Binomial Theorem Test

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Binomial Theorem Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

If sum of all the coefficients in the expansion of $$(x^{3/2} + x^{1/3})^n$$ is 128, then the coefficient of $$x^5$$ is

A
35

B
45

C
7

D
none of these

Solution

Substituting $$x=1$$, we get the sum of the coefficients as
$$(2)^{n}=128$$
$$\therefore n=7$$
Hence writing the general term, we get
$$T_{r+1}=\:^{7}C_{r}x^{\frac{63-11r}{6}}$$
Hence for the coefficient of $$x^{5}$$
$$63-11r=6(5)$$
$$63-11r=30$$
$$33-11r=0$$
$$\therefore r=3$$
Hence coefficient is $$\:^{7}C_{3}$$ $$=35$$.
Question 2
1 / -0

The sum of the coefficients in the expansion of $$(x + 2y + z)^{10}$$ is

A
$$2^{10}$$

B
$$4^{10}$$

C
$$3^{10}$$

D
$$1$$

Solution

Given expression is $$(x+2y+z)^{10}$$
Substituting $$x=y=z=1$$, we get the sum of the coefficients as
$$(1+2+1)^{10}$$
$$=4^{10}$$
Question 3
1 / -0

In the expansion of $$\displaystyle \left [ 7^{1/3}+11^{1/9} \right ]^{6561}$$, the number of terms free from radicals is

A
$$730$$

B
$$729$$

C
$$725$$

D
$$750$$

Solution

The general term of a binomial expansion is $$ { T }_{ r+1 }={ _{ }^{ n }{ C } }_{ r }{ x }^{ n-r }{ a }^{ r }$$

From the above question $${ T }_{ r+1 }={ _{ }^{ 6561 }{ C } }_{ r }{ \left( { 7 }^{ \cfrac { 1 }{ 3 } } \right) }^{ 6561-r }\times { \left( { 11 }^{ \frac { 1 }{ 9 } } \right) }^{ r }={ _{ }^{ 6561 }{ C } }_{ r }{ 7 }^{ \frac { 6561 }{ 3 } -\frac { r }{ 3 } }{ 11 }^{ \frac { r }{ 9 } }.$$

Number of terms free from radicals are those which makes the general term an integer

$${ T }_{ r+1 }$$ will be integer when $$ \dfrac { 6561-r }{ 3 } $$ and $$ \dfrac { r }{ 9 } $$ are positive integers for $$0\le r\le 6561$$.

Take $$ r=0,{ \dfrac { 6561-0 }{ 3 } =2187 ,\dfrac { 6561 }{ 9 } =729 }$$ is an integer.

For $$r=9$$, $$ { T }_{ r+1 }$$ is an integer

Thus, $$r=0,9,18,27,..6561$$ are the multiples of $$9$$, which makes $$ { T }_{ r+1 }$$ is an integer.

Hence,$$\dfrac { 6561 }{ 9 } =729$$ There are $$729$$ terms excluding $$0$$

$$\therefore$$ Number of terms with free radicals$$=729+1=730$$ terms (including the term 0)
Question 4
1 / -0

In the expression of $$\displaystyle \left ( 3-\sqrt{\frac{17}{4}+3\sqrt{2}} \right )^{15}$$, the 11th term is a

A
positive integer

B
positive irrational number

C
negative integer

D
negative irrational number

Solution

General term$$={ T }_{ r+1 }={ _{ }^{ n }{ C } }_{ r }{ x }^{ n-r }{ a }^{ r }$$ where $$ n=15$$,$$ x=3$$
$$a=\sqrt { \frac { 17 }{ 4 } +3\sqrt { 2 } } \\ { T }_{ 11 }={ T }_{ 10+1 }={ _{ }^{ 15 }{ C } }_{ 10 }{ (3) }^{ 15-10 }{ \left( \frac { 17 }{ 4 } +3\sqrt { 2 } \right) }^{ \frac { 1 }{ 2 } \times 10 }\\ ={ _{ }^{ 15 }{ C } }_{ 10 }{ 3 }^{ 5 }\left( \frac { 17 }{ 4 } +3\sqrt { 2 } \right) ^{ 5 }$$
Note:$${ (3\sqrt { 2 } ) }^{ 5 }=972\sqrt { 2 } $$ is an irrational number.
Sum of a rational number and an irrational number is irrational number.
Hence,11th term is a positive irrational number.
Question 5
1 / -0

The total number of distinct terms in the expansion of $$\displaystyle \left ( x+a \right )^{100}+\left ( x-a \right )^{100}$$ after simplification is

A
$$50$$

B
$$202$$

C
$$51$$

D
$$none\ of\ these$$

Solution

Number of terms in a binomial expansion is$$=(n+1)$$
When n is even,in the expansion$$ { (x+a) }^{ n }+{ (x-a) }^{ n }=2\left\{ { _{ }^{ n }{ C } }_{ 0 }{ x }^{ n }+{ _{ }^{ n }{ C } }_{ 2 }{ x }^{ n-2 }{ a }^{ 2 }+...+{ a }^{ n } \right\}$$
Thus,the odd number of terms get cancelled and even number of terms get added except the first term.
Therefore,total number of terms is=$$\left( \frac { n }{ 2 } +1 \right)$$terms
Hence,$${ (x+a) }^{ 100 }+{ (x-a) }^{ 100 }$$ has
$$\left( \frac { 100 }{ 2 } +1 \right) terms=51$$ terms therefore Number of terms$$=51$$terms
Question 6
1 / -0

If $$p+q=1$$, then the value of $$\displaystyle \sum _{ r=0 }^{ 15 }{ { _{ }^{ 15 }{ C } }_{ r } } { p }^{ 15-r }{ q }^{ r }$$

A
$$1$$

B
$$15$$

C
$$18$$

D
$$20$$

Solution

It is given that $$p+q=1$$
Therefore $$q=1-p$$
Hence $$\sum_{r=0} ^{15} \:^{15}C_{r}p^{15-r}q^{r}$$
$$\quad =(p+q)^{15}$$
Now $$q=1-p$$
Hence $$(p+q)^{15}$$
$$=(p+1-p)^{15}$$
$$=1$$.
Question 7
1 / -0

The value of $$ \sum _{ r=1 }^{ 10 }{ { r }^{ 2 }.\cfrac { { _{ }^{ 26 }{ C } }_{ r } }{ { _{ }^{ 26 }{ C } }_{ r-1 } } } $$ is-

A
$$1100$$

B
$$1300$$

C
$$900$$

D
$$1800$$

Solution

$$\sum _{ r=1 }^{ 10 }{ { r }^{ 2 } } .\left( \cfrac { 27-r }{ r } \right) =\sum _{ r=1 }^{ 10 }{ \left( 27r-{ r }^{ 2 } \right) } $$

$$=\cfrac { 27r(r+1) }{ 2 } -\cfrac { r(r+1).(2r+1) }{ 6 } $$

put $$r=10$$

$$=27.5.11-5.11.7=1100$$
Question 8
1 / -0

If $${ \left( { x }^{ 2 }+\cfrac { 1 }{ { x }} \right) }^{ n }$$ has exactly one middle term which is equal to $$\alpha.{x}^{3}$$ then the value of $$(\alpha+n)$$ is- ($$n\in N$$)

A
$$18$$

B
$$21$$

C
$$24$$

D
$$26$$

Solution

$$\because$$ Only one middle term $$\Rightarrow$$ $$n$$ is even middle term $$=\alpha.{x}^{3}$$
$${ _{ }^{ n }{ C } }_{ n/2 }{ \left( { x }^{ 2 } \right) }^{ n/2 }.{ \left( \cfrac { 1 }{ x } \right) }^{ n/2 }=\alpha .{ x }^{ 3 }$$
$${ _{ }^{ n }{ C } }_{ n/2 }x^{n/2}=\alpha .{ x }^{ 3 }\Rightarrow \cfrac { n }{ 2 } =3$$
and $$\alpha ={ _{ }^{ n }{ C } }_{ n/2 }$$
$$\Rightarrow$$ $$n=6$$ and $$\alpha={ _{ }^{ 6 }{ C } }_{ 3 }=20$$
$$\therefore \alpha +n=20+6=26$$
Question 9
1 / -0

If the second term of the expansion $$\displaystyle \left [ a^{1/13}+\frac{a}{\sqrt{a^{-1}}} \right ]^{n}\: \: is\: \: 14a^{5/2}$$, then the value of $$\displaystyle \frac{^{n}{C}_{3}}{^{n}{C}_{2}}$$ is

A
$$4$$

B
$$3$$

C
$$12$$

D
$$6$$

Solution

Using the formula $${ T }_{ r+1= }{ _{ }^{ n }{ C } }_{ r }{ { x }^{ n-r }{ a }^{ r } }$$
in the above question,we have $${ T }_{ 2 }={ _{ }^{ n }{ C } }_{ 1 }{ \left( { a }^{ \frac { 1 }{ 13 } } \right) }^{ n-1 }(a\sqrt { a } )=14{ a }^{ \frac { 5 }{ 2 } }$$

$$\Rightarrow { T }_{ 2 }={ _{ }^{ n }{ C } }_{ 1 }{ a }^{ \tfrac { n-1 }{ 13 } }{ a }^{ 1+\frac { 1 }{ 2 } }=14{ a }^{ \tfrac { 5 }{ 2 } }$$

$$\Rightarrow { _{ }^{ n }{ C } }_{ 1 }{ a }^{ \frac { n-1 }{ 13 } +\frac { 3 }{ 2 } }=14{ a }^{ \frac { 5 }{ 2 } }$$

Comparing L.H.S and R.H.S,

$$\Rightarrow { a }^{ \tfrac { 2n+37 }{ 26 } }={ a }^{ \tfrac { 5 }{ 2 } }$$

Since bases are same,we can equate the powers

Thus, $$\cfrac { 2n+37 }{ 26 } =\cfrac { 5 }{ 2 } $$

On solving , $$2(2n+37)=5\times 26\Rightarrow 2n+37=65$$

$$\therefore 2n=65-37=28$$

Hence, $$ n=\cfrac { 28 }{ 2 } =14$$

Using the formula, $$\cfrac { { _{ }^{ n }{ C } }_{ r } }{ { _{ }^{ n }{ C } }_{ r-1 } } =\cfrac { n-r+1 }{ r }$$ where $$r=3$$ and $$n=14$$

$$\cfrac { { _{ }^{ n }{ C } }_{ 3 } }{ { _{ }^{ n }{ C } }_{ 2 } } =\cfrac { 14-3+1 }{ 3 } =\cfrac { 12 }{ 3 } =4$$
Question 10
1 / -0

The value of the expression $$\displaystyle \left ( \sum_{r=0}^{10} \ ^{10}C_{r} \right )\left ( \sum_{k=0}^{10}\left ( -1 \right )^{k}\dfrac{^{10}\textrm{C}_{k}}{2^{k}} \right )$$ is:

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

B
$$\displaystyle 2^{20}$$

C
$$1$$

D
$$\displaystyle 2^{5}$$

Solution

Given
$$\left ( \sum_{r=0}^{10} \displaystyle^{10}C_{r} \right ) \left ( (-1)^k \dfrac{\sum_{k=0}^{10} \displaystyle^{10}C_{k}}{2^k} \right )$$
We konw that
$$(1+x)^n=C_{0}+C_{1}x+C_{2}x^2+..............+C_{n}x^n------(1)$$

When putting $$x=1$$ we get
$$2^n=C_{0}+C_{1}+C_{2}+..............+C_{n}$$

Replace x with $$\dfrac{-1}{x}$$ in eq (1) we get

$$\left(1-\dfrac{1}{x}\right)^n=C_{0}-C_{1}\dfrac{1}{x}+\dfrac{C_{2}}{x^2}-\dfrac{C_{3}}{x^3}+..............+(-1)^n\dfrac{ C_{n}}{x^n}$$

Putting $$x=2$$ in above we get
$$\left(\dfrac{2-1}{2}\right)^n=C_{0}-C_{1}\dfrac{1}{2}+\dfrac{C_{2}}{2^2}-\dfrac{C_{3}}{2^3}+..............+(-1)^n\dfrac{ C_{n}}{2^n}$$

$$\dfrac{1}{2^n}=C_{0}-C_{1}\dfrac{1}{2}+\dfrac{C_{2}}{2^2}-\dfrac{C_{3}}{2^3}+..............+(-1)^n\dfrac{ C_{n}}{2^n}$$

Hence

$$\left ( \sum_{r=0}^{10} \displaystyle^{10}C_{r} \right ) \left ( (-1)^k \dfrac{\sum_{k=0}^{10} \displaystyle^{10}C_{k}}{2^k} \right )=2^{10} \left ( \dfrac{1}{2^{10}} \right )=1$$

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Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 34

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Binomial Theorem Test - 34

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SHARING IS CARING

Question 1

35

45

7

none of these

Solution

Question 2

$$2^{10}$$

$$4^{10}$$

$$3^{10}$$

$$1$$

Solution

Question 3

$$730$$

$$729$$

$$725$$

$$750$$

Solution

Question 4

positive integer

positive irrational number

negative integer

negative irrational number

Solution

Question 5

$$50$$

$$202$$

$$51$$

$$none\ of\ these$$

Solution

Question 6

$$1$$

$$15$$

$$18$$

$$20$$

Solution

Question 7

$$1100$$

$$1300$$

$$900$$

$$1800$$

Solution

Question 8

$$18$$

$$21$$

$$24$$

$$26$$

Solution

Question 9

$$4$$

$$3$$

$$12$$

$$6$$

Solution

Question 10

$$\displaystyle 2^{10}$$

$$\displaystyle 2^{20}$$

$$1$$

$$\displaystyle 2^{5}$$

Solution

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