Binomial Theorem Test

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Binomial Theorem Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

The sum of the series $$\displaystyle\sum_{r=0}^{n}\left ( ^{n+1}C_{r} \right ) $$ equals

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

B
$$\displaystyle \frac{1}{2}\frac{2n!}{n!n!}$$

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

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

Solution

Let $$\sum_{r=0}^{n} (^{n+1}C_r)$$ be $$f$$.

$$f=\sum_{r=0}^n (^{n+1}C_r)$$

Consider the expansion of $$(1+x)^{n+1}$$.
$$(1+x)^{n+1}=\sum_{k=0}^{n+1} (^{n+1}C_r 1^{n+1-k} x^{k}) $$
Substituting $$x=1$$ in above equation
$$(1+1)^{n+1}=\sum_{k=0}^{n+1} (^{n+1}C_k) = f + ^{n+1}C_{n+1}$$

$$\therefore f=2^{n+1}-1$$

The value of asked expression is $$2^{n+1}-1$$.
Question 2
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the $$7th$$ term in $$\displaystyle { \left( \frac { 1 }{ y } +{ y }^{ 2 } \right) }^{ 10 }$$, when expanded in descending power of $$y$$, is

A
$$\displaystyle \frac { 210 }{ { y }^{ 2 } } $$

B
$$\displaystyle \frac { { y }^{ 2 } }{ 210 } $$

C
$$210{ y }^{ 2 }$$

D
None of these

Solution

When $$\displaystyle { \left( \frac { 1 }{ y } +{ y }^{ 2 } \right) }^{ 10 }$$ is expanded, the power of $$y$$ go on increasing as the terms proceed.
Hence it is expanded in ascending powers of $$y$$.
So $$\displaystyle { \left( { y }^{ 2 }+\frac { 1 }{ y } \right) }^{ 10 }$$, when expanded will be in descending power of $$y$$.
Hence $$\displaystyle { t }_{ 7 }=_{ }^{ 10 }{ { C }_{ 6 }^{ } }{ \left( { y }^{ 2 } \right) }^{ 4 }{ \left( \frac { 1 }{ y } \right) }^{ 6 }=\frac { 10\times 9\times 8\times 7 }{ 4\times 3\times 2\times 1 } { y }^{ 2 }=210{ y }^{ 2 }$$
Question 3
1 / -0

Number of rational term is the expansion of $$\left ( 7^{1/3}+11^{1/9} \right )^{729}$$

A
81

B
82

C
730

D
None of these

Solution

Since 7 and 11 are prime numbers, hence application of general formula for number of rational terms will be
$$=1+\dfrac{729}{L.C.M(3,9)}$$

$$=1+\dfrac{729}{9}$$

$$=1+81$$
$$=82$$ rational terms.
Question 4
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The values of $$x$$ in the expansion $$\displaystyle \left ( x+x^{log_{10}x} \right )^{5}$$ , if the third term in the expansion is $$10,00,000$$

A
$$\displaystyle 10$$

B
$$\displaystyle 10^{2}$$

C
$$\displaystyle 10^{3}$$

D
None of these

Solution

$$T_{3}$$ $$=\:^{5}C_{2}x^{3+2log_{10}(x)}$$
$$=10^6$$
$$x^{3+2log_{10}(x)}=10^{5}$$
Now x has to be in powers of $$10$$.
Let $$x=10^{k}$$
Therefore, $$10^{k(3+2k)}=10^{6}$$
$$2k^{2}+3k=5$$
$$2k^{2}+3k-5=0$$
$$2k^{2}+5k-2k-5=0$$
$$(k-1)(2k+5)=0$$
$$k=1$$ and $$k=\frac{-5}{2}$$
Therefore, $$x=10$$ and $$x=10^{\frac{-5}{2}}$$
Question 5
1 / -0

If the sum of the coefficients in the expansion of $$\left ( 1+2x^{} \right )^{m}$$ and $$\left ( 2+x \right )^{n}$$ are respectively $$6561$$ and $$243$$, then the position of the point $$\left ( m,n \right )$$ with respect to circle $$x^{2}+y^{2}-4x-6y-32=0$$

A
is on the circle

B
is outside the circle

C
is inside the circle

D
circle can not be fixed

Solution

Substituting $$x=1$$, we get
$$3^{m}=6561$$
$$\Rightarrow 3^{m}=3^{8}$$
$$\Rightarrow m=8$$
And $$3^{n}=243$$
$$3^{n}=3^{5}$$
$$ \Rightarrow n=5$$
Substituting the point $$(m,n)$$ in the given equation of circle we get
$$8^{2}+5^{2}-4(8)-6(5)-32$$
$$=64+25-32-30-32$$
$$=25-30$$
$$=-5$$
$$-5<0$$ hence the point is inside the circle.
Question 6
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If the sum of the coefficients in the expansion of $$(1+2x)^{n}$$ is $$2,187$$, the greatest term in the expansion, if $$\displaystyle x = \frac{1}{2}$$ is/are

A
$$5^{th}$$

B
$$4^{th}$$ and $$5^{th}$$

C
$$4^{th}$$

D
None of these

Solution

Substituting $$x=1$$ for the sum of coefficients of the given expansion.
$$3^{n}=2187$$
$$\log_{3}(n)=\log_{3}(2187)$$
$$n=7$$
Hence the expansion will be of the form $$(1+2x)^{7}$$
Now there will be total of 8 terms, and for $$x=\frac{1}{2}$$ the value of a term will only depend on its binomial coefficient.
Therefore the greatest terms will be the set of two middle terms. which are the 4th term and the 5th term.
Question 7
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If $$\displaystyle\left ( 1+x \right )^{n}=\sum_{r=0}^{n}C_{r}x^{r}$$ and $$\sum { \sum _{ 0\le i<j\le n }{ { C }_{ i }\times { C }_{ j } } } $$ represent the products of the $$C_{i}$$'s taken two at a time, then its value equals

A
$$\displaystyle2^{2n-1}-\frac{\left (2n\right)!}{\left (n!\right)^{2}}$$

B
$$\displaystyle2^{2n-1}+\frac{2n!}{n!n!}$$

C
$$\displaystyle2^{2n-1}-\frac{2n!}{2\cdot n!n!}$$

D
None of these

Solution

$$A=\sum { \sum _{ 0\le i<j\le 1 }{ { C }_{ i }\times { C }_{ j } } } $$
so, $$A=\cfrac { 1 }{ 2 } \times ({ (\sum _{ r=0 }^{ n }{ { C }_{ i }) } }^{ 2 })-\sum _{ r=0 }^{ n }{ { { C }_{ i } }^{ 2 } } )$$
therefore our required answer is option C.
$$\sum _{ i=0 }^{ n }{ { ({ C }_{ i }) }^{ 2 }= } ^{ 2n }{ { C }_{ n } }$$
$${ (\sum _{ i=0 }^{ n }{ { C }_{ i } } ) }^{ 2 }={ 2 }^{ 2n }$$
Question 8
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Sum of the coefficients of the terms of degree $$m$$ in the expansion of
$${ (1+x) }^{ n }{ (1+y) }^{ n }{ (1+z) }^{ n }$$ is

A
$${ ({ _{ }^{ n }{ C } }_{ m }) }^{ 3 }$$

B
$$3({ _{ }^{ n }{ C } }_{ m })$$

C
$${ _{ }^{ n }{ C } }_{ 3m }$$

D
$${ _{ }^{ 3n }{ C } }_{ m }$$

Solution

The Coefficient of $$x^{m}$$ $$=$$ Number of ways of choosing $$m$$ balls out of $$n$$ black balls, $$n$$ green balls and $$n$$ blue ball.
Hence total number of balls $$=3n$$.
Required is $$m$$.
Hence required combination is $$\:^{3n}C_{m}$$.
Hence the coefficient of $$x^{m}$$ in $$(1+x)^{n}(1+y)^{n}(1+z)^{n}$$
$$=\:^{3n}C_{m}$$.
Question 9
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Value of the expression $${ C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }+.....+(n+1){ C }_{ n }^{ 2 }$$ is

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

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

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

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

Solution

Let $$S = { C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }+.....n{ C }_{ n-1 }^{ 2 }+(n+1){ C }_{ n }^{ 2 }\quad ......(1)$$
using $${ C }_{ r }={ C }_{ n-r }$$, we write $$(1)$$ in the reverse order to obtain
$$S=(n+1){ C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+....2{ C }_{ n-1 }^{ 2 }+{ C }_{ n }^{ 2 }\quad ......(2)$$
Adding $$(1)$$ and $$(2)$$, we get
$$2S=(n+2)\left[

{ C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+......+{ C }_{ n }^{ 2 } \right]

\quad =(n+2)\quad ({ _{ }^{ 2n }{ C } }_{ n })$$
$$\Rightarrow \quad S=\left( \cfrac { n }{ 2 } +1 \right) ({ _{ }^{ 2n }{ C } }_{ n })$$
Question 10
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Directions For Questions

If a, b are prime numbers, $$ \displaystyle n\in N $$ then the rational terms in the expansion of $$ \displaystyle \left ( a^{\dfrac 1p}+a^{\dfrac 1q} \right )^{n} $$ are the terms in which indices of a & b are integral numbers.
On the basis of above information answer the following questions.

...view full instructions

In the expansion of $$ \displaystyle \left ( \sqrt{2} + \sqrt[5]{3}\right )^{120} $$ the number of irrational terms is

A
$$12$$

B
$$13$$

C
$$108$$

D
$$54$$

Solution

Total number of rational terms is
$$\dfrac{120}{L.C.M(5,2)}+1$$
$$=\dfrac{120}{10}+1$$
$$=13$$
Hence total number of irrational terms are
$$=121-13$$
$$=108$$

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Binomial Theorem Test - 28

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Binomial Theorem Test - 28

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Question 1

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

$$\displaystyle \frac{1}{2}\frac{2n!}{n!n!}$$

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

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

Solution

Question 2

$$\displaystyle \frac { 210 }{ { y }^{ 2 } } $$

$$\displaystyle \frac { { y }^{ 2 } }{ 210 } $$

$$210{ y }^{ 2 }$$

None of these

Solution

Question 3

81

82

730

None of these

Solution

Question 4

$$\displaystyle 10$$

$$\displaystyle 10^{2}$$

$$\displaystyle 10^{3}$$

None of these

Solution

Question 5

is on the circle

is outside the circle

is inside the circle

circle can not be fixed

Solution

Question 6

$$5^{th}$$

$$4^{th}$$ and $$5^{th}$$

$$4^{th}$$

None of these

Solution

Question 7

$$\displaystyle2^{2n-1}-\frac{\left (2n\right)!}{\left (n!\right)^{2}}$$

$$\displaystyle2^{2n-1}+\frac{2n!}{n!n!}$$

$$\displaystyle2^{2n-1}-\frac{2n!}{2\cdot n!n!}$$

None of these

Solution

Question 8

$${ ({ _{ }^{ n }{ C } }_{ m }) }^{ 3 }$$

$$3({ _{ }^{ n }{ C } }_{ m })$$

$${ _{ }^{ n }{ C } }_{ 3m }$$

$${ _{ }^{ 3n }{ C } }_{ m }$$

Solution

Question 9

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

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

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

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

Solution

Question 10

$$12$$

$$13$$

$$108$$

$$54$$

Solution

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