Binomial Theorem Test

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Binomial Theorem Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

$$9^{th}$$ term in the expansion of $$\left (\dfrac {x}{a}-\dfrac {3a}{x^2}\right )^{12}$$ is:

A
$$^{12}C_9\cdot 3^9x^{-12}a^6$$

B
$$^{12}C_6\cdot 3^8x^{-16}a^6$$

C
$$^{12}C_4\cdot 3^8x^{-12}a^4$$

D
none of the above

Solution

Given
$$\left ( \dfrac{x}{a}-\dfrac{3a}{x^2} \right )^{12}$$
General term of given expansion is
$$T_{r+1}=\sum_{r=0}^{12}\displaystyle^{12}C_{r}\left ( \dfrac{x}{a} \right )^{12-r}\left( \dfrac{3a}{x^2} \right )^{r} (-1)^r$$
We need $$9^{th}$$ term hence put $$r=8$$
$$T_{8+1}=\displaystyle^{12}C_{8}\left ( \dfrac{x}{a} \right )^{12-8}\left( \dfrac{3a}{x^2} \right )^{8} (-1)^8$$

$$T_{9}=\displaystyle^{12}C_{8}\left ( \dfrac{x}{a} \right )^{4}\left( \dfrac{3a}{x^2} \right )^{8} $$

$$T_{9}=\displaystyle^{12}C_{8} x^{4-16}\cdot 3^8 \cdot a^{8-4} $$

$$T_{9}=\displaystyle^{12}C_{8}\cdot 3^8 x^{-12} \cdot a^{4} $$
But we know that $$\displaystyle^{n}C_{r}=\displaystyle^{n}C_{n-r}$$
$$\displaystyle^{12}C_{8}=\displaystyle^{n}C_{4}$$
So
$$T_{9}=\displaystyle^{12}C_{4}\cdot 3^8 x^{-12} \cdot a^{4} $$
Question 2
1 / -0

The middle term in the expansion of $$(x + 4)^4$$ is:

A
$$96x^3$$

B
$$96x^2$$

C
$$-96x^2$$

D
none of the above

Solution

Given
$$(x+4)^4$$
Here the middle terms can be calculated by
middle term$$=\dfrac{n}{2}+1$$

$$=\dfrac{4}{2}+1$$

$$=3^{th} term$$

$=\displaystyle^4C_{2}x^2 4^2$$

$$=\dfrac{4!}{2!2!}x^2\times 16$$

$$=\dfrac{4\times 3}{2\times 1}x^2\times 16$$

$$=96x^2$$
Question 3
1 / -0

Find the term which has the exponent of x as 8 in the expansion of $$\displaystyle\left(x^{\displaystyle\frac{5}{2}}-\frac{3}{x^3\sqrt{x}}\right)^{10}$$

A
$$T_2$$

B
$$T_3$$

C
$$T_4$$

D
Does not exist

Solution

Given, $$\left(x^{\dfrac{5}{2}}-\dfrac{3}{x^3\sqrt{x}}\right)^{10}$$

General term,

$$T_{r+1}={^{n}C_r}x^{n-r}y^r$$

$$T_{r+1}={^{10}C_r}(x^{\dfrac{5}{2}})^{10-r}\left(-\dfrac{3}{x^3\sqrt{x}}\right)^r$$

$$T_{r+1}={^{10}C_r}(x)^{25-\dfrac{5r}{2}}\dfrac{(-3)^r}{x^{\dfrac{7}{2}r}}$$

$$T_{r+1}={^{10}C_r}(-3)^r(x)^{25-6r}$$

given the exponent of x as $$8$$

$$25-6r=8$$

$$25-8=6r$$

$$6r=17$$

$$r=\dfrac{17}{6}$$

since r is a fraction number which is not possible. Hence, the term which has the exponents of x as $$8$$ in the expansion of $$\left(x^{\dfrac{5}{2}}-\dfrac{3}{x^3\sqrt{x}}\right)^{10}$$
does not exist.
Question 4
1 / -0

Find the coefficient of the term independent of x in the expansion of $$\displaystyle\left(6x^3-\frac{5}{x^6}\right)^{12}$$.

A
$$^{12}C_45^86^4$$

B
$$^{12}C_55^76^5$$

C
$$^{12}C_46^85^4$$

D
None of these

Solution

Given, $$\left(6x^3-\dfrac{5}{x^6}\right)^{12}$$

General term,
$$T_{r+1}={^{n}C_r}x^{n-r}y^r$$

To find coefficient of the term independent of x
$$T_{r+1}={^{12}C_r}(6x^3)^{12-x}\left(\dfrac{-5}{x^6}\right)^r$$
$$T_{r+1}={^{12}C_r}(6)^{12-r}(x)^{36-3r}\dfrac{(-5)^r}{x^{6r}}$$
$$T_{r+1}={^{12}C_r}(6)^{12-x}(-5)^r(x)^{36-9r}$$ ......$$(1)$$

Since, term is independent of x.
$$36-9r=0$$
$$9r=36$$
$$r=4$$

Putting $$r=4$$ in $$(1)$$
$$T_{r+1}={^{12}C_4}(6)^{12-4}(-5)^4(x)^0$$
$$T_{r+1}={^{12}C_4}(6)^8(5)^4$$
So, the coefficient of the term independent of x in the expansion of $$\left(6x^3-\dfrac{5}{x^6}\right)^{12}$$ is $$^{12}C_46^85^4$$.
Question 5
1 / -0

If 'p' and 'q' are the coefficients of $$x^a$$ and $$x^b$$ respectively in $$(1+x)^{a+b}$$, then

A
$$2p=q$$

B
$$p+q=0$$

C
$$p=q$$

D
$$p=2q$$

Solution

Given $$(1+x)^{a+b}$$
General term
$$T_{r+1}={^{n}C_r}x^{n-r}y^r$$
$$T_{r+1}={^{n}C_r}x^{n-r}y^r$$
$$T_{r+1}={^{a+b}C_r}(1)^{a+b-r}(x)^r$$
given 'p' the coefficeint of $$x^a$$
$$r=a$$
$$T_{r+1}={^{a+b}C_a}(1)^{a+b-a}(x)^a$$
$$T_{r+1}={^{a+b}C_a}(1)^b(x)^a$$
given 'q' the coefficient of $$x^b$$
$$r=b$$
$$T_{r+1}={^{a+b}C_b}(1)^{a+b-b}(x)^b$$
$$T_{r+1}={^{a+b}C_b}(1)^a(x)^b$$
$$p={^{a+b}C_a}$$
$$q={^{a+b}C_b}={^{a+b}C_{a+b-b}}={^{a+b}C_a}$$
Using $${^{n}C_r}={^{n}C_{n-r}}$$
Hence $$p=q$$.
Question 6
1 / -0

If sum of the first 3 coefficients is $$16$$ in the expansion $$\displaystyle\left(x+\frac{1}{x^3}\right)^n$$, then find n.

A
$$10$$

B
$$8$$

C
$$5$$

D
$$4$$

Solution

Given $$\left(x+\dfrac{1}{x^3}\right)^n$$

General term. $$T_{r+1}={^{n}C_r}x^{n-r}y^r$$
$$T_{r+1}={^{n}C_r}(x)^{n-r}\left(\dfrac{1}{x^3}\right)^r$$
$$T_{r+1}={^{n}C_r}(x)^{n-r-3r}$$
$$T_{r+1}={^{n}C_r}(x)^{n-4r}$$

Given, sum of the first three coefficients is $$16$$
$$T_1=T_{0+1}={^{n}C_0}(x)^n$$
$$T_2=T_{1+1}={^{n}C_1}(x)^{n-4}$$
$$T_3=T_{2+1}={^{n}C_2}(x)^{n-8}$$

$${^{n}C_0}+{^{n}C_1}+{^{n}C_2}=16$$
$$\dfrac{n!}{n!}+\dfrac{n!}{1!(n-1)!}+\dfrac{n!}{2!(n-2)!}=16\\$$
$$\dfrac{n(n-1)!}{(n-1)!}+\dfrac{n(n-1)(n-2)!}{2(n-2)!}=15\\$$
$$n+\dfrac{n(n-1)}{2}=15\\$$
$$\dfrac{2n+n^2-n}{2}=15\\$$
$$n^2+n=30\\$$
$$n^2+n-30=0\\$$
$$n^2+6n-5n-30=0\\$$
$$n(n+6)-5(n+6)=0\\$$
$$(n+6)(n-5)=0\\$$
$$\therefore$$ Since n cannot be negative
so, $$n=5$$
Hence, option C is correct.
Question 7
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The value of the sum $${ \left( _{ }^{ n }{ { C }_{ 1 } } \right) }^{ 2 }+{ \left( _{ }^{ n }{ { C }_{ 2 } } \right) }^{ 2 }+{ \left( _{ }^{ n }{ { C }_{ 3 } } \right) }^{ 2 }+\dots +{ \left( _{ }^{ n }{ { C }_{ n } } \right) }^{ 2 }$$ is

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

B
$$^{ 2n }{ { Cn }_{ n } }$$

C
$$^{ 2n }{ { C }_{ n } }+1$$

D
$$^{ 2n }{ { C }_{ n } }-1$$

Solution

We know that
$${ \left( 1+x \right) }^{ n }=\ ^{ n }{ C }_{ 0 }+\ ^{ n }{ C }_{ 1 }x+\ ^{ n }{ C }_{ 2 }{ x }^{ 2 }+\dots +\ ^{ n }{ C }_{ n }{ x }^{ n }$$ .....(i)
and
$${ \left( x+1 \right) }^{ n }=\ ^{ n }{ C }_{ 0 }{ x }^{ n }+\ ^{ n }{ C }_{ 1 }{ x }^{ n-1 }+\ ^{ n }{ C }_{ 2 }{ x }^{ n-2 }+\dots +\ ^{ n }{ C }_{ n }$$ .....(ii)
On multiplying equations (i) and (ii), we get
$${ \left( 1+x \right) }^{ 2n }=\left( ^{ n }{ C }_{ 0 }+^{ n }{ C }_{ 1 }x+^{ n }{ C }_{ 2 }{ x }^{ 2 }+\dots +^{ n }{ C }_{ n }{ x }^{ n } \right) \times \left( ^{ n }{ C }_{ 0 }{ x }^{ n }+^{ n }{ C }_{ 1 }{ x }^{ n-1 }+^{ n }{ C }_{ 2 }{ x }^{ n-2 }+\dots +^{ n }{ C }_{ n } \right) $$
Coefficient of $${ x }^{ n }$$ in right hand side $$={ \left( ^{ n }{ C }_{ 0 } \right) }^{ 2 }+{ \left( ^{ n }{ C }_{ 1 } \right) }^{ 2 }+\dots +{ \left( ^{ n }{ C }_{ n } \right) }^{ 2 }$$
and
coefficient of $${ x }^{ n }$$ in left hand side $$=\ ^{ 2n }{ C }_{ n }$$
$$\therefore { \left( ^{ n }{ C }_{ 0 } \right) }^{ 2 }+{ \left(\ ^{ n }{ C }_{ 1 } \right) }^{ 2 }+\dots +{ \left( ^{ n }{ C }_{ n } \right) }^{ 2 }=\dfrac { 2n! }{ n!n! } $$
$$\Rightarrow { \left( ^{ n }{ C }_{ 1 } \right) }^{ 2 }+\dots +{ \left( ^{ n }{ C }_{ n } \right) }^{ 2 }=\dfrac { \left( 2n \right) ! }{ n!n! } -1$$ $$=\ ^{ 2n }{ C }_{ n }-1$$
Question 8
1 / -0

The coefficient of $$x^{10}$$ in the expansion of $$1 + (1 + x) + ...............................+(1 + x)^{20}$$ is

A
$$^{19}C_{9}$$

B
$$^{20}C_{10}$$

C
$$^{21}C_{11}$$

D
$$^{22}C_{12}$$

Solution

The given series is in GP. Hence, its sum
$$S = \dfrac {1\left \{(1 + x)^{20 + 1} - 1\right \}}{(1 + x) - 1} = \dfrac {(1 + x)^{21} - 1}{x}$$
Therefore, the required coefficient of $$x^{10}$$ in the expansion of $$\dfrac {(1 + x)^{21} - 1}{x}$$
$$=$$ Coefficient of $$x^{11}$$ in the expansion of $$(1 + x)^{21} - 1$$
$$=\ ^{21}C_{11}$$
Question 9
1 / -0

The middle term in the expansion of $$(1 + x)^{2n}$$ is

A
$$\dfrac {1.3.5...(2n - 1)}{n}x^{n}$$

B
$$\dfrac {1.3.5...(2n - 1)}{n!}2^{n - 1}x^{n}$$

C
$$\dfrac {1.3.5...(2n - 1)}{n!}x^{n}$$

D
$$\dfrac {1.3.5...(2n - 1)}{n!}2^{n}x^{n}$$

Solution

Middle term in the expansion of $$(1 + x)^{2n}$$; is $$= t_{n + 1}$$

$$= ^{2n}C_{n}.1^{2n - n}.x^{n}$$

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

$$= \dfrac {\overline {2n}(2n - 1)(\overline {2n - 2})(2n - 3) .... \overline {4}\times 3\times \overline {2} \times 1}{(n!)(n!)} . x^{n}$$

$$= \dfrac {2^{n}[n(n - 1)(n - 2) ....\times 2\times 1][(2n - 1)(2n - 3) .... 3\times 1]}{(n!)(n!)} x^{n} $$

$$=\dfrac {(2n - 1)(2n - 3) ....3\times 1}{n!} 2^{n}x^{n}$$
Question 10
1 / -0

The number of irrational terms in the binomial expansion of $$(3^{1/5}+7^{1/3})^{100}$$ is

A
$$90$$

B
$$88$$

C
$$94$$

D
$$95$$

Solution

General term of $$(3^{1/5}+7^{1/3})^{100}$$ is given by,
$$T_{r+1}=\ ^{100}C_r(3^{1/5})^{100-r}(7^{1/3})^r$$
$$\displaystyle =\ ^{100}C_r\cdot 3^{\frac{100-r}{5}}\cdot 7^{\frac{r}{3}}$$
For a rational term,
$$\dfrac{100-r}{5}$$ and $$\dfrac{r}{3}$$ must be integer.
Hence,
$$r=0, 15, 30, 45, 60, 75, 90$$
$$\therefore$$ There are seven rational terms.
Hence, number of irrational terms $$=101-7=94$$

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Binomial Theorem Test - 36

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Binomial Theorem Test - 36

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Question 1

$$^{12}C_9\cdot 3^9x^{-12}a^6$$

$$^{12}C_6\cdot 3^8x^{-16}a^6$$

$$^{12}C_4\cdot 3^8x^{-12}a^4$$

none of the above

Solution

Question 2

$$96x^3$$

$$96x^2$$

$$-96x^2$$

none of the above

Solution

Question 3

$$T_2$$

$$T_3$$

$$T_4$$

Does not exist

Solution

Question 4

$$^{12}C_45^86^4$$

$$^{12}C_55^76^5$$

$$^{12}C_46^85^4$$

None of these

Solution

Question 5

$$2p=q$$

$$p+q=0$$

$$p=q$$

$$p=2q$$

Solution

Question 6

$$10$$

$$8$$

$$5$$

$$4$$

Solution

Question 7

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

$$^{ 2n }{ { Cn }_{ n } }$$

$$^{ 2n }{ { C }_{ n } }+1$$

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

Solution

Question 8

$$^{19}C_{9}$$

$$^{20}C_{10}$$

$$^{21}C_{11}$$

$$^{22}C_{12}$$

Solution

Question 9

$$\dfrac {1.3.5...(2n - 1)}{n}x^{n}$$

$$\dfrac {1.3.5...(2n - 1)}{n!}2^{n - 1}x^{n}$$

$$\dfrac {1.3.5...(2n - 1)}{n!}x^{n}$$

$$\dfrac {1.3.5...(2n - 1)}{n!}2^{n}x^{n}$$

Solution

Question 10

$$90$$

$$88$$

$$94$$

$$95$$

Solution

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