Binomial Theorem Test

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Binomial Theorem Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

If the fourth term in the expansion of $$\displaystyle { \left( \sqrt { \frac { 1 }{ { x }^{ \log { x } +1 } } } +{ x }^{ 1/12 } \right) }^{ 6 }$$ is equal to $$200$$ and $$x>1$$, then $$x$$ is equal to

A
$${ 10 }^{ \sqrt { 2 } }$$

B
$$10$$

C
$${10}^{4}$$

D
None of these

Solution

Given $${ T }_{ 4 }=200$$
$$\displaystyle \Rightarrow ^{ 6 }{ { C }_{ 3 } }{ \left( \sqrt { \frac { 1 }{ { x }^{ \log _{ 10 }{ x+1 } } } } \right) }^{ 3 }{ \left( { x }^{ \frac { 1 }{ 12 } } \right) }^{ 3 }=200$$
$$\displaystyle \Rightarrow 20\times { x }^{ \frac { 3 }{ 2\left( \log _{ 10 }{ x+1 } \right) } +\frac { 1 }{ 4 } }=200\Rightarrow { x }^{ \left( \frac { 3 }{ 2\left( \log _{ 10 }{ x+1 } \right) } +\frac { 1 }{ 4 } \right) }=10$$
$$\displaystyle \Rightarrow \left( \frac { 3 }{ 2\left( \log _{ 10 }{ x+1 } \right) } +\frac { 1 }{ 4 } \right) =\log _{ x }{ 10 } =\frac { 1 }{ \log _{ 10 }{ x } } $$
$$\displaystyle \Rightarrow \frac { 3 }{ 2\left( y+1 \right) } +\frac { 1 }{ 4 } =\frac { 1 }{ y } $$ where $$y=\log _{ 10 }{ x } $$
$$\Rightarrow y=-4$$ or $$y=1$$ $$\Rightarrow \log _{ 10 }{ x } =-4$$ or $$1$$
$$\Rightarrow x={ 10 }^{ -4 }$$ or $$10$$
$$\Rightarrow x=10\quad \left( \because x>1 \right) $$
Question 2
1 / -0

If $$A$$ and $$B$$ are coefficients of $${x}^{n}$$ in the expansions of $${ (1+x) }^{ 2n }$$ and $${ (1+x) }^{ 2n-1 }$$ respectively, then $$\dfrac AB$$ is equal to

A
$$1$$

B
$$2$$

C
$$\dfrac 12$$

D
$$\dfrac 1n$$

Solution

We know that coefficient of $${x}^{r}$$ in the expansion of $${ (1+x) }^{ m }$$ is $${ _{ }^{ m }{ C } }_{ r }$$
Thus $$A={ _{ }^{ 2n }{ C } }_{ n }\quad ;\quad B={ _{ }^{ 2n-1 }{ C } }_{ n }$$
We have
$$\cfrac { A }{ B } =\cfrac { { _{ }^{ 2n }{ C } }_{ n } }{ { _{ }^{ 2n-1 }{ C } }_{ n } } =\cfrac { (2n)! }{ n!n! } \cfrac { (n!)(n-1)! }{ (2n-1)! } =\cfrac { 2n }{ n } =2$$
Question 3
1 / -0

The expression $${ C }_{ 0 }+2{ C }_{ 1 }+3{ C }_{ 2 }+......+(n+1){ C }_{ n }$$ is equal to

A
$${ 2 }^{ n-1 }$$

B
$$n({ 2 }^{ n-1 })$$

C
$$n({ 2 }^{ n-1 })+{ 2 }^{ n }$$

D
$$(n+1){ 2 }^{ n }$$

Solution

Let $$E={ C }_{ 0 }+2{ C }_{ 1 }+3{ C }_{ 2 }+......n{ C }_{ n-1 }+(n+1){ C }_{ n }\quad ........(1)$$
Using $${ C }_{ r }={ C }_{ n-r }\quad E=(n+1){ C }_{ 0 }+n{ C }_{ 1 }+(n-1){ C }_{ 2 }+.....+2{ C }_{ n-1 }+{ C }_{ n }\quad ........(2)$$
Adding $$(1)$$ and $$(2)$$, we get
$$2E=(n+2){ C }_{ 0 }+(n+2){ C }_{ 1 }+(n+2){ C }_{ 2 }+....+(n+2){ C }_{ n }$$
$$=(n+2)\left\{ { C }_{ 0 }+{ C }_{ 1 }+.....+{ C }_{ n } \right\} =(n+2){ 2 }^{ n }$$
$$\Rightarrow \quad E=(n+2){ 2 }^{ n-1 }$$
Question 4
1 / -0

The number of irrational terms in the expansion of $${ ({ 5 }^{\tfrac 16 }+{ 2 }^{ \tfrac 18 }) }^{ 100 }$$ is

A
$$96$$

B
$$97$$

C
$$98$$

D
$$99$$

Solution

$$(r+1)$$th in the expansion of $${ ({ 5 }^{\tfrac 16 }+{ 2 }^{\tfrac 18 }) }^{ 100 }$$ is
$${ t }_{ r+1 }={ _{ }^{ 100 }{ C } }_{ r }{ ({ 5 }^{\tfrac 16 }) }^{ 100-r }{ ({ 2 }^{\tfrac 18 }) }^{ r }$$
As $$5$$ and $$2$$ are relatively prime, $${t}_{r+1}$$ will be rational if $$\cfrac { 100-r }{ 6 } $$ and $$\cfrac { r }{ 8 } $$ are both integers. i.e., if $$100-r$$ is a multiple of $$6$$ and $$r$$ is a multiple of $$8$$. As $$0\le r \le 100$$, multiples of $$8$$ upto $$100$$ and corresponding value of $$100-r$$ are
$$r=0,8,16,24,.....88,96$$
$$100-r=100,92,84,76,....12,4$$
Out of values of $$100-r$$, multiples of $$6$$ are $$84,60,36,12$$
$$\therefore$$ there are just four rational terms.
$$\Rightarrow$$ number of irrational terms is $$101-4=97$$
Question 5
1 / -0

If $$f(n)=\sum _{ k=1 }^{ n }{ \sum _{ j=k }^{ n }{ ({ _{ }^{ n }{ C } }_{ j } } ) } ({ _{ }^{ j }{ C } }_{ k })$$, find $$f(n)$$.

A
$${ 3 }^{ n }-{ 2 }^{ n }$$

B
$${ 3 }^{ n }+{ 2 }^{ n }$$

C
$${ 3 }^{ n-1 }-{ 2 }^{ n -1}$$

D
$${ 3 }^{ n+1 }-{ 2 }^{ n+1 }$$

Solution

Simplifying the above expression we get
$$\:^{n}C_{1}(\:^{1}C_{1})+\:^{n}C_{2}(\:^{2}C_{1}+\:^{2}C_{2})+...+\:^{n}C_{n}(\:^{n}C_{1}+\:^{n}C_{2}+...\:^{n}C_{n})$$
$$=(2^{1}-1)\:^{n}C_{1}+(2^{2}-1)\:^{n}C_{2}+...\:^{n}C_{n}(2^{n}-1)$$
$$=2^{1}\:^{n}C_{1}+2^{2}\:^{n}C_{2}+...2^{n}\:^{n}C_{n}-[\:^{n}C_{1}+\:^{n}C_{2}+...\:^{n}C_{n}]$$
$$=(1+2)^{n}-1-[2^{n}-1]$$
$$=3^{n}-2^{n}$$.
Question 6
1 / -0

Value of $$S=1\times 2\times 3\times 4+2\times 3\times 4\times 5+.......+n(n-1)(n+2)(n+3)$$ is

A
$$\cfrac { 1 }{ 5 } n(n+1)(n+2)(n+3)(n+4)$$

B
$$\cfrac { 1 }{ 5! } ({ _{ }^{ n+3 }{ C } }_{ 5 })$$

C
$$\cfrac { 1 }{ 5 }{ _{ }^{ n+4 }{ C } }_{ 4 }$$

D
None of these

Solution
Question 7
1 / -0

If the third term in the expansion of
$${ \left( \cfrac { 1 }{ x } +{ x }^{ \log _{ 2 }{ x } } \right) }^{ 5 }$$ is $$40$$ then $$x$$ equals

A
$$\dfrac {1}{\sqrt{2}},2$$

B
$$\sqrt{2},4$$

C
$$\dfrac {1}{\sqrt{2}},4$$

D
$$\sqrt{2},\dfrac{1}{\sqrt{2}}$$

Solution

The middle term of $$\displaystyle { \left( \frac { 1 }{ x } +x\log _{ 2 }{ x } \right) }^{ 5 }$$ is $$\displaystyle ^{ 5 }{ { C }_{ 2 } }{ \left( \frac { 1 }{ x } \right) }^{ 3 }{ \left( { x }^{ \log _{ 2 }{ x } } \right) }^{ 2 }=10{ x }^{ 2\log _{ 2 }{ x-3 } }=40$$
$$\therefore{ x }^{2 \log _{ 2 }{ x-3 } }=4={ 2 }^{ 2 }$$
Taking logarithm to the base $$2$$ of both the sides, we get $$(2t-3)t=2$$ where $$t=\log _{ 2 }{ x } $$
$$\Rightarrow 2{ t }^{ 2 }-3t-2=0$$ or $$\left( 2t+1 \right) \left( t-2 \right) =0$$
$$\displaystyle \Rightarrow t=-\frac { 1 }{ 2 } , 2\Rightarrow x=\frac { 1 }{ \sqrt { 2 } },4 $$
Question 8
1 / -0

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

A
$$5$$

B
$$7$$

C
$$11$$

D
$$14$$

Solution

For the terms to be in A.P, it must follow
$$(n-2r)^{2}=n+2$$
In the above case $$r=2$$
Substituting in the equation, we get
$$(n-4)^{2}=n+2$$
$$n^{2}-8n+16=n+2$$
$$n^{2}-9n+14=0$$
$$(n-2)(n-7)=0$$
$$n=2$$ and $$n=7$$
However for $$n=2$$ we will have only $$3$$ terms.
Hence the required answer is $$7.$$
Question 9
1 / -0

If the third term in the expansion $${ \left( x+{ x }^{ \log _{ 5 }{ x } } \right) }^{ 5 }$$ is $$2$$, then $$x$$ equals

A
$$1/5,5$$

B
$$1/5,1/\sqrt {5}$$

C
$$\sqrt{5},5$$

D
$$1/\sqrt{5},5$$

Solution

$$T_{2+1}$$
$$=\:^{5}C_{2}x^{3+2log_{5}(x)}$$
$$=10x^{3+2log_{5}(x)}$$
$$=2$$
$$x^{3+2log_{5}(x)}=\dfrac{1}{5}$$
$$log_{5}(x)(3+2log_{5}(x))=-1$$
$$2log(x)^{2}+3log(x)+1=0$$
$$2log(x)^{2}+2log(x)+log(x)+1=0$$
$$(2log(x)+1)(log(x)+1)=0$$
$$log(x)=-1$$
$$x=\dfrac{1}{5}$$ and $$log(x)=\dfrac{-1}{2}$$
$$x=\dfrac{1}{\sqrt{5}}$$
Question 10
1 / -0

Value of
$$\sum _{ k=1 }^{ \infty } \sum _{ r=0 }^{ k }{ \cfrac { 1 }{ { 3 }^{ k } } } ({ _{ }^{ k }{ C } }_{ r })$$ is

A
$$2$$

B
$$\cfrac{2}{3}$$

C
$$\cfrac{1}{3}$$

D
none of these

Solution

$$\sum \dfrac{1}{3^{k}}\:^{k}C_{r}$$
$$=\dfrac{1}{3^{k}}\sum\:^{k}C_{r}$$
$$=\dfrac{2^{k}}{3^{k}}$$
This is a G.P
Therefore, the sum of the series will be
$$S=\dfrac{\frac{2}{3}}{1-\frac{2}{3}}$$ $$=2$$

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Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 29

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Binomial Theorem Test - 29

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SHARING IS CARING

Question 1

$${ 10 }^{ \sqrt { 2 } }$$

$$10$$

$${10}^{4}$$

None of these

Solution

Question 2

$$1$$

$$2$$

$$\dfrac 12$$

$$\dfrac 1n$$

Solution

Question 3

$${ 2 }^{ n-1 }$$

$$n({ 2 }^{ n-1 })$$

$$n({ 2 }^{ n-1 })+{ 2 }^{ n }$$

$$(n+1){ 2 }^{ n }$$

Solution

Question 4

$$96$$

$$97$$

$$98$$

$$99$$

Solution

Question 5

$${ 3 }^{ n }-{ 2 }^{ n }$$

$${ 3 }^{ n }+{ 2 }^{ n }$$

$${ 3 }^{ n-1 }-{ 2 }^{ n -1}$$

$${ 3 }^{ n+1 }-{ 2 }^{ n+1 }$$

Solution

Question 6

$$\cfrac { 1 }{ 5 } n(n+1)(n+2)(n+3)(n+4)$$

$$\cfrac { 1 }{ 5! } ({ _{ }^{ n+3 }{ C } }_{ 5 })$$

$$\cfrac { 1 }{ 5 }{ _{ }^{ n+4 }{ C } }_{ 4 }$$

None of these

Solution

Question 7

$$\dfrac {1}{\sqrt{2}},2$$

$$\sqrt{2},4$$

$$\dfrac {1}{\sqrt{2}},4$$

$$\sqrt{2},\dfrac{1}{\sqrt{2}}$$

Solution

Question 8

$$5$$

$$7$$

$$11$$

$$14$$

Solution

Question 9

$$1/5,5$$

$$1/5,1/\sqrt {5}$$

$$\sqrt{5},5$$

$$1/\sqrt{5},5$$

Solution

Question 10

$$2$$

$$\cfrac{2}{3}$$

$$\cfrac{1}{3}$$

none of these

Solution

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