Binomial Theorem Test

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Binomial Theorem Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

The sum of coefficient of integral powers of $$x$$ in the binomial expansion of $$(1-2\sqrt x)^{50}$$ is :

A
$$\dfrac {1}{2}(3^{50}+1)$$

B
$$\dfrac {1}{2}(3^{50})$$

C
$$\dfrac {1}{2}(3^{50}-1)$$

D
$$\dfrac {1}{2}(2^{50}+1)$$

Solution

$$(1-2\sqrt{x})^{50}= ^{50}C_0-^{50}C_1(2\sqrt{x})+^{50}C_2(2\sqrt{x})^2.....$$

we need to find sum of even forms, so

Put $$\sqrt{x}=1 \Longrightarrow (1-2)^{50}=1=^{50}C_0-^{50}C_1(2)+^{50}C_2(2)^2.........(1)$$

and $$\sqrt{x}=-1 \Longrightarrow (1+2)^{50}=3^{50}=^{50}C_0+^{50}C_12+^{50}C_2(2)^2.......(2)$$

Adding (1) and (2) we get,

$$ 3^{50}+1=2(^{50}C_0+^{50}C_2(2)^2+........)$$

$$\Rightarrow (^{50}C_0+^{50}C_2(2)^2+........)=\cfrac{1}{2}(1+3^{50})$$
Question 2
1 / -0

The value of $$\displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {^{15}C_{r}}{^{15}C_{r - 1}}\right )$$ is equal to:

A
$$680$$

B
$$1085$$

C
$$560$$

D
$$1240$$

Solution

$$\displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {^{15}C_{r}}{^{15}C_{r - 1}}\right )$$

$$\Rightarrow \left (\dfrac {^{n}C_{r}}{^{n}C_{r - 1}}\right ) = \dfrac{n-r+1}{r}$$

$$\displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {^{15}C_{r}}{^{15}C_{r - 1}}\right ) = \displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {n-r+1}{r}\right )\\$$

$$\sum_{r = 1}^{15} (nr-r^2+r) = \sum_{r = 1}^{15} ((n+1)r-r^2)$$

$$ \sum_{r = 1}^{15} (nr-r^2+r) = \dfrac{n(n+1)^2}{2}- \dfrac{n(n+1)(2n+1)}{6}$$

Put $$n=15,$$ we get

$$\displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {^{15}C_{r}}{^{15}C_{r - 1}}\right ) = 680$$
Question 3
1 / -0

Let $$(x + 10)^{50} + (x - 10)^{50} = a_0 + a_1x + a_2x^2 + .... + a_{50} x^{50}$$, for all $$x \in R$$ then $$\dfrac{a_2}{a_0}$$ is equal to:-

A
$$12.50$$

B
$$12.00$$

C
$$12.75$$

D
$$12.25$$

Solution

Given:- $${\left( x + 10 \right)}^{50} + {\left( x - 10 \right)}^{50} = {a}_{0} + {a}_{1} x + {a}_{2} {x}^{2} + ........... + {a}_{50} {x}^{50}$$
To find:- $$\cfrac{{a}_{2}}{{a}_{0}} = \cfrac{\text{coefficient of } {x}^{2}}{\text{Coefficient of } {x}^{0}}$$
As we know that, the general term in an expansion $${\left( a + b \right)}^{n}$$ is given as,
$${T}_{r + 1} = {^{n}{C}_{r}} {\left( a \right)}^{n - r} {\left( b \right)}^{r}$$
Now,
General term of $${\left( x + 10 \right)}^{50}$$-
Here,
$$a = x, b = 10$$
$${T}_{r + 1} = {^{50}{C}_{r}}{\left( x \right)}^{50 - r} {\left( 10 \right)}^{r}$$
For coefficient of $${x}^{2}$$-
$$50 - r = 2 \Rightarrow r = 48$$
$${T}_{48 + 1} = {^{50}{C}_{48}} {\left( x \right)}^{50 - 48} {\left( 10 \right)}^{48}$$
$${T}_{49} = {^{50}{C}_{48}} {\left( 10 \right)}^{48} {x}^{2}$$
For coeficient of $${x}^{0}$$-
$$50 - r = 0 \Rightarrow r = 50$$
$${T}_{50 + 1} = {^{50}{C}_{50}} {\left( x \right)}^{50 - 50} {\left( 10 \right)}^{50}$$
$${T}_{51} = {^{50}{C}_{50}} {\left( 10 \right)}^{50} {x}^{0}$$
Now,
General term of $${\left( x + \left( -10 \right) \right)}^{50}$$-
Here,
$$a = x, b = -10$$
$${T}_{r + 1} = {^{50}{C}_{r}}{\left( x \right)}^{50 - r} {\left( -10 \right)}^{r}$$
For coeficient of $${x}^{2}$$-
$$50 - r = 2 \Rightarrow r = 48$$
$${T}_{48 + 1} = {^{50}{C}_{48}} {\left( x \right)}^{50 - 48} {\left( -10 \right)}^{48}$$
$${T}_{49} = {^{50}{C}_{48}} {\left( 10 \right)}^{48} {x}^{2}$$
For coeficient of $${x}^{0}$$-
$$50 - r = 0 \Rightarrow r = 50$$
$${T}_{50 + 1} = {^{50}{C}_{50}} {\left( x \right)}^{50 - 50} {\left( -10 \right)}^{50}$$
$${T}_{51} = {^{50}{C}_{50}} {\left( 10 \right)}^{50} {x}^{0}$$
Now from the given expansion,
$${a}_{2} = {^{50}{C}_{48}} {\left( 10 \right)}^{48} + {^{50}{C}_{48}} {\left( 10 \right)}^{48} = {^{50}{C}_{48}} \left( {\left( 10 \right)}^{48} + {\left( 10 \right)}^{48} \right)$$
$${a}_{0} = {^{50}{C}_{50}} {\left( 10 \right)}^{50} + {^{50}{C}_{50}} {\left( 10 \right)}^{50} = {^{50}{C}_{50}} \left( {\left( 10 \right)}^{50} + {\left( 10 \right)}^{50} \right)$$
Now,
$$\cfrac{{a}_{2}}{{a}_{0}} = \cfrac{{^{50}{C}_{48}} \left( {\left( 10 \right)}^{48} + {\left( 10 \right)}^{48} \right)}{{^{50}{C}_{50}} \left( {\left( 10 \right)}^{50} + {\left( 10 \right)}^{50} \right)}$$
As we know that,
$${^{n}{C}_{r}} = \cfrac{n!}{r! \left( n - r \right)!}$$
Therefore,
$$\cfrac{{a}_{2}}{{a}_{0}} = \cfrac{50 \times 49}{2} \times \left( \cfrac{{10}^{48}}{{10}^{50}}\right)$$
$$\Rightarrow \cfrac{{a}_{2}}{{a}_{0}} = \cfrac{49}{4}= 12.25$$
Question 4
1 / -0

The coefficient of $$x^{18}$$ in the product $$(1+x)(1-x)^{10}(1+x+x^2)^9$$ is?

A
$$-84$$

B
$$84$$

C
$$126$$

D
$$-126$$

Solution

$$(1+x)(1-x)^{10}(1+x+x^2)^9$$
$$(1-x^2)(1-x^3)^9$$
$${^{9}C_6}=84$$.
Question 5
1 / -0

Let $$s_n = 1 + q + q^2 + ................. + q^n$$ and
$$T_n = 1 + \left(\dfrac{q + 1}{2}\right) + \left(\dfrac{q + 1}{2}\right)^2 +......... \left(\dfrac{q + 1}{2}\right)^n$$
where q is a real number and q $$\neq$$ 1.
If $$^{101}C_1 + ^{101}{C_2}.S_{1} +...... + ^{101}C_{101}.S_{100} = \alpha T_{100}, then\,\,\, \alpha$$ is equal to :-

A
$$2^{100}$$

B
$$200$$

C
$$2^{99}$$

D
$$202$$

Solution
Question 6
1 / -0

If $$(2+\dfrac {x}{3})^{55}$$ is expanded in the ascending powers of $$x$$ and the coefficients of powers of $$x$$ in two consecutive terms of the expansion are equal, then these terms are :

A
$$7^{th}$$ and $$8^{th}$$

B
$$8^{th}$$ and $$9^{th}$$

C
$$28^{th}$$ and $$29^{th}$$

D
$$27^{th}$$ and $$28^{th}$$

Solution

Coefficient of $$x^r $$ and $$x^{r+1} $$ are equal.

Coefficient of $$T_{r+1} =$$ Coefficient of $$T_{r+2} $$

$$\Rightarrow\displaystyle { ^{ 55 }C }_{ r }{ (2) }^{ 55-r }{ \left(\dfrac { 1 }{ 3 } \right) }^{ r }={ ^{ 55 }C }_{ r+1 }{ (2) }^{ 54-r }{ \left(\dfrac { 1 }{ 3 } \right) }^{ r+1 }$$
.
$$\dfrac { 55! }{ (55-r)!r! } (2)=\dfrac { 55! }{ (54-r)!(r+1)! } \dfrac { 1 }{ 3 } $$

$$\Rightarrow 6(r+1) = 55-r$$

$$\Rightarrow r =7$$

So, the terms will be $$8^{th}$$ and $$9^{th}$$
Question 7
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If the sum of odd numbered terms and the sum of even numbered terms in the expansion of $$(x + a)^{n}$$ are $$A$$ and $$B$$ respectively, then the value of $$(x^{2} - a^{2})^{n}$$ is

A
$$A^{2} - B^{2}$$

B
$$A^{2} + B^{2}$$

C
$$4AB$$

D
None

Solution

$$(x + a)^{n} = ^{n}C_{0} x^{n} + ^{n}C_{1} x^{n - 1}a + ^{n}C_{2} x^{n - 2}a^{2} + ^{n}C_{3} x^{n - 3} a^{3} + ^{n}C_{4}x^{n - 4} a^{4} + .....$$
$$= (^{n}C_{0}x^{n} + ^{n}C_{2}x^{n - 2}a^{2} + ^{n}C_{4} x^{n - 4} a^{4} + .....) + (^{n}C_{1}x^{n - 1}a + ^{n}C_{3} x^{n - 3}a^{3} + ^{n}C_{5} x^{n - 5}a^{5}) + ....$$
$$= A + B$$ .... (1)
Similarly, $$(x - a)^{n} = A - B$$ .... (2)
Multiplying eqns. (1) and (2), we get
$$(x^{2} - a^{2})^{n} = A^{2} - B^{2}$$
Question 8
1 / -0

The coefficients of the $$ (3r)^{ th}$$ term and the $$(r + 2)^{ th}$$ term in the expansion $$(1 + x)^{2n}$$ are equal, then :

A
$$n = 2r $$

B
$$n = 3r$$

C
$$n = 2r + 1$$

D
none of these

Solution

The coefficients of $$(3r)$$ and $$(r + 2)$$ terms in the expansion $$(1 + x)^{2n}$$ are same.
So, $$^{2n}C_{(3r - 1)} = ^{2n}C_{(r + 1)}$$
As we know that in general, $$^{n}C_{r} = ^{n}C_{n - r}, $$
$$\Rightarrow 3r - 1 + r + 1 = 2n$$
i.e. $$4r = 2n$$ or $$n = 2r$$
Question 9
1 / -0

The number of rational terms in the expansion of $$(9^{1/4} + 8^{1/6})^{1000}$$ is:

A
$$500$$

B
$$400$$

C
$$501$$

D
none of the above

Solution

The general term in the expansion of $$(9^{1/4} + 8^{1/6})^{1000}$$ is
$$T_{r+1}={}^{1000}C_r\left (9^{\frac {1}{4}}\right )^{1000-r}\left (8^{\frac {1}{6}}\right )^r$$
$$={}^{1000}C_r 3^{\frac {1000-r}{2}} 2^{\frac {1}{2}}$$
The above term will be rational if exponent of 3 and 2 are integers.
It means $$\dfrac {1000-r}{2}$$ and $$\dfrac {r}{2}$$ must be integers
The possible set of values of $$r$$ is $$\left \{0, 2, 4, ...., 1000\right \}$$
Hence, number of rational terms is $$501$$.
Question 10
1 / -0

In the expansion of $$(\sqrt 2+\sqrt [3]{5})^{20}$$ the number of rational terms will be:

A
$$3$$

B
$$10$$

C
$$4$$

D
$$8$$

Solution

Given
$$(\sqrt{2}+\sqrt[3]{5})^{20}$$

General term of given expamsion
$$T_{r+1}=\displaystyle^{20}C_{r}(\sqrt{2})^{20-r}(\sqrt[3]{5})^r$$

$$T_{r+1}=\displaystyle^{20}C_{r}(2)^{\dfrac{20-r}{2}}(5)^{\dfrac{r}{3}}$$

Now, $$T_{r+1}$$ be rational if $$\dfrac{20-r}{2}$$ and $$\dfrac{r}{3}$$ are integers
Hence $$20-r$$ Should be multiple of $$2$$ and $$r$$ should be multiple of $$3$$ where $$0\le r\le 20$$
Hence possible value of $$r=0,3,6,9,12,15,18$$
$$20-r=20,17,14,11,8,5,2$$

$$20-r$$ should be multiple of $$2$$
Hence $$20-r=20,14,8,2$$
Hence possible rational terms are $$4$$

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Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 9

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Binomial Theorem Test - 9

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SHARING IS CARING

Question 1

$$\dfrac {1}{2}(3^{50}+1)$$

$$\dfrac {1}{2}(3^{50})$$

$$\dfrac {1}{2}(3^{50}-1)$$

$$\dfrac {1}{2}(2^{50}+1)$$

Solution

Question 2

$$680$$

$$1085$$

$$560$$

$$1240$$

Solution

Question 3

$$12.50$$

$$12.00$$

$$12.75$$

$$12.25$$

Solution

Question 4

$$-84$$

$$84$$

$$126$$

$$-126$$

Solution

Question 5

$$2^{100}$$

$$200$$

$$2^{99}$$

$$202$$

Solution

Question 6

$$7^{th}$$ and $$8^{th}$$

$$8^{th}$$ and $$9^{th}$$

$$28^{th}$$ and $$29^{th}$$

$$27^{th}$$ and $$28^{th}$$

Solution

Question 7

$$A^{2} - B^{2}$$

$$A^{2} + B^{2}$$

$$4AB$$

None

Solution

Question 8

$$n = 2r $$

$$n = 3r$$

$$n = 2r + 1$$

none of these

Solution

Question 9

$$500$$

$$400$$

$$501$$

none of the above

Solution

Question 10

$$3$$

$$10$$

$$4$$

$$8$$

Solution

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