Binomial Theorem Test

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Binomial Theorem Test - 46

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Question 1
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The ratio of coefficient of $$x^{3}andx^{4}$$ in expansion $$(1+x)^{12}$$is:

A
$$\frac{4}{9}$$

B
$$\frac{1}{3}$$

C
$$\frac{2}{3}$$

D
$$\frac{1}{9}$$

Solution

$${\left( 1 + x \right)}^{12}$$
As we know general term in the expansion $${\left( a + b \right)}^{n}$$ is given as-
$${T}_{r+1} = {^{n}{C}_{r}} {\left( a \right)}^{n-r} {\left( b \right)}^{r}$$
For expansion $${\left( 1 + x \right)}^{12}$$,
$$a = 1$$
$$b = x$$
$$n = 12$$
Therefore,
$${T}_{r+1} = {^{12}{C}_{r}} {\left( 1 \right)}^{12 - r} {\left( x \right)}^{r}$$
For coefficient of $${x}^{3}$$-
$$r = 3$$
Therefore,
$${T}_{3+1} = {^{12}{C}_{3}} {\left( 1 \right)}^{12 - 3} {\left( x \right)}^{3}$$
As we know that,
$${^{n}{C}_{r}} = \cfrac{n!}{r! \left( n - r \right)!}$$
$$\therefore {T}_{4} = 220 {x}^{3} ..... \left( 1 \right)$$
Similaraly,
For coefficient of $${x}^{4}$$
$$r = 4$$
Therefore,
$${T}_{4+1} = {^{12}{C}_{4}} {\left( 1 \right)}^{12-4} {\left( x \right)}^{4}$$
$$\Rightarrow {T}_{5} = 495 {x}^{4} ..... \left( 2 \right)$$
From $${eq}^{n} \left( 1 \right) \& \left( 2 \right)$$, we have
$$\cfrac{\text{Coefficient of } {x}^{3}}{\text{Coefficient of } {x}^{4}} = \cfrac{220}{495} = \cfrac{4}{9}$$
Hence the ratio of coefficient of $${x}^{3}$$ and $${x}^{4}$$ is $$\cfrac{4}{9}$$.
Hence the correct answer is $$\left( A \right) \cfrac{4}{9}$$.
Question 2
1 / -0

The first $$3$$ terms in the expansion of $$(1+ax)^{n}(n\neq 0)$$ are $$1, 6x$$ and $$16x^{2}$$. Then the value of $$a$$ and $$n$$ are respectively

A
$$2$$ and $$9$$

B
$$3$$ and $$2$$

C
$$2/3$$ and $$9$$

D
$$3/2$$ and $$6$$

Solution

$$(1+ax)^n={^{n}C_o1}+{^{n}C_1}ax+{^{n}C_2}(ax)^2$$
$$\therefore {^{n}C_1}(ax)=6x$$
$$\therefore {^{n}C_2}(ax)^2=16x^2$$
$$\therefore \dfrac{n!}{(n-1)!}\times a=6$$
$$\therefore xa=6$$
$$\therefore \dfrac{n!a^2}{(n-2)!2!}=16$$
$$\therefore (n)(n-1)a^2=32$$
$$\therefore \dfrac{1}{(n-1)(a)}=\dfrac{3}{16}$$
$$\therefore \dfrac{1}{na-a}=\dfrac{3}{16}$$
$$\therefore 6-a=\dfrac{16}{3}$$
$$\therefore a=\dfrac{6-16}{3}=\dfrac{2}{3}$$
$$\therefore x=\dfrac{6}{2}\times 3=9$$
$$\therefore x=9, a=\dfrac{2}{3}$$.
Question 3
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For $$r=0,1,2,,....10$$ let $${A}_{r},{B}_{r}$$ and $${C}_{r}$$ denote respectively the coefficient of $${x}^{r}$$ in the expansions of $${(1+x)}^{10},{(1+x)}^{20}$$ and $${(1+x)}^{30}$$. Then $$\sum _{ r=1 }^{ 10 }{ { A }_{ r }\left( { B }_{ 10 }{ B }_{ r }-{ C }_{ 10 }{ A }_{ r } \right) } $$ is equal to

A
$${B}_{10}-{C}_{10}$$

B
$${A}_{10}({B}_{10}^{2}-{C}_{10}{A}_{10})$$

C
$$0$$

D
$${C}_{10}-{B}_{10}$$

Solution

$$A_{r}=$$ Coefficient of $$x^{r}$$ in $$(1+r)^{10}=10\ C_{r}$$
$$B_{r}=$$ Coefficient of $$x^{r}$$ in $$(1+r)^{20}=20\ C_{r}$$
$$C_{r}=$$ Coefficient of $$x^{r}$$ in $$(1+r)^{30}=30\ C_{r}$$
$$\displaystyle\sum_{r=1}^{10}Ar (B_{10}B_{r}-C_{10}A_{r})=\sum_{r=1}^{10}A_{r}B_{10}B_{r}-\sum_{r=1}^{10}A_{r}C_{10}A_{r}$$
$$=\displaystyle\sum_{r=1}^{10} \_{}^{10}C_{r} \_{}^{20}C_{10} \_{}^{20}C_{r}-\sum_{r=1}^{10} \_{}^{10}C_{r} \_{}^{30}C_{10} \_{}^{10}C_{r}$$
$$=\displaystyle\sum_{r=1}^{10} \_{}^{10}C_{r} \_{}^{20}C_{10} \_{}^{20}C_{r}-\sum_{r=1}^{10} \_{}^{10}C_{r} \_{}^{30}C_{10} \_{}^{10}C_{r}$$
$$= \_{}^{20}C_{10}\displaystyle\sum_{r=1}^{10} \_{}^{10}C_{10-r} \_{}^{20}C_{r}- \_{}^{30}C_{10}\sum_{r=1}^{10} \_{}^{10}C_{10-r} \_{}^{10}C_{r}$$
$$=_{}^{20}C_{10}\left(_{}^{30}C_{10}-1\right)-_{}^{30}C_{10}\left(_{}^{20}C_{10}-1\right)$$
$$=_{}^{30}C_{10}-_{}^{20}C_{10}=C_{10}-B_{10}$$
$$C_{10}-B_{10}$$
Question 4
1 / -0

The numerical value of middle terms in $${ \left( 1-\cfrac { 1 }{ x } \right) }^{ n }{(1-x)}^{n}$$ is

A
$${ _{ }^{ 2n }{ C } }_{ n }$$

B
$${ _{ }^{ n }{ C } }_{ n }$$

C
$$\left( { _{ }^{ 2n }{ C } }_{ n } \right) $$

D
$$\left( { _{ }^{ n }{ C } }_{ n } \right) $$

Solution

$$(1-\dfrac{1}{x})^{n}(1-x)^{n}$$

$$=\dfrac{(x-1)^{n}}{x^{n}}(1-x)^{n}$$

$$=\dfrac{(-1)^{n}}{x^{n}}(1-x)^{2n}$$

middle term = $$T_{n+1}=\dfrac{(-1)^{n}}{x^{n}}.^{2n}C_{n}(1)^{2n-n}(-x)^{n}$$

$$=\dfrac{(-1)^{n}}{x^{n}}.^{2n}C_{n}.1.(-1)^{n}.x^{n}$$

middle term = $$(-1)^{2n}.^{2n}C_{n}$$
Question 5
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The value of
$$\left( { _{ }^{ 7 }{ C } }_{ 0 }+{ _{ }^{ 7 }{ C } }_{ 1 } \right) +\left( { _{ }^{ 7 }{ C } }_{ 1 }+{ _{ }^{ 7 }{ C } }_{ 2 } \right) +.....\left( { _{ }^{ 7 }{ C } }_{ 6 }+{ _{ }^{ 7 }{ C } }_{ 7 } \right) $$ is

A
$${2}^{7}-1$$

B
$${2}^{8}-2$$

C
$${2}^{8}-1$$

D
$${2}^{8}$$

Solution

$$(^{7}C_{0}+^{7}C_{1}+...^{7}C_{6})+(^{7}C_{1}+^{7}C_{2}+...^{7}C_{7})$$
$$(2^{7}-^{7}C_{7})+(2^{7}-^{7}C_{0})$$
$$(2^{7}-1)+(2^{7}-1)$$
$$2.2^{7}-2$$
$$2^{8}-2$$
Question 6
1 / -0

The $$r$$th term of series $$2\dfrac{1}{2} + 1\dfrac{7}{{13}} + 1\dfrac{1}{9} + \dfrac{{20}}{{23}} + .....$$ is

A
$$\dfrac{{20}}{{5r + 3}}$$

B
$$\dfrac{{20}}{{5r - 3}}$$

C
$$20\left( {5r + 3} \right)$$

D
$$\dfrac{{20}}{{5{r^2} + 3}}$$

Solution
Question 7
1 / -0

If $$r$$th term is middle term in $${ \left( { x }^{ 2 }-\cfrac { 1 }{ 2x } \right) }^{ 20 }$$ then $$(r+3)$$th term is:

A
$$\cfrac { { _{ }^{ 20 }{ C } }_{ 7 }x }{ { 2 }^{ 13 } } $$

B
$$-\left( \cfrac { { _{ }^{ 20 }{ C } }_{ 5 }x }{ { 4 }^{ 13 } } \right) $$

C
$$-\left( \cfrac { { _{ }^{ 20 }{ C } }_{ 7 }x }{ { 2 }^{ 13 } } \right) $$

D
$$-\left( \cfrac { { _{ }^{ 20 }{ C } }_{ 14 }x }{ { 4 }^{ 13 } } \right) $$
Question 8
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If the rth term in the expansion of $${\left( {{x \over 3} - {2 \over {{x^2}}}} \right)^{10}}$$ contains $${x^4}$$ then r is equal to

A
2

B
3

C
4

D
5

Solution
Question 9
1 / -0

If the constant term in the expansion of $$\left(x^{2}-\dfrac{1}{x}\right)^{n}$$ is $$15$$ then the value of $$n$$ is

A
$$6$$

B
$$9$$

C
$$12$$

D
$$15$$

Solution
Question 10
1 / -0

If $$(1-x-x^2)^{20}$$ = $$\sum _{ r=0 }^{ 40 }{ a_4,x^x },$$ then
$$a_1+3a_3+5a_5+........+39a_{39}=$$

A
$$40$$

B
$$-40$$

C
$$80$$

D
$$-80$$

Solution