Binomial Theorem Test

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Binomial Theorem Test - 57

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Question 1
1 / -0

If the fourth term of $${ \left( \sqrt { { x }^{ \left( \cfrac { 1 }{ 1+\log { x } } \right) } } +\sqrt [ 12 ]{ x } \right) }^{ 6 }$$ is equal to 200 and $$x>1$$, then $$x$$ is equal to

A
$$10\sqrt { 2 } $$

B
$$10$$

C
$${ 10 }^{ 4 }$$

D
$$10/\sqrt { 2 } $$

Solution

$${ \left( \sqrt { { x }^{ \left( \cfrac { 1 }{ 1+\log { x } } \right) } } +\sqrt [ 12 ]{ x } \right) }^{ 6 }$$
$$={ \left( { x }^{ \cfrac { 1 }{ 2(1+\log { x) } } }+{ x }^{ \frac { 1 }{ 12 } } \right) }^{ 6 }$$

Now, $$T_{4}=T_{3+1}=^6C_{3}{ x }^{ \cfrac { 3 }{ 2(1+\log { x) } } } {x}^{\cfrac{1}{4}}$$
$$\Rightarrow \displaystyle 200=20{ x }^{ \cfrac { 3 }{ 2(1+\log { x) } } +\frac { 1 }{ 4 } }$$
$$\Rightarrow 10={ x }^{ \cfrac { 3 }{ 2(1+\log { x) } } +\frac { 1 }{ 4 } }$$
Taking log on both sides,

$$\Rightarrow \displaystyle 1=\left[{ \cfrac { 3 }{ 2(1+\log { x) } } +\frac { 1 }{ 4 }}\right]log_{10} x$$
$$\Rightarrow \displaystyle \frac{1}{log x}-\frac{1}{4}=\frac{3}{2(1+\log x)}$$
$$\Rightarrow (\log x)^2+3\log x-4=0$$
$$\Rightarrow (\log x+4)(\log x -1)=0$$
$$\Rightarrow \log _{10}x=1$$ ($$\because \log x=-4$$, not possible )
$$\Rightarrow x=10$$
Question 2
1 / -0

The number of irrational terms in the expansion of $$(\sqrt[8]{5}+\sqrt[6]{2})^{100}$$ is

A
$$97$$

B
$$98$$

C
$$96$$

D
$$99$$

Solution

$$T_{r+1}=\:^{100}C_{r}5^\left({\tfrac{100-r}{8}}\right)2^\left({\tfrac{r}{6}}\right)$$
For rational terms, $$100 -r$$ should be a multiple of $$8$$ and $$r$$ should be a multiple of $$6$$
Hence we get rational terms for
$$r=12, 36, 60$$ and $$84$$
Hence there will be $$4$$ rational terms.
Therefore, total number of irrational terms will be
$$101-4$$
$$=97$$ terms.
Question 3
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Directions For Questions

If $${ C }_{ r }={ _{ }^{ n }{ C } }_{ r }$$ then to evaluate the expression
$$P= \sum _{ 0\le }^{ }{ \sum _{ r<s\le n }^{ }{ { C }_{ r } } { C }_{ s } } $$, we make use of $${ C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+....+{ C }_{ n }^{ 2 }={ _{ }^{ 2n }{ C } }_{ n }$$ and expansion of $${ \left( { C }_{ 0 }+{ C }_{ 1 }+....+{ C }_{ n } \right) }^{ 2 }$$.

...view full instructions

Value of $$P=\sum _{ 0\le }^{ }{ \sum _{ r<s\le n }^{ }{ { C }_{ r } } { C }_{ s } } $$ is

A
$${ 2 }^{ 2n }-\cfrac { 1 }{ 2 } ({ _{ }^{ 2n }{ C } }_{ n } )$$

B
$${ 2 }^{ 2n-1 }-\cfrac { 1 }{ 2 } ({ _{ }^{ 2n }{ C } }_{ n })$$

C
$${ 2 }^{ 2n }-\cfrac 12 ({ _{ }^{ 2n }{ C } }_{ n })$$

D
None of these

Solution
Question 4
1 / -0

If $$n$$ is even, then value of the expression
$${ C }_{ 0 }-\cfrac { 1 }{ 2 } { C }_{ 1 }^{ 2 }+\cfrac { 1 }{ 3 } { C }_{ 2 }^{ 2 }-.....+\cfrac { { \left( -1 \right) }^{ n } }{ n+1 } { C }_{ n }^{ 2 }$$
where
$${ C }_{ r }={ _{ }^{ n }{ C } }_{ r }$$ is

A
$$\cfrac { { \left( -1 \right) }^{ n }n! }{ (n+1){ (n/2)! }^{ 2 } } $$

B
$$\cfrac { { \left( -1 \right) }^{ n-1 }n! }{ (n+1){ (n/2)! }^{ 2 } } $$

C
$$\cfrac { { \left( -1 \right) } }{ (n+1){ (n/2)! }^{ 2 } } $$

D
$$\cfrac { { \left( -1 \right) }^{ n/2 }n! }{ (n+1){ (n/2)! }^{ 2 } } $$

Solution

We have
$${ C }_{ 0 }-\cfrac { 1 }{ 2 } { C }_{ 1 }x+\cfrac { 1 }{ 3 } { C }_{ 2 }{ x }^{ 2 }-......+{ \left( -1 \right) }^{ n }\cfrac { 1 }{ n+1 } { C }_{ n }{ x }^{ n }\quad \cfrac { 1 }{ (n+1)x } \left[ 1-{ (1-x) }^{ n+1 } \right] ........(1)$$
and
$${ C }_{ 0 }+\quad { C }_{ 1 }\cfrac { 1 }{ x } +{ C }_{ 2 }\cfrac { 1 }{ { x }^{ 2 } } +........+{ C }_{ n }{ \left( \cfrac { 1 }{ x } \right) }^{ n }={ (1+x) }^{ n }\quad .....(2)$$
Note that
$$S={ C }_{ 0 }^{ 2 }-\cfrac { 1 }{ 2 } { C }_{ 1 }^{ 2 }+\cfrac { 1 }{ 3 } { C }_{ 2 }^{ 2 }-........+{ \left( -1 \right) }^{ n }\cfrac { 1 }{ n+1 } { C }_{ n }^{ 2 }\quad $$
$$=$$ Coefficient of constant term in
$$\left[ { C }_{ 0 }-\cfrac { 1 }{ 2 } { C }_{ 1 }x+\cfrac { 1 }{ 3 } { C }_{ 2 }{ x }^{ 2 }-......+{ \left( -1 \right) }^{ n }\cfrac { 1 }{ n+1 } { C }_{ n }{ x }^{ n } \right] \times \left[ { C }_{ 0 }+\quad { C }_{ 1 }\cfrac { 1 }{ x } +{ C }_{ 2 }\cfrac { 1 }{ { x }^{ 2 } } +........+{ C }_{ n }{ \left( \cfrac { 1 }{ { x }^{ n } } \right) } \right] $$
$$=$$ coefficient of constant term in
$$=\cfrac { 1 }{ n+1 } \left[ \cfrac { 1-{ (1-x) }^{ n+1 } }{ x } \right] { \left( 1+\cfrac { 1 }{ x } \right) }^{ n }\quad $$
$$=$$ coefficient of $${c}^{n+1}$$ in
$$\cfrac { 1 }{ n+1 } \left[ { (1+x) }^{ n }-{ (1+{ x }^{ 2 }) }^{ n }(1-x) \right] $$
$$=-\cfrac { 1 }{ n+1 } \quad coefficient\quad of\quad { x }^{ n+1 }\quad in\quad { (1+{ x }^{ 2 }) }^{ n }(1-x)$$
As $$n$$ is even, let $$n=2m$$, then
$$S=\cfrac { 1 }{ 2m+1 } \quad coefficient\quad of\quad { x }^{ 2m }\quad in\quad { (1+{ x }^{ 2 }) }^{ 2m }\quad $$
$$=\cfrac { 1 }{ 2m+1 } { \left( -1 \right) }^{ m }\left( { _{ }^{ 2m }{ C } }_{ m } \right) $$
$$=\cfrac { { \left( -1 \right) }^{ n } }{ n+1 } ({ _{ }^{ n }{ C } }_{ n/2 })\quad =\cfrac { { \left( -1 \right) }^{ n/2 }n! }{ (n+1){ (n/2)! }^{ 2 } } $$
Question 5
1 / -0

Directions For Questions

If $$ \displaystyle C_{r}=^{n}C_{r} $$ then to evaluate the expansion $$ \displaystyle A=\sum \sum_{0\leq r\leq s\leq n}^{} $$ $$ \displaystyle C_{r}C_{s} $$, We make use of $$ \displaystyle \sum_{r=0}^{n}C_{r}^{2}=^{2n}C_{n} $$ and the expansion of $$ \displaystyle \left ( \sum_{r=0}^{n}C_{r} \right )^{2} $$. On the basis of above information answer the following questions.

...view full instructions

The value of $$ \displaystyle B=\sum_{0\leq r\leq s\leq n}^{} $$ $$ \displaystyle \sum \left ( C_{r}-C_{s} \right )^{2} $$ is

A
$$ \displaystyle \left ( n+1 \right )^{n}-2^{2n} $$

B
$$ \displaystyle \left ( n+1 \right )\displaystyle ^{2n}C_{n}-2^{n} $$

C
$$ \displaystyle \left ( n+1 \right )\displaystyle ^{2n}C_{n}-2^{2n} $$

D
$$ \displaystyle 2^{2n}-2^{n} $$

Solution

Given $$\displaystyle A=\sum \sum _{ 0\leq r\leq s\leq n } C_{ r }C_{ s }$$ ....(1)

and $$\displaystyle \sum _{ r=0 }^{ n } C_{ r }^{ 2 }=^{ 2n }C_{ n }$$ ....(2)

Given expansion can be written as
$$\left( \sum _{ r=0 }^{ n } C_{ r } \right) ^{ 2 }=(C_{ 0 }+C_{ 1 }+C_{ 2 }+....+C_{ n })^{ 2 }$$

Consider, $$(C_{ 0 }+C_{ 1 }+C_{ 2 }+....+C_{ n })^{ 2 }$$
$$\Rightarrow \displaystyle (C_{ 0 }+C_{ 1 }+C_{ 2 }+....+C_{ n })^{ 2 }=\sum _{ r=0 }^{ n } C_{ r }^{ 2 }+2\sum \sum _{ 0\leq r\leq s\leq n } C_{ r }C_{ s }$$

We know that
$$C_0+C_1+C_2+....+C_n = 2^n$$

$$\Rightarrow \displaystyle (2^n)^2=^{2n}C_{n}+2A$$ (by (1), (2))
$$\Rightarrow \displaystyle 2^{2n}=^{2n}C_n+2A$$
$$\Rightarrow \displaystyle 2A= 2^{2n}-^{2n}C_n $$ .....(3)

Now, consider $$ \displaystyle B=\sum_{0\leq r\leq s\leq n} \displaystyle \sum \left ( C_{r}-C_{s} \right )^{2} $$

$$\displaystyle ={ ({ C }_{ 0 }-{ C }_{ 1 }) }^{ 2 }+{ ({ C }_{ 0 }-{ C }_{ 2 }) }^{ 2 }+{ ({ C }_{ 0 }-{ C }_{ 3 }) }^{ 2 }+.....+{ ({ C }_{ 0 }-{ C }_{ n }) }^{ 2 }$$
$$+ { ({ C }_{ 1 }-{ C }_{ 2 }) }^{ 2 }+{ ({ C }_{ 1 }-{ C }_{ 3 }) }^{ 2 }+....+{ ({ C }_{ 1 }-{ C }_{ n }) }^{ 2 }$$
$$+{ ({ C }_{ 2 }-{ C }_{ 3 }) }^{ 2 }+{ ({ C }_{ 2 }-{ C }_{ 4 }) }^{ 2 }+...+{ ({ C }_{ 2 }-{ C }_{ n }) }^{ 2 }$$
$$+....+{ ({ C }_{ n-1 }-{ C }_{ n }) }^{ 2 }$$

$$={ { (C }_{ 0 } }^{ 2 }+{ { C }_{ 1 } }^{ 2 }-2{ C }_{ 0 }{ { C }_{ 1 } })+{ { (C }_{ 0 } }^{ 2 }+{ { C }_{ 2 } }^{ 2 }-2{ C }_{ 0 }{ { C }_{ 2 } })+{ { (C }_{ 0 } }^{ 2 }+{ { C }_{ 3 } }^{ 2 }-2{ C }_{ 0 }{ { C }_{ 3 } })+..+{ { (C }_{ 0 } }^{ 2 }+{ { C }_{ n } }^{ 2 }-2{ C }_{ 0 }{ { C }_{ n } })+$$
$${ { (C }_{ 1 } }^{ 2 }+{ { C }_{ 2 } }^{ 2 }-2{ C }_{ 1 }{ { C }_{ 2 } })+{ { (C }_{ 1 } }^{ 2 }+{ { C }_{ 3 } }^{ 2 }-2{ C }_{ 1 }{ { C }_{ 3 } })+....+{ { (C }_{ 1 } }^{ 2 }+{ { C }_{ n } }^{ 2 }-2{ C }_{ 1 }{ { C }_{ n } })+$$
$${ { (C }_{ 2 } }^{ 2 }+{ { C }_{ 3 } }^{ 2 }-2{ C }_{ 2 }{ { C }_{ 3 } })+{ { (C }_{ 2 } }^{ 2 }+{ { C }_{ 4 } }^{ 2 }-2{ C }_{ 2 }{ { C }_{ 4 } })+...+{ { (C }_{ 2 } }^{ 2 }+{ { C }_{ n } }^{ 2 }-2{ C }_{ 2 }{ { C }_{ n } })+$$
$$....+{ { (C }_{ n-1 } }^{ 2 }+{ { C }_{ n } }^{ 2 }-2{ C }_{ n-1 }{ { C }_{ n } })$$

$$\Rightarrow \displaystyle B=n{ { (C }_{ 0 } }^{ 2 }+{ { C }_{ 1 } }^{ 2 }+...+{ { C }_{ n } }^{ 2 })-2\sum \sum _{ 0\leq r\leq s\leq n } C_{ r }C_{ s }$$

$$\Rightarrow B=n(^{2n}C_n)-2A$$ (by(2))
$$\Rightarrow B=n(^{2n}C_n)-2^{2n}+^{2n}C_n$$ (by (3))
$$\Rightarrow B=(n+1)^{2n}C_n -2^{2n}$$
Question 6
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The value of the expression $$\displaystyle \frac{1 + 4\:.\:343 + 7\:.\:4 + 2\:.\:3\:.\:49 + 7\:.\:343}{16 + 2^6\:.\:3^1 + 2^{5}\:.\:3^{3} + 2^{6}\: . \: 3^{3} + 2^{4}\:.\:3^{4}}$$ equal

A
$$\displaystyle 1$$

B
$$\displaystyle 2$$

C
$$\displaystyle 4$$

D
$$\displaystyle 3$$

Solution

$$\displaystyle \frac{1 + 4\:.\:343 + 7\:.\:4 + 2\:.\:3\:.\:49 +

7\:.\:343}{16 + 2^6\:.\:3^1 + 2^{5}\:.\:3^{3} + 2^{6}\: . \: 3^{3}

+ 2^{4}\:.\:3^{4}}$$

Above expression can be written as

$$\dfrac{\>^4C_0\cdot1^4 + \>^4C_1\cdot1^3\cdot7 + \>^4C_2\cdot 1^2\cdot 7^2 + \>^4C_3\cdot 1\cdot 7^3 + \>^4C_4\cdot7^4}{\>^4C_0\cdot2^4 + \>^4C_1\cdot2^3\cdot6 + \>^4C_2\cdot 2^2\cdot 6^2 + \>^4C_3\cdot 2\cdot 6^3 + \>^4C_4\cdot6^4}$$

$$= \dfrac{(1+7)^4}{(2+6)^4} = 1$$
Question 7
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If in the expansion of $${ \left( { x }^{ 3 }-\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ n }$$,
$$n\in N$$, sum of coefficient of $${ x }^{ 5 }$$ and $${ x }^{ 10 }$$ is $$0$$, then value of $$n$$ is

A
$$5$$

B
$$10$$

C
$$15$$

D
none of these

Solution

$$(r+1)$$th term in the expansion of
$${ \left( { x }^{ 3 }-\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ n }$$ is
$${ _{ }^{ n }{ C } }_{ r }{ \left( { x }^{ 3 } \right) }^{ n-r }{ \left( -\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ r }={ _{ }^{ n }{ C } }_{ r }{ (-1) }^{ r }{ x }^{ 3n-5r }\quad $$
For coefficient of $${ x }^{ 5 }$$, we set $$3n-5r=5$$ $$\Rightarrow \quad r=\cfrac { 3n-5 }{ 5 }=p \quad $$ (say) and for coefficient of $${ x }^{ 10 }$$, we set
$$3n-5r=10$$ $$\Rightarrow \quad r=\cfrac { 3n-10 }{ 5 } =q$$ (say)
Note that $$p-q=1$$. We are given
$${ _{ }^{ n }{ C } }_{ p }{ (-1) }^{ p }+{ _{ }^{ n }{ C } }_{ q }{ (-1) }^{ q }=0$$
$$\Rightarrow { _{ }^{ n }{ C } }_{ p }{ (-1) }^{ p }+{ _{ }^{ n }{ C } }_{ p-1 }{ (-1) }^{ p-1 }=0$$
$$\Rightarrow { _{ }^{ n }{ C } }_{ p }={ _{ }^{ n }{ C } }_{ p-1 }\quad \Rightarrow \quad n-p=p-1$$
$$\Rightarrow n=2p-1\quad \Rightarrow \quad p=\cfrac { n+1 }{ 2 } $$
Thus, $$\cfrac { n+1 }{ 2 } =\cfrac { 3n-5 }{ 5 } \quad \Rightarrow 5n+5=6n-10\quad $$
$$\Rightarrow 15=n$$
Question 8
1 / -0

If $${ S }_{ n }=1+q+{ q }^{ 2 }+{ q }^{ 3 }+...+{ q }^{ n }$$ and $$\displaystyle { S' }_{ n }=1+\left( \frac { q+1 }{ 2 } \right) +{ \left( \frac { q+1 }{ 2 } \right) }^{ 2 }+...+{ \left( \frac { q+1 }{ 2 } \right) }^{ n },q\neq 1$$ then $$^{ n+1 }{ { C }_{ 1 } }+^{ n+1 }{ { C }_{ 2 } }.{ S }_{ 1 }+^{ n+1 }{ { C }_{ 3 } }.{ S }_{ 2 }+...+^{ n+1 }{ { C }_{ n+1 } }.{ S }_{ n }=$$

A
$${ 2 }^{ n-1 }.{ S' }_{ n }$$

B
$${ 2 }^{ n }.{ S' }_{ n }$$

C
$${ 2 }^{ n+1 }.{ S' }_{ n }$$

D
None of these

Solution

$$S_n$$=$$\dfrac{q^{n+1}-1}{q-1}$$ summation of g.p
$$^{n+1}{C}_{1}$$+$$^{n+1}{C}_{2}$$$$\dfrac{q^2-1}{q-1}$$+$$^{n+1}{C}_{3}$$$$\dfrac{q^3-1}{q-1}$$+..................+$$^{n+1}{C}_{n+1}$$$$\dfrac{q^{n+1}-1
}{q-1}$$
$$\dfrac{^{n+1}{C}_{0}}{q-1}$$-$$^{n+1}{C}_{1}$$$$\dfrac{q-1}{q-1}$$+$$^{n+1}{C}_{2}$$$$\dfrac{q^2-1}{q-1}$$+$$^{n+1}{C}_{3}$$$$\dfrac{q^3-1}{q-1}$$+..................+$$^{n+1}{C}_{n+1}$$$$\dfrac{q^{n+1}-1
}{q-1}$$-$$\dfrac{^{n+1}{C}_{0}}{n-1}$$
($$\dfrac{^{n+1}{C}_{0}}{q-1}$$+$$^{n+1}{C}_{1}$$$$\dfrac{q}{q-1}$$+$$^{n+1}{C}_{2}$$$$\dfrac{q^2}{q-1}$$+$$^{n+1}{C}_{3}$$$$\dfrac{q^3}{q-1}$$+..................+$$^{n+1}{C}_{n+1}$$$$\dfrac{q^{n+1}
}{q-1}$$)-($$\dfrac{^{n+1}{C}_{0}}{q-1}$$+$$^{n+1}{C}_{1}$$$$\dfrac{}{q-1}$$+$$^{n+1}{C}_{2}$$$$\dfrac{}{q-1}$$+$$^{n+1}{C}_{3}$$$$\dfrac{}{q-1}$$+.................. +$$^{n+1}{C}_{n+1}$$$$\dfrac{}{q-1}$$)
=$$\dfrac{{(1+q)}^{n+1}}{q-1}$$-$$\dfrac{2^{n+1}}{q-1}$$
=$$2^n(\dfrac{{(\dfrac{1+q}{2}})^{n+1}}{\dfrac{q-1}{2}}$$-$$\dfrac{2}{q-1})$$
=$$2^n(\dfrac{{(\dfrac{1+q}{2}})^{n+1}}{\dfrac{q-1}{2}}$$-$$\dfrac{1}{\dfrac{q-1}{2}})$$
=$$2^n$$$$S'_n$$
and$$S'_n$$=$$(\dfrac{{(\dfrac{1+q}{2}})^{n+1}}{\dfrac{q-1}{2}}$$$$-$$$$\dfrac{1}{\dfrac{q-1}{2}})$$ summation of g.p
Question 9
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$$\left( _{ }^{ m }{ { C }_{ 0 }^{ } }+^{ m }{ { C }_{ 1 }^{ } }-^{ m }{ { C }_{ 2 }^{ } }-^{ m }{ { C }_{ 3 }^{ } } \right) +\left( ^{ m }{ { C }_{ 4 }^{ } }+^{ m }{ { C }_{ 5 }^{ } }-^{ m }{ { C }_{ 6 }^{ } }-^{ m }{ { C }_{ 7 }^{ } } \right) +...=0$$ if and only if for some positive integer $$k, m=$$

A
$$4k$$

B
$$4k+1$$

C
$$4k-1$$

D
$$4k+2$$

Solution

Consider $${ \left( \cos { \theta } -i\sin { \theta } \right) }^{ m }=^{ m }{ { C }_{ 0 }\cos ^{ m }{ \theta } - }^{ m }{ { C }_{ 1 }\cos ^{ m-1 }{ \theta } i }\sin { \theta } +...+^{ m }{ { C }_{ m } }{ \left( -i\sin { \theta } \right) }^{ m }$$ ....(1)
$${ \left( \cos { \theta } +i\sin { \theta } \right) }^{ m }=^{ m }{ { C }_{ 0 }\cos ^{ m }{ \theta } + }^{ m }{ { C }_{ 1 }\cos ^{ m-1 }{ \theta } i }\sin { \theta } +...+^{ m }{ { C }_{ m } }{ \left( i\sin { \theta } \right) }^{ m }$$ ....(2)
Adding (1) and (2), we get
$$2\cos { m\theta =2\left[ ^{ m }{ { C }_{ 0 }\cos ^{ m }{ \theta } -^{ m }{ { C }_{ 2 }\cos ^{ m-2 }{ \theta } \sin ^{ 2 }{ \theta ... } } } \right] } $$ ...(3)
Subtracting (3) and (4), we get
$$2i\sin { m\theta } =2i\left[ ^{ m }{ { C }_{ 1 }\cos ^{ m-1 }{ \theta } i }\sin { \theta } -^{ m }{ { C }_{ 3 }\cos ^{ m-3 }{ \theta } \sin ^{ 3 }{ \theta ... } } \right] $$ ....(4)
Adding (3) and (4), we get
$$\cos { m\theta } +\sin { m\theta } \\ =\left[ ^{ m }{ { C }_{ 0 }\cos ^{ m }{ \theta } + }^{ m }{ { C }_{ 1 }\cos ^{ m-1 }{ \theta } i }\sin { \theta } -^{ m }{ { C }_{ 2 }\cos ^{ m-2 }{ \theta } \sin ^{ 2 }{ \theta - } }^{ m }{ { C }_{ 3 }\cos ^{ m-3 }{ \theta } \sin ^{ 3 }{ \theta ... } } \right] $$
$$\displaystyle \Rightarrow \sqrt { 2 } \sin { \left( m\theta +\frac { \pi }{ 4 } \right) } \\ =\left[ ^{ m }{ { C }_{ 0 }\cos ^{ m }{ \theta } + }^{ m }{ { C }_{ 1 }\cos ^{ m-1 }{ \theta } i }\sin { \theta } -^{ m }{ { C }_{ 2 }\cos ^{ m-2 }{ \theta } \sin ^{ 2 }{ \theta - } }^{ m }{ { C }_{ 3 }\cos ^{ m-3 }{ \theta } \sin ^{ 3 }{ \theta ... } } \right] $$
Putting $$\displaystyle \theta =\frac { \pi }{ 4 } $$, we get
$$\displaystyle \sqrt { 2 } \sin { \left( \frac { \left( m+1 \right) \pi }{ 4 } \right) } =\frac { 1 }{ { 2 }^{ \frac { m }{ 2 } } } [\left( ^{ m }{ { C }_{ 0 } }+^{ m }{ { C }_{ 1 }-^{ m }{ { C }_{ 2 }-^{ m }{ { C }_{ 3 } } } } \right) \\ +\left( ^{ m }{ { C }_{ 4 }+^{ m }{ { C }_{ 5 }-^{ m }{ { C }_{ 6 } }-^{ m }{ { C }_{ 7 } } } } \right) +...+\left( ^{ m }{ { C }_{ m-3 }+^{ m }{ { C }_{ m-2 }-^{ m }{ { C }_{ m-1 }-^{ m }{ { C }_{ m } } } } } \right) ]$$
Hence $$m+1=4k$$, for given quantity to be $$0$$.
$$\Rightarrow m=4k-1$$ where $$k\in N$$
Question 10
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The value of $$\displaystyle ^{20}C_{0}+ ^{20}C_{1}+^{20}C_{2}+^{20}C_{4}+^{20}C_{12}+^{20}C_{13}+^{20}C_{14}+^{20}C_{15} $$ is

A
$$\displaystyle 2^{19}-\frac{(^{20}C_{10}+^{20}C_{9})}{2}$$

B
$$\displaystyle 2^{19}-\frac{(^{20}C_{10}+^2 \times {20}C_{9})}{2}$$

C
$$\displaystyle 2^{19}-\frac{^{20}C_{10}}{2}$$

D
$$2^{20}-(20_{C_{10}}+20_{C_{9}})$$

Solution

$$ ^{20}C_{0}+^{20}C_{1}+^{20}C_{2}+^{20}C_{4}+^{20}C_{12}+^{20}C_{13}+^{20}C_{14}+^{20}C_{15} $$
$$ \because ^{n}C_{x} = ^{n}C_{n-x} $$
$$ \because ^{20}C_{15} = ^{20}C_{5} $$
$$ ^{20}C_{14} = ^{20}C_{6} $$
$$ ^{20}C_{13} = ^{20}C_{7} $$
$$ ^{20}C_{12} = ^{20}C_{8} $$
$$ \therefore $$ we get
$$ ^{20}C_{0}+C_{1}+C_{2}+C_{4}+C_{5}+C_{6}+C_{7}+C_{8} $$
$$ \because ^{20}C_{0}+C_{1}+...C_{8}+C_{9}+C_{11}+...^{20}C_{20} = 2^{20} $$
$$ 2\left \{ C_{0}+C_{1}+...C_{9} \right \}+C_{10} = 2^{20} $$
$$ 2\left \{ C_{0}+C_{1}+...C_{8} \right \}+^{2}C_{9}+C_{10} = 2^{20} $$
$$ \therefore C_{0}+C_{1}....C_{8} = \dfrac{2^{20}-(^{2}C_{9}+C_{10})}{2} $$
$$ = 2^{19}-\dfrac{(^{20}C_{10}+2^{20}C_{9})}{2} $$
Ans. (B)

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Binomial Theorem Test - 57

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Binomial Theorem Test - 57

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SHARING IS CARING

Question 1

$$10\sqrt { 2 } $$

$$10$$

$${ 10 }^{ 4 }$$

$$10/\sqrt { 2 } $$

Solution

Question 2

$$97$$

$$98$$

$$96$$

$$99$$

Solution

Question 3

$${ 2 }^{ 2n }-\cfrac { 1 }{ 2 } ({ _{ }^{ 2n }{ C } }_{ n } )$$

$${ 2 }^{ 2n-1 }-\cfrac { 1 }{ 2 } ({ _{ }^{ 2n }{ C } }_{ n })$$

$${ 2 }^{ 2n }-\cfrac 12 ({ _{ }^{ 2n }{ C } }_{ n })$$

None of these

Solution

Question 4

$$\cfrac { { \left( -1 \right) }^{ n }n! }{ (n+1){ (n/2)! }^{ 2 } } $$

$$\cfrac { { \left( -1 \right) }^{ n-1 }n! }{ (n+1){ (n/2)! }^{ 2 } } $$

$$\cfrac { { \left( -1 \right) } }{ (n+1){ (n/2)! }^{ 2 } } $$

$$\cfrac { { \left( -1 \right) }^{ n/2 }n! }{ (n+1){ (n/2)! }^{ 2 } } $$

Solution

Question 5

$$ \displaystyle \left ( n+1 \right )^{n}-2^{2n} $$

$$ \displaystyle \left ( n+1 \right )\displaystyle ^{2n}C_{n}-2^{n} $$

$$ \displaystyle \left ( n+1 \right )\displaystyle ^{2n}C_{n}-2^{2n} $$

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

Solution

Question 6

$$\displaystyle 1$$

$$\displaystyle 2$$

$$\displaystyle 4$$

$$\displaystyle 3$$

Solution

Question 7

$$5$$

$$10$$

$$15$$

none of these

Solution

Question 8

$${ 2 }^{ n-1 }.{ S' }_{ n }$$

$${ 2 }^{ n }.{ S' }_{ n }$$

$${ 2 }^{ n+1 }.{ S' }_{ n }$$

None of these

Solution

Question 9

$$4k$$

$$4k+1$$

$$4k-1$$

$$4k+2$$

Solution

Question 10

$$\displaystyle 2^{19}-\frac{(^{20}C_{10}+^{20}C_{9})}{2}$$

$$\displaystyle 2^{19}-\frac{(^{20}C_{10}+^2 \times {20}C_{9})}{2}$$

$$\displaystyle 2^{19}-\frac{^{20}C_{10}}{2}$$

$$2^{20}-(20_{C_{10}}+20_{C_{9}})$$

Solution

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