Binomial Theorem Test

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Binomial Theorem Test - 20

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Question 1
1 / -0

If $${ C }_{ 0 },{ C }_{ 1 },{ C }_{ 2 },...,{ C }_{ n }$$ are the coefficients of the expansion of $${ \left( 1+x \right) }^{ n }$$, then the value of $$\displaystyle \sum _{ 0 }^{ n }{ \frac { { C }_{ k } }{ k+1 } } $$ is

A
$$0$$

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

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

D
None of these

Solution

Here $$\displaystyle { t }_{ r+1 }=\frac { ^{ n }{ { C }_{ r } } }{ r+1 } =\frac { 1 }{ r+1 } .^{ n }{ { C }_{ r } }=\frac { 1 }{ n+1 } .^{ n+1 }{ { C }_{ r+1 } }$$
Putting $$r=0,1,2,,...n$$ and adding we get,
$$\displaystyle \sum _{ 0 }^{ n }{ \frac { ^{ n }{ { C }_{ k } } }{ k+1 } } =\frac { 1 }{ n+1 } \left\{ ^{ n+1 }{ { C }_{ 1 } }+^{ n+1 }{ { C }_{ 2 } }+^{ n+1 }{ { C }_{ 3 } }+...+^{ n+1 }{ { C }_{ n+1 } } \right\} $$
$$\displaystyle =\frac { 1 }{ n+1 } \left\{ { 2 }^{ n+1 }-^{ n+1 }{ { C }_{ 0 } } \right\} =\frac { { 2 }^{ n+1 }-1 }{ n+1 } $$
Question 2
1 / -0

If $$n$$ is an integer between $$0$$ and $$21$$, then the minimum value of $$n!(21 - n)!$$ is attained for $$n=$$

A
$$1$$

B
$$10$$

C
$$12$$

D
$$20$$

Solution

Let us consider $$(1+x)^{21}$$
Hence $$T_{r+1}=\:^{21}C_{r}x^{r}$$
We know that both the middle terms will have the highest value of binomial coefficient.
The middle terms in thew above binomial expansion are $$11^{th}$$ and $$12^{th}$$ terms.
$$T_{10+1}=\:^{21}C_{10}x^{10}=\dfrac{21!}{10!(11)!}x^{10}$$

$$T_{11+1}=\:^{21}C_{11}x^{11}=\dfrac{21!}{11!(10)!}x^{11}$$

Hence $$n!(21-n)!$$ is least for $$n=0, n=21$$ followed by $$n=11$$ and $$n=10$$
Question 3
1 / -0

If the coefficient of $$x^7$$ in $$\left[ax + \left(\displaystyle\frac{1}{bx}\right)\right]^{11} is\ 55 a^{11}$$, then $$a$$ and $$b$$ satisfy the relation

A
$$a + b = 1$$

B
$$a - b = 1$$

C
$$ab = 1$$

D
$$\displaystyle\frac{a}{b} = 1$$

Solution

General term of $$\left[ax + \left(\displaystyle\frac{1}{bx}\right)\right]^{11}$$ is,
$$\displaystyle\quad T_{r+1} = ^{11}C_r (ax)^r\cdot \left(\frac{1}{bx} \right)^{11-r} = ^{11}C_ra^rb^{r-11}x^{2r-11}$$
Let $$T_{r+1}$$ posses $$x^7\Rightarrow 2r-11=7=> r=9$$
Hence coefficient of $$x^7$$ is $$=^{11}C_9a^9b^{-2}=55a^9b^{-2} = 55a^{11}$$ (given)
$$\therefore a^2b^2=1\Rightarrow ab = \pm 1$$
Question 4
1 / -0

If $${ \left( 1+x \right) }^{ n }={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }+...+{ C }_{ n }{ x }^{ n }$$, then $$\displaystyle 2{ C }_{ 0 }+{ 2 }^{ 2 }.\frac { { C }_{ 1 } }{ 2 } +{ 2 }^{ 3 }.\frac { { C }_{ 2 } }{ 3 } +...+{ 2 }^{ n+1 }.\frac { { C }_{ n } }{ n+1 } =$$

A
$$\displaystyle \frac { { 3 }^{ n+1 }-1 }{ n+1 } $$

B
$$\displaystyle \frac { { 3 }^{ n }-1 }{ n } $$

C
$$\displaystyle \frac { { 3 }^{ n+2 }-1 }{ n+2 } $$

D
None of these

Solution

We have,
$$\displaystyle { t }_{ r+1 }={ 2 }^{ r+1 }\frac { ^{ n }{ { C }_{ r } } }{ r+1 } ={ 2 }^{ r+1 }\frac { 1 }{ n+1 } .^{ n+1 }{ { C }_{ r+1 } }$$
Putting $$r=0,1,2,...,n$$ and adding, we get the required sum
$$\displaystyle =\frac { 1 }{ n+1 } \left\{ 2.^{ n+1 }{ { C }_{ 1 } }+{ 2 }^{ 2 }.^{ n+1 }{ { C }_{ 2 } }+...+{ 2 }^{ n+1 }.^{ n+1 }{ { C }_{ r+1 } } \right\} $$
$$\displaystyle =\frac { 1 }{ n+1 } \left\{ { \left( 1+2 \right) }^{ n+1 }-^{ n+1 }{ { C }_{ 0 } } \right\} =\frac { { 3 }^{ n+1 }-1 }{ n+1 } $$.
Question 5
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The value of $$\displaystyle \frac { { _{ }^{ n }{ C } }_{ 1 }^{ } }{ 2 } +\frac { { _{ }^{ n }{ C } }_{ 3 }^{ } }{ 4 } +\frac { { _{ }^{ n }{ C } }_{ 5 }^{ } }{ 6 } +...$$ is

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

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

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

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

Solution

The $$rth$$ term in the given expression is
$$\displaystyle { T }_{ r }=\frac { { _{ }^{ n }{ C } }_{ 2r-1 }^{ } }{ 2r } $$

Since $$\displaystyle \frac { 1 }{ r+1 } .{ _{ }^{ n }{ C } }_{ r }=\frac { 1 }{ n+1 } .{ _{ }^{ n+1 }{ C } }_{ r+1 }$$

$$\displaystyle \therefore \quad { T }_{ r }=\frac { { _{ }^{ n }{ C } }_{ 2r-1 } }{ 2r } =\frac { 1 }{ n+1 } .{ _{ }^{ n+1 }{ C } }_{ 2r }$$

$$\displaystyle \therefore \quad \frac { { _{ }^{ n }{ C } }_{ 1 } }{ 2 } +\frac { { _{ }^{ n }{ C } }_{ 3 } }{ 4 } +\frac { { _{ }^{ n }{ C } }_{ 5 } }{ 6 } +....=\frac { 1 }{ n+1 } \left[ { _{ }^{ n+1 }{ C } }_{ 2 }+{ _{ }^{ n+1 }{ C } }_{ 4 }+... \right] $$

$$\displaystyle =\frac { 1 }{ n+1 } \left[ { 2 }^{ n+1-1 }-{ _{ }^{ n }{ C } }_{ 0 } \right] =\frac { { 2 }^{ n }-1 }{ n+1 } $$
Question 6
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Find the sum 1 $$\times$$ 2 $$\times$$ C$$_1$$ + 2 $$\times$$ 3 C$$_2$$ + n (n+1)C$$_{n'}$$ where C$$_r$$ = $$^n$$C$$_r$$.

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

B
$$n(n+3)2^{n-2}$$

C
$$2^n({^{2n}}C_n)$$

D
None of these

Solution

Let $$n=1$$
Hence $$(n)(n+1)C_{n}$$
$$=(1)(2)C_{1}$$
$$=2$$.
This is satisfied by Option A and Option B
Let $$n=2$$
Hence $$2\:^{2}C_{1}+(2)(3)\:^{2}C_{2}$$
$$=4+6$$
$$=10$$
This is satisfied only by Option B
Hence answer is Option B.
Question 7
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If$$(1+2x+x^2)^n = \displaystyle \sum_{r=0}^{2n} a_r x^r$$, then $$a_r=$$

A
$$(^nC_r)^2$$

B
$$^nC_r\cdot^nC_{r+1}$$

C
$$^{2n}C_r$$

D
$$^{2n}C_{r+1}$$

Solution

$$(x^2+2x+1)^{n}$$
$$=((1+x)^{2})^{n}$$
$$=(1+x)^{2n}$$
Hence
$$T_{r+1}=\:^{2n}C_{r}x^{r}$$
Hence $$a_{r}=\:^{2n}C_{r}$$
Question 8
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Find the ratio of the coefficient of $${ x }^{ 10 }$$ in $${ \left( 1-{ x }^{ 2 } \right) }^{ 10 }$$ and the term independent of $$x$$ in the expansion of $${ \left( x-\cfrac { 2 }{ x } \right) }^{ 10 }$$

A
$$1:8$$

B
$$1:16$$

C
$$1:24$$

D
$$1:32$$

Solution

The general term $$(r+1)^{th}$$ in the expansion of $$(a+b)^{n}$$ is $${T}_{r+1}={}^{n}{C}_{r}a^{n-r}b^{r}$$
In expansion of $$\left(1-x^{2}\right)^{10}$$
$${T}_{r+1}=\quad^{10}{C}_{r}(1)^{10-r}(-x^{2})^{r}$$
$$\Rightarrow {T}_{r+1}=\quad^{10}{C}_{r}(-1)^{r}x^{2r}$$
$$\therefore x^{2r}= x^{10}$$
$$\Rightarrow 2r=10$$
$$\Rightarrow r=5$$
$$coeff({T}_{5+1})={}^{10}{C}_{5}(-1)^{5}= {}^{-10}{C}_{5}$$ (equation $$1$$)
In expansion of $$\left(x-\cfrac{2}{x} \right)^{10}$$
$${T}_{r+1}= \quad^{10}{C}_{r}(x)^{10-r}(\cfrac{-2}{x})^{r}$$
$$\Rightarrow {T}_{r+1}= \quad^{10}{C}_{r} x^{10-2r}(-2)^{r}$$
$$\therefore x^{10-2r}=x^{0}$$
$$\Rightarrow r=5$$
$$coeff({T}_{5+1}) = {}^{10}{C}_{5}(-2)^{5}= {}^{-10}{C}_{5}(32)$$ (equation $$2$$)
By dividing equation $$(1)$$ and $$(2)$$ we get,
$$=\cfrac{^{-10}{C}_{5}}{^{-10}{C}_{5}32}= \cfrac{1}{32}$$
Question 9
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The sum of the coefficients in the expansion of $${ \left( 1+5x-7{ x }^{ 3 } \right) }^{ 3165 }$$ is

A
$$1$$

B
$${ 2 }^{ 3165 }$$

C
$${ 2 }^{ 3164 }$$

D
$$-1$$

Solution

To get sum of coefficient put $$x=1$$
Hence sum of the coefficients in the expansion of $${ \left( 1+5x-7{ x }^{ 3 } \right) }^{ 3165 }$$ is,
$$\quad =(1+5-7)^{3165}=(-1)\cdot (-1)^{3164}=-1$$
Question 10
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Which term in the expansion of $${ \left[ \sqrt [ 3 ]{ \left( \frac { a }{ \sqrt { b } } \right) } +\sqrt { \left( \frac { b }{ \sqrt [ 3 ]{ a } } \right) } \right] }^{ 21 }$$ contains $$a$$ and $$b$$ to one and same power.

A
$$8^{th}$$ term

B
$$9^{th}$$ term

C
$$14^{th}$$ term

D
$$15^{th}$$ term

Solution

For the above question
$$T_{r+1}=\:^{21}C_{r}a^{7-\frac{r}{3}+\frac{r}{6}}b^{\frac{7}{2}-\frac{r}{6}+\frac{r}{2}}$$
$$=\:^{21}C_{r}a^{7-\frac{r}{6}}b^{\frac{7}{2}+\frac{r}{3}}$$
For the powers of a and b to be same
$$7-\frac{r}{6}=\frac{7}{2}+\frac{r}{3}$$
$$\frac{7}{2}=\frac{3r}{6}$$
$$r=7$$
This implies that the term number is 8th.

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Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 20

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Binomial Theorem Test - 20

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SHARING IS CARING

Question 1

$$0$$

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

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

None of these

Solution

Question 2

$$1$$

$$10$$

$$12$$

$$20$$

Solution

Question 3

$$a + b = 1$$

$$a - b = 1$$

$$ab = 1$$

$$\displaystyle\frac{a}{b} = 1$$

Solution

Question 4

$$\displaystyle \frac { { 3 }^{ n+1 }-1 }{ n+1 } $$

$$\displaystyle \frac { { 3 }^{ n }-1 }{ n } $$

$$\displaystyle \frac { { 3 }^{ n+2 }-1 }{ n+2 } $$

None of these

Solution

Question 5

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

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

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

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

Solution

Question 6

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

$$n(n+3)2^{n-2}$$

$$2^n({^{2n}}C_n)$$

None of these

Solution

Question 7

$$(^nC_r)^2$$

$$^nC_r\cdot^nC_{r+1}$$

$$^{2n}C_r$$

$$^{2n}C_{r+1}$$

Solution

Question 8

$$1:8$$

$$1:16$$

$$1:24$$

$$1:32$$

Solution

Question 9

$$1$$

$${ 2 }^{ 3165 }$$

$${ 2 }^{ 3164 }$$

$$-1$$

Solution

Question 10

$$8^{th}$$ term

$$9^{th}$$ term

$$14^{th}$$ term

$$15^{th}$$ term

Solution

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