Binomial Theorem Test

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Binomial Theorem Test - 30

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Question 1
1 / -0

Value of $$\sum { \sum _{ 0\le r<s\le n }^{ }{ { (r+s)\left( { C }_{ r }+{ C }_{ s } \right) }^{ 2 } } } $$ is

A
$$n\left[ (n-1)({ _{ }^{ 2n }{ C } }_{ n })+{ 2 }^{ 2n } \right] $$

B
$$n\left[ (n+1)({ _{ }^{ 2n }{ C } }_{ n })+{ 2 }^{ 2n } \right] $$

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

D
None of these

Solution

$$S=\sum _{ 0\le r }^{ }{ \sum _{ <s\le n }^{ }{ { (n-r+n-s)({ C }_{ n-r }-{ C }_{ n-s }) }^{ 2 } } } $$
$$=2n\sum _{ 0\le r }^{ }{ \sum _{ <s\le n }^{ }{ { ({ C }_{ r }+{ C }_{ s }) }^{ 2 } } } -S$$
$$\Rightarrow S=n\left[ (n-1)({ _{ }^{ 2n }{ C } }_{ n })+{ 2 }^{ 2n } \right] $$
Question 2
1 / -0

If $$A=^{ 2n }{ { C }_{ 0 } }.^{ 2n }{ { C }_{ 1 } }+^{ 2n }{ { C }_{ 1 } }^{ 2n-1 }{ { C }_{ 1 } }+^{ 2n }{ { C }_{ 2 } }^{ 2n-2 }{ { C }_{ 1 } }+...$$ then $$A$$ is

A
$$0$$

B
$$n.2^{2n}$$

C
$${ 2 }^{ 10 }-2$$

D
$$1$$

Solution

Given $$A=^{2n}C_0.^{2n}C_1+^{2n}C_1.^{2n-1}C_1+^{2n}C_2.^{2n-2}C_1+........$$

we can write $$A$$ as summation .
$$\Rightarrow A=\sum_{r=0}^{2n} .^{2n}C_r.^{2n-r}C_1=\sum_{r=0}^{2n}.\dfrac{(2n)!}{r!(2n-r)!}.\dfrac{(2n-r)!}{(2n-r-1)!1!}=\sum_{r=0}^{2n}\dfrac{(2n)!}{r!(2n-r-1)!}=2n\sum_{r=0}^{2n}.^{2n-1}C_r$$
$$= 2n.2^{2n-1}=n.2^{2n}$$.

$$\therefore A=n.2^{2n}$$
Question 3
1 / -0

The coefficient of $${ x }^{ 9 }$$ in the expansion of $${ \left( { x }^{ 3 }+\cfrac { 1 }{ { 2 }^{ t } } \right) }^{ 11 }$$, where $$t=\log _{ \sqrt { 2 } }{ { (x }^{ \tfrac 32 } } )$$,

A
$$-5$$

B
$$330$$

C
$$520$$

D
$$5+\log _{ \sqrt { 2 } }{ (3 } )$$

Solution

$$2^{ t }= { 2 }^{\displaystyle \log _{ \sqrt { 2 } }{ { x }^{\tfrac 32 } } }= { 2 }^{\displaystyle \log _{ 2 }{ { x }^{ 3 } } } = x^{ 3 }$$
The general term in the expansion of $$\left( x^{ 3 }+\displaystyle\frac { 1 }{ x^{ 3 } } \right) ^{ 11 }$$ is
$$ T_{ r+1 }= ^{ 11 }C_{ r }\left( x^{ 3 } \right) ^{ 11-r }\displaystyle\frac { 1 }{ x^{ 3r } } = ^{ 11 }C_{ r }x^{ 33-6r }$$
To find the coefficient of $$ x^{ 9 }$$
Hence, $$ 33-6r = 9$$
$$ \Rightarrow r= 4$$
Hence, the coefficient of $$x^{ 9 }$$ is $$^{ 11 }C_{ 4 } = 330$$.
Question 4
1 / -0

Value of $$\sum { \sum _{ 0\le i<j\le n }^{ }{ { \left( { C }_{ i }+{ C }_{ j } \right) }^{ 2 } } } $$ is

A
$${ 2 }^{ 2n }(n+1)({ _{ }^{ 2n }{ C } }_{ n })$$

B
$${ 2 }^{ 2n }+(n+1)({ _{ }^{ 2n }{ C } }_{ n })$$

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

D
none of these

Solution

$$\displaystyle\sum _{ 0\le i }^{ }{ \sum _{ \le j\le n }^{ }{ { \left( { C }_{ i }+{ C }_{ j } \right) }^{ 2 } } } i=0,1,2,...,\left( n-1 \right) $$ where $$j=1,2,3,...,n$$ and $$i<j$$
$$\displaystyle=n\left( { C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+...+{ C }_{ n }^{ 2 } \right) +2\sum { \sum { { C }_{ i }{ C }_{ j }0\le i<j\le n } } $$
$$\displaystyle=n.^{ 2n }{ { C }_{ n } }+\left[ { \left( { C }_{ 0 }+{ C }_{ 1 }+...+{ C }_{ n } \right) }^{ 2 }-\left( { C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+...+{ C }_{ n }^{ 2 } \right) \right] $$
$$\displaystyle=n.^{ 2n }{ { C }_{ n } }+\left( 2n \right) ^{ 2 }-^{ 2n }{ { C }_{ n } }=\left( n-1 \right) .^{ 2n }{ { C }_{ n } }+{ 2 }^{ 2n }.$$
Question 5
1 / -0

Value(s) of $$x$$ for which the fourth term in the expansion of
$${ \left( { \sqrt { x } }^{ 1/(\log _{ 2 }{ x+1 } ) }+{ x }^{ 1/2 } \right) }^{ 6 }$$ is $$40$$ is (are)

A
$$1/8$$

B
$$2$$

C
$$1/16,2$$

D
$$1/8,4$$

Solution

$${ t }_{ 4 }={ _{ }^{ 6 }{ C } }_{ 3 }\left[ { x }^{ 1/2(\log _{ 2 }{ x } +1) } \right] { \left( \cfrac { 1 }{ { x }^{ 12 } } \right) }^{ 3 }=40\quad $$
$$\Rightarrow \left( \cfrac { 3 }{ 2t+2 } +\cfrac { 1 }{ 4 } \right) t=1$$ where $$t=\log _{ 2 }{ x } $$
$$\quad \Rightarrow \quad 7t+{ t }^{ 2 }=4t+4\quad \Rightarrow \quad t=-4,1$$
$$\quad \therefore \quad x=1/16,2$$
Question 6
1 / -0

If $${x}^{2r}$$ occurs in $${ \left( x+\cfrac { 2 }{ { x }^{ 2 } } \right) }^{ n }$$, then $$n-2r$$ must be of the form

A
$$3k$$

B
$$3k-1$$

C
$$3k+1$$

D
$$4k\pm 1$$

Solution

Writing the general term
$$T_{k+1}=\:^{n}C_{r}x^{n-3k}2^{k}$$
For $$x^{2r}$$
$$n-3k=2r$$
$$n-2r=3k$$
Where K is a positive integer $$\geq 0$$
Question 7
1 / -0

If the middle term of $${ \left( x+\cfrac { 1 }{ x } \sin ^{ -1 }{ x } \right) }^{ 8 }$$ is $$\cfrac { 35{ \pi }^{ 4 } }{ 8 } $$, then value of $$x$$ can be

A
$$\dfrac12$$

B
$$\dfrac{\sqrt {3}}2$$

C
$$\dfrac1{\sqrt {2}}$$

D
$$1$$

Solution

Middle term is $$T_{5}$$
$$T_5=\ ^{8}C_{4}(\sin^{-1}(x))^{4}$$

$$=70(\sin^{-1}(x))^{4}$$

$$=\dfrac{35}{8}\pi^{4}$$

Hence, $$70(\sin^{-1}(x))^{4}=\dfrac{70}{16}\pi^{4}$$

$$\Rightarrow(\sin^{-1}(x))^{4}=\dfrac{\pi^{4}}{16}$$

$$\Rightarrow(\sin^{-1}(x)=\pm\dfrac{\pi}{2}$$

$$\Rightarrow x=\pm1$$
Question 8
1 / -0

Value of the expression
$$\quad { C }_{ 0 }+({ C }_{ 0 }+{ C }_{ 1 })+({ C }_{ 0 }+{ C }_{ 1 }+{ C }_{ 2 })+....+({ C }_{ 0 }+{ C }_{ 1 }+....+{ C }_{ n-1 })$$ is

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

B
$$n{2}^{n-2}$$

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

D
$$2n$$

Solution

Simplifying, we get
$$n\:^{n}C_{0}+(n-1)\:^{n}C_{1}+(n-2)\:^{n}C_{2}+...(n-n)\:^{n}C_{n}$$
$$=n[\:^{n}C_{0}+\:^{n}C_{1}+\:^{n}C_{2}+...\:^{n}C_{n}]-[\:^{n}C_{1}+2\:^{n}C_{2}+3\:^{n}C_{3}+...n\:^{n}C_{n}]$$
$$=n2^{n}-\cfrac{d(1+x)^{n}}{dx}|_{x=1}$$
$$=n2^{n}-n2^{n-1}$$
$$=n2^{n-1}$$
Question 9
1 / -0

If $${ S }_{ n }=\cfrac { 1 }{ n+1 } \sum _{ i=0 }^{ n }{ \begin{pmatrix} n \\ i \end{pmatrix} } $$, then $$2{ S }_{ n+1 }-{ S }_{ n }$$ equals

A
$$=2^{n}[\dfrac{3n-2}{(n+1)(n+2)}]$$

B
$$=2^{n}[\dfrac{3n+2}{(n+1)(n+2)}]$$

C
$$=2^{n}[\dfrac{3n+2}{(n-1)(n+2)}]$$

D
0

Solution

$$\sum \:^{n}C_{i}=2^{n}$$ ....sum of binomial coefficients of $$(1+x)^{n}$$

Hence

$$S_{n}=\dfrac{2^{n}}{n+1}$$

Now

$$S_{n+1}=\dfrac{2^{n+1}}{n+2}$$

$$2S_{n+1}-S_{n}$$

$$=\dfrac{2^{n+2}}{n+2}-\dfrac{2^{n}}{n+1}$$

$$=2^{n}[\dfrac{4}{n+2}-\dfrac{1}{n+1}]$$

$$=2^{n}[\dfrac{4n+4-n-2}{(n+1)(n+2)}]$$

$$=2^{n}[\dfrac{3n+2}{(n+1)(n+2)}]$$
Question 10
1 / -0

The number of irrational terms in the expansion of $${ \left( { 4 }^{ 1/5 }+{ 7 }^{ 1/10 } \right) }^{ 45 }$$ is

A
40

B
5

C
41

D
none of these

Solution

Total number of terms in the expansion of $$\displaystyle { \left( { 4 }^{ \frac { 1 }{ 5 } }+{ 7 }^{ \frac { 1 }{ 10 } } \right) }^{ 45 }$$ is $$45+1=46$$
The general term in the expansion is
$$\displaystyle { T }_{ r+1 }=^{ 45 }{ { C }_{ r } }\times { 4 }^{ \frac { 45-r }{ 5 } }\times { 7 }^{ \frac { r }{ 10 } }$$
$${ T }_{ r+1 }$$ is rational if $$r=0,10,20,30,40$$
$$\therefore$$ Number of rational terms $$=5$$
$$\therefore$$ Number of irrational terms $$=46-5=41$$

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Binomial Theorem Test - 30

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Binomial Theorem Test - 30

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Question 1

$$n\left[ (n-1)({ _{ }^{ 2n }{ C } }_{ n })+{ 2 }^{ 2n } \right] $$

$$n\left[ (n+1)({ _{ }^{ 2n }{ C } }_{ n })+{ 2 }^{ 2n } \right] $$

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

None of these

Solution

Question 2

$$0$$

$$n.2^{2n}$$

$${ 2 }^{ 10 }-2$$

$$1$$

Solution

Question 3

$$-5$$

$$330$$

$$520$$

$$5+\log _{ \sqrt { 2 } }{ (3 } )$$

Solution

Question 4

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

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

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

none of these

Solution

Question 5

$$1/8$$

$$2$$

$$1/16,2$$

$$1/8,4$$

Solution

Question 6

$$3k$$

$$3k-1$$

$$3k+1$$

$$4k\pm 1$$

Solution

Question 7

$$\dfrac12$$

$$\dfrac{\sqrt {3}}2$$

$$\dfrac1{\sqrt {2}}$$

$$1$$

Solution

Question 8

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

$$n{2}^{n-2}$$

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

$$2n$$

Solution

Question 9

$$=2^{n}[\dfrac{3n-2}{(n+1)(n+2)}]$$

$$=2^{n}[\dfrac{3n+2}{(n+1)(n+2)}]$$

$$=2^{n}[\dfrac{3n+2}{(n-1)(n+2)}]$$

0

Solution

Question 10

40

5

41

none of these

Solution

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