+1 \end{matrix} \right) $$,
$$\Longrightarrow \sum _{ 0 }^{ n-1 }{ { \left( \frac { \left( \begin{matrix} n \\ r \end{matrix} \right) }{ \left( \begin{matrix} n+1 \\ r+1 \end{matrix} \right) } \right) }^{ 3 } } =\sum _{ 0 }^{ n-1 }{ { \left( \frac { r+1 }{ n+1 } \right) }^{ 3 } } $$
$$\\ \Longrightarrow \frac { { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+...+{ n }^{ 3 } }{ { \left( n+1 \right) }^{ 3 } } \\ \Longrightarrow \frac { { n }^{ 2 }{ \left( n+1 \right) }^{ 2 } }{ 4{ \left( n+1 \right) }^{ 3 } } \\ \Longrightarrow \frac { { n }^{ 2 } }{ 4{ \left( n+1 \right) } } $$
Given that $$\frac { { n }^{ 2 } }{ 4{ \left( n+1 \right) } } =\frac { 4 }{ 5 } $$
$$\Longrightarrow 5{ n }^{ 2 }-16n-16=0\\ \Longrightarrow \left( n-4 \right) \left( 5n+4 \right) =0$$
As $$n$$ is a natural number
$$\Longrightarrow n=4$$
Question 6
1 / -0

Sum of the last $$30$$ coefficients in the expansion of $${ \left( 1+x \right) }^{ 59 }$$, when expanded in ascending power of $$x$$ is

A
$${ 2 }^{ 59 }$$

B
$${ 2 }^{ 58 }$$

C
$${ 2 }^{ 30 }$$

D
$${ 2 }^{ 29 }$$

Solution

$${ (1+x) }^{ n }=^{ n }C_{ 0 }+^{ n }C_{ 1 }x+^{ n }C_{ 2 }{ x }^{ 2 }.........^{ n }C_{ n }{ x }^{ n }\\ \Rightarrow ^{ n }C_{ 0 }+^{ n }C_{ 1 }+^{ n }C_{ 2 }.........^{ n }C_{ n }={ 2 }^{ n }$$
For the given expansion $${ (1+x) }^{ 59 }$$ let sum of last $$30$$ coefficients be $$S$$

$$S=^{ 59 }C_{ 30 }+^{ 59 }C_{ 31 }+^{ 59 }C_{ 32 }..........^{ 59 }C_{ 59 }$$ ....(i)

As $$^{ n }C_{ r }=^{ n }C_{ n-r }$$

$$\Rightarrow S=^{ 59 }C_{ 29 }+^{ 59 }C_{ 28 }+^{ 59 }C_{ 27 }..........^{ 59 }C_{ 0 }$$ ....(ii)

Adding (i) and (ii), we get

$$\Rightarrow 2S=^{ 59 }C_{ 0 }+^{ 59 }C_{ 1 }+^{ 59 }C_{ 2 }........^{ 59 }C_{ 59 }\\ \Rightarrow 2S={ 2 }^{ 59 }\\ \Rightarrow S={ 2 }^{ 58 }$$

So, option B is correct.
Question 7
1 / -0

If $$(1+x)^{10} = a_0 + a_1x + a_2x^2 + ..... + a_{10}x^{10}$$, then value of $$(a_0 -a_2 + a_4 - a_6 + a_8 - a_{10})^2 + (a_1 -a_3 + a_5 - a_7 + a_9)^2$$ is

A
$$2^{10}$$

B
$$2$$

C
$$2^{20}$$

D
$$2^{30}$$

Solution

Here $${ a }_{ i }=^{ 10 }{ C }_{ i }$$
ie,$${ ({ a }_{ 0 }-{ a }_{ 2 }+{ a }_{ 4 }-{ a }_{ 6 }+{ a }_{ 8 }-{ a }_{ 10 }) }^{ 2 }+{ ({ a }_{ 1 }-{ a }_{ 3 }+{ a }_{ 5 }-{ a }_{ 7 }+{ a }_{ 9 }) }^{ 2 }$$
$$={ (^{ 10 }{ C }_{ 0 }-^{ 10 }{ C }_{ 2 }+^{ 10 }{ C }_{ 4 }-^{ 10 }{ C }_{ 6 }+^{ 10 }{ C }_{ 8 }-^{ 10 }{ C }_{ 10 }) }^{ 2 }+{ (^{ 10 }{ C }_{ 1 }-^{ 10 }{ C }_{ 3 }+^{ 10 }{ C }_{ 5 }-^{ 10 }{ C }_{ 7 }+^{ 10 }{ C }_{ 9 }) }^{ 2 }$$
$$\Rightarrow { ((^{ 10 }{ C }_{ 0 }-^{ 10 }{ C }_{ 10 })+(^{ 10 }{ C }_{ 8 }-^{ 10 }{ C }_{ 2 })+(^{ 10 }{ C }_{ 4 }-^{ 10 }{ C }_{ 6 })) }^{ 2 }+{ ( (-1){ (-2) }^{ \frac { 10 }{ 2 } }) }^{ 2 }={ 2 }^{ 10 }$$
Since $$ \sum _{ i=0 }^{ \left[ n/2 \right] }{ ^{ n }{ C }_{ 2i } } =\begin{cases} 0,\quad \text{If}\quad \dfrac { n+2 }{ 4 } \in \quad \text{Integers} \\ { (-1) }^{ \left[ (n+2)/4 \right] }{ 2 }^{ \left[ n/2 \right] } \end{cases}$$
and $$\sum _{ i=0 }^{ \left[ n/2 \right] }{ ^{ n }{ C }_{ 2i+1 } } =\begin{cases} 0,\quad \text{If}\quad \dfrac { n }{ 4 } \in \text{Integers} \\ { (-1) }^{ [n/4] }{ 2 }^{ [n/2] } \end{cases}$$
Option A is correct
Question 8
1 / -0

The value of $$x$$ in the expression $${ \left( x+{ x }^{ \log _{ 10 }{ x } } \right) }^{ 5 }$$, if the third term in the expansion is $$1,000,000$$, is

A
$$10,{ 10 }^{ { -3 }/{ 2 } }$$

B
$$100$$ or $${ 10 }^{ { -3 }/{ 2 } }$$

C
$$10$$ or $${ 10 }^{ { -5 }/{ 2 } }$$

D
None of these

Solution

$$\log { x } $$ is defined only when $$x>0$$
Now, the $${ 3 }^{ rd }$$ term in the expansion
$${ T }_{ 2+1 }=^{ 5 }{ { C }_{ 2 } }\cdot { x }^{ 5-2 }\cdot { \left( { x }^{ \log _{ 10 }{ x } } \right) }^{ 2 }=1,000,000$$ ....(Given)
$$\Rightarrow { x }^{ 3+2\log _{ 10 }{ x } }={ 10 }^{ 5 }$$
Taking logarithm of both sides, we get
$$\left( 3+2\log _{ 10 }{ x } \right) \cdot \log _{ 10 }{ x } =5$$
$$\Rightarrow 2{ y }^{ 2 }+3y-5=0$$
where $$\log _{ 10 }{ x } =y$$
$$\Rightarrow \left( y-1 \right) \left( 2y+5 \right) =0$$
$$\Rightarrow y=1$$ or $${ -5 }/{ 2 }$$
$$\Rightarrow \log _{ 10 }{ x } =1$$ or $${ -5 }/{ 2 }$$
$$\Rightarrow x={ 10 }^{ 1 }=10$$ or $${ 10 }^{ { -5 }/{ 2 } }$$
Question 9
1 / -0

If there is a term containing $$x^{2r}$$ in $$\left( x + \dfrac{1}{x^2} \right )^{n - 3}$$, then

A
n - 2r is a positive integral multiple of 3.

B
n - 2r is even

C
n - 2r is odd

D
None of the above

Solution

Suppose $$(s + 1)^{th}$$ term conains $$x^{2r}$$
Then, we have
$$T_{s + 1} = ^{n - 3}C_s x^{n - 3- s} \left( \dfrac{1}{x^2} \right )^s$$
$$= ^{n - 3}C_s x^{n - 3 - 3s}$$
This will contain $$x^{2r}$$, if
$$n - 3 - 3 s = 2r$$
$$\Rightarrow s = \dfrac{n - 3 - 2r}{3}$$
$$\Rightarrow s = \dfrac{n - 2r}{3} - 1$$
$$\Rightarrow s + 1 = \dfrac{n - 2r}{3}$$
$$\Rightarrow n - 2r = 3 ( s + 1)$$
Hence, $$(n - 2r) $$ is a positive integral multiple of $$3$$.
Question 10
1 / -0

Coefficient of $$x^n$$ in the expansion of $$\left(\displaystyle 1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^n}{n!}\right)^2$$ is?

A
$$\displaystyle\frac{2^n}{n!}$$

B
$$\displaystyle\frac{2^{n-1}}{n!}$$

C
$$\displaystyle\frac{2^{n+1}}{n!}$$

D
None of these

Solution

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Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 59

Result

Binomial Theorem Test - 59

Score

-

Rank

-

SHARING IS CARING

Question 1

18

20

24

22

Solution

Question 2

$$291$$

$$242$$

$$252$$

$$284$$

Solution

Question 3

$$-1$$

$$-2$$

$$1$$

$$2$$

Solution

Question 4

$$1$$

$$^{10}C_{1}$$

$$\dfrac {5}{12}$$

None of these

Solution

Question 5

$$4$$

$$6$$

$$8$$

None of these

Solution

Question 6

$${ 2 }^{ 59 }$$

$${ 2 }^{ 58 }$$

$${ 2 }^{ 30 }$$

$${ 2 }^{ 29 }$$

Solution

Question 7

$$2^{10}$$

$$2$$

$$2^{20}$$

$$2^{30}$$

Solution

Question 8

$$10,{ 10 }^{ { -3 }/{ 2 } }$$

$$100$$ or $${ 10 }^{ { -3 }/{ 2 } }$$

$$10$$ or $${ 10 }^{ { -5 }/{ 2 } }$$

None of these

Solution

Question 9

n - 2r is a positive integral multiple of 3.

n - 2r is even

n - 2r is odd

None of the above

Solution

Question 10

$$\displaystyle\frac{2^n}{n!}$$

$$\displaystyle\frac{2^{n-1}}{n!}$$

$$\displaystyle\frac{2^{n+1}}{n!}$$

None of these

Solution

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