Binomial Theorem Test

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Binomial Theorem Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

In the expansion of $$\left (x + \dfrac {1}{x}\right )^{n}$$, then the coefficient of the term indepenent of x is

A
$$\dfrac {n!}{(r!)^{2}}$$

B
$$\dfrac {n!}{(r + 1)! (r - 1)!}$$

C
$$\dfrac {n!}{\left (\dfrac {n + r}{2}\right )! \left (\dfrac {n - r}{2}\right )!}$$

D
$$\dfrac {n!}{\left [\left (\dfrac {n}{2}\right )!\right ]^{2}}$$

Solution

In the expansion of $$\left (x + \dfrac {1}{x}\right )^{n}$$, $$r^{th}$$ term is

$$t_{r+1}={}^{n}C_{r} x^{n-r} \left(\dfrac{1}{x}\right)^{r}$$

$$\implies t_{r+1}={}^{n}C_{r} x^{n-r} x^{-r}$$

$$\implies t_{r+1}={}^{n}C_{r} x^{n-2r}$$ ... $$(i)$$

We need to find the coefficient of $$x^{0}$$

Then substitute $$n-2r=0\implies r=\dfrac{n}{2}$$

Substitute $$r$$ in RHS of $$(i)$$, we get

$$t_{r+1}={}^{n}C_{\frac{n}{2}}x$$

Then, coefficient of $$x$$ is $$\displaystyle ={}^{n}C_{\frac{n}{2}}=\dfrac{n!}{\left(\dfrac{n}{2}\right)! \left(n-\dfrac{n}{2}\right)!}=\dfrac{n!}{\left(\dfrac{n}{2}\right)!\left(\dfrac{n}{2}\right)!}=\dfrac{n!}{\left[\left(\dfrac{n}{2}\right)!\right]^{2}}$$

Hence, coefficient of $$x^{0}=x$$ is $$\dfrac{n!}{\left[\left(\dfrac{n}{2}\right)!\right]^{2}}$$
Question 2
1 / -0

The sum of the series $$^{20}C_0 - \,^{20}C_1+\,^{20}C_2-\,^{20}C_3+...+\,^{20}C_{10}$$ is

A
$$-^{20}C_{10}$$

B
$$\dfrac{1}{2}\,^{20}C_{10}$$

C
$$0$$

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

Solution

On putting $$x=-1$$ in
$$(1+x)^{20}=^{20}C_0 + ^{20}C_1x+... +{20}C_{10}x^{10} + ... + \,^{20} C_{20} x^{20}$$
We get,
$$0=\,^{20}C_0 - \,^{20}C_1+... - \,^{20}C_9+\,^{20}C_{10} - \, ^{20}C_{11}+ ... + \,^{20}C_{20}$$
$$\Rightarrow 0 = \,^{20}C_0 - \,^{20}C_1+ ... - \,^{20}C_9 + \,^{20}C_{10}-\,^{20}C_9+...+\,^{20}C_0$$ ........ $$[\because {}^{n}C_{r}={}^{n}C_{n-r}]$$
$$\Rightarrow 0 = 2(^{20}C_0-\,^{20}C_1+ ... -\,^{20}C_9)+\,^{20}C_{10}$$
$$\Rightarrow \,^{20}C_{10} = 2(^{20}C_0 - \,^{20}C_1+ ... + \,^{20}C_{10})$$ .... [Adding $${}^{20}C_{10}$$ on both the sides]
$$\Rightarrow \,^{20}C_0 - \,^{20}C_1+...+\,^{20}C_{10}=\dfrac{1}{2}\,^{20}C_{10}$$
Question 3
1 / -0

If $$T_r=^{2016}C_rx^{2016-r}$$, for $$r=0, 1, ,....2016$$, then $$(T_0 - T_2+T_4....+T_{2016})^2+(T_1-T_3+T_5....T_{2015})^2$$ is equal to -

A
$$(X^2-1)^{1008}$$

B
$$(X+1)^{2016}$$

C
$$(X^2-1)^{2016}$$

D
$$(X^2+1)^{2016}$$

Solution

Let $$i$$ represent iota.
$$(T_0-T_2+T_4+\dots+T_{2016}) = (T_0+i^2T_2+i^4T_4+\dots+i^{2016}T_{2016})\dots(1)$$
Now, multiply $$(T_1-T_3+T_5+\dots-T_{2015})$$ with $$-i$$ to get $$-(iT_1+i^3T_3+i^3T_5+\dots+i^{2015}T_{2015})\dots(2)$$
$$\implies (T_0-T_2+T_4+\dots+T_{2016})^2+(T_1-T_3+T_5+\dots-T_{2015})^2$$
$$= (T_0+i^2T_2+i^4T_4+\dots+i^{2016}T_{2016})^2 + i^4(T_1-T_3+T_5+\dots-T_{2015})^2$$
$$ = (T_0+i^2T_2+i^4T_4+\dots+i^{2016}T_{2016})^2 -(iT_1+i^3T_3+i^3T_5+\dots+i^{2015}T_{2015})^2$$
$$ = (T_0+iT_1+i^2T_2+\dots+i^{2016}T_{2016}) (T_0-iT_1+i^2T_2-i^3T_3\dots)$$ (As $$a^2-b^2 = (a+b)(a-b)$$)
Now, consider these terms seperately
$$(T_0+iT_1+i^2T_2+\dots)=(^{2016}C_0x^{2016}i^0 +^{2016}C_1x^{2015}i^1\dots) = (i+x)^{2016}$$
$$(T_0-iT_1+i^2T_2-i^3T_3\dots) = (^{2016}C_0(ix)^{2016} +^{2016}C_1(ix)^{2015}\dots) = (1+ix)^{2016}$$
$$\therefore (T_0+iT_1+i^2T_2+\dots+i^{2016}T_{2016}) (T_0-iT_1+i^2T_2-i^3T_3\dots)$$
$$= (i+x)^{2016}\times(1+ix)^{2016}$$
$$= [(i+x)(1+ix)]^{2016}$$
$$= [i-x+x+ix^2]^{2016}$$
$$= (1+x^2)^{2016}$$
Question 4
1 / -0

If $$C_{0}, C_{1}, C_{2}, ...., C_{n}$$ are binomial coefficients of order $$n$$, then the value of $$\dfrac {C_{1}}{2} + \dfrac {C_{3}}{4} + \dfrac {C_{5}}{6} + .... =$$

A
$$\dfrac {2^{n} + 1}{n + 1}$$

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

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

D
$$\dfrac {2^{n}}{n + 1}$$

Solution

$$(1+x)^{ n }=C_{ 0 }+C_{ 1 }x+C_{ 2 }x^{ 2 }+C_{ 3 }x^{ 3 }+......C_{ n }x^{ n }$$
Integrating, We have
$$ \dfrac { (1+x)^{ n+1 } }{ n+1 } =C_{ 0 }x+C_{ 1 }\dfrac { x^{ 2 } }{ 2 } +C_{ 2 }\dfrac { x^{ 3 } }{ 3 } +.....C_{ n }\dfrac { x^{ n+1 } }{ n+1 } $$
Putting, $$x=1$$
$$ \dfrac { (2)^{ n+1 } }{ n+1 } =C_{ 0 }+C_{ 1 }\dfrac { 1 }{ 2 } +C_{ 2 }\dfrac { 1 }{ 3 } +....$$
Putting, $$x=-1$$
$$ \dfrac { (0)^{ n+1 } }{ n+1 } =-C_{ 0 }+C_{ 1 }\dfrac { 1 }{ 2 } +C_{ 2 }\dfrac { (-1) }{ 3 } +....$$
Adding above two equations we have,
$$ \dfrac { (2)^{ n+1 } }{ n+1 } =2\left(\dfrac { C_{ 1 } }{ 2 } +\dfrac { { C }_{ 3 } }{ 4 } .....\right)$$
$$\implies \dfrac { (2)^{ n } }{ n+1 } =\dfrac { C_{ 1 } }{ 2 } +\dfrac { { C }_{ 3 } }{ 4 } .....$$
Hence, option D is correct.
Question 5
1 / -0

The total number of terms in the expansion of $$(x + a)^{47} - (x - a)^{47}$$ after simplification is

A
24

B
47

C
48

D
96

Solution

$${ \left( x+a \right) }^{ 47 }-{ \left( x-a \right) }^{ 47 }$$
When we expand the above equation using binomial expansion
$$ (x +y)^{n} = \displaystyle (\sum_{k=0}^{n} {^{n}C_k} x^{k}y^{n-k}) $$
So the above equation becomes
$$ (x +a)^{47} = \displaystyle (\sum_{k=0}^{47} {^{47}C_k} x^{k}a^{47-k}) $$
$$ (x -a)^{47} = \displaystyle (\sum_{k=0}^{47} {^{47}C_k} x^{k}(-a)^{47-k}) $$
$${ \left( x+a \right) }^{ 47 }\Rightarrow $$There are 48 terms in the expansion and all are positive
$${ \left( x-a \right) }^{ 47 }\Rightarrow $$There are 48 terms in the expansion
The terms with odd powers of a will be cancelled and those with even powers of a will add up.
24 terms will be positive and 24 negative in the expansion of $$ (x-a)^{47} $$
48 terms positive-[24 terms negative and 24 terms positive]
$$=48\quad terms\quad positive +24\quad terms\quad negative+24\quad terms\quad positive$$
$$=24$$ terms
Question 6
1 / -0

Let $$((1 + x) + x^{2})^{9} = a_{0} + a_{1}x + a_{2}x^{2} + ..... + a_{18}x^{18}$$. Then

A
$$a_{0} + a_{2} + ..... + a_{18} = a_{1} + a_{3} + ..... + a_{17}$$

B
$$a_{0} + a_{2} + ..... + a_{18}$$ is even

C
$$a_{0} + a_{2} + ..... + a_{18}$$ is divisible by $$9$$

D
$$a_{0} + a_{2} + ..... + a_{18}$$ is divisible by $$3$$ but not by $$9$$

Solution

Given : $$((1 + x) + x^{2})^{9} = a_{0} + a_{1}x + a_{2}x^{2} + ..... + a_{18}x^{18}$$ ..... $$(i)$$
Put $$x=1$$ in $$(i)$$, we get
$$(1+1+1)^{9}=a_{0}+a_{1}+a_{2}+.....+a_{18}$$
$${3}^{9}=a_{0}+a_{1}+a_{2}+.....+a_{18}$$ .... $$(ii)$$
Put $$x=-1$$ in $$(i)$$, we get
$$(1-1+1)^{9}=a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-..... -a_{17}+a_{18}$$
$$1=a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-..... -a_{17}+a_{18}$$ ..... $$(iii)$$
Adding $$(ii)$$ and $$(iii)$$, we get
$$3^{9}+1=a_{0}+a_{1}+....+a_{18}+a_{0}-a_{1}+a_{2}-.....-a_{17}+a_{18}$$
$$=2a_{0}+2a_{2}+2a_{4}+....+2a_{18}$$
$$\implies 2(a_{0}+a_{2}+a_{4}+....+a_{18})=3^{9} + 1$$
$$\implies a_{0} + a_{2} + ..... + a_{18} = \dfrac {3^{9} + 1}{2} \rightarrow even$$
Hence, $$a_{0}+a_{2}+a_{4}+....+a_{18}$$ is even.
Question 7
1 / -0

Let $$n \ge 5$$ and $$b \neq 0$$. In the binomial expansion of $${ \left( a-b \right) }^{ n }$$, the sum of the 5th and 6th terms is zero then $${ a }/{ b }$$ equals

A
$$\dfrac { 5 }{ n-4 } $$

B
$$\dfrac { 1 }{ 5\left( n-4 \right) } $$

C
$$\dfrac{n-5}{6}$$

D
$$\dfrac{n-4}{5}$$

Solution

Solution:
Given that:
$$n\ge5,b\neq0,$$ given expansion is $$(a-b)^n$$ and
$$t_5+t_6=0$$
To find: $$\cfrac ab=?$$
Solution:
$$t_5={}^nC_4a^4(-b)^{(n-4)}$$ and
$$t_6={}^nC_5a^5(-b)^{(n-5)}$$
$$\because t_5+t_6=0$$
$$\therefore {}^nC_4a^4(-b)^{(n-4)}$$$$+{}^nC_5a^5(-b)^{(n-5)}=0$$
$$\Longrightarrow {}^nC_5a={}^nC_4b$$
$$\Longrightarrow \cfrac ab=\cfrac{{}^nC_4}{{}^5C_5}=\cfrac{\cfrac{n!}{4!(n-4)!}}{\cfrac{n!}{5!(n-5)!}}=\cfrac{5}{n-4}$$
Hence, A is the correct answer.
Question 8
1 / -0

Given $$(1-2x+5x^2-10x^3)(1+x)^n=1+a_1x+a_2x^2+...$$ and that $$a_1^2=2a_2$$ then the value of $$n$$ is-

A
6

B
2

C
5

D
3

Solution

Given that $$\left( 1-2x+5{ x }^{ 2 }-10{ x }^{ 3 } \right) { \left( 1+x \right) }^{ n }=1+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+...$$
Given that $${ { a }_{ 1 } }^{ 2 }=2{ a }_{ 2 }$$
Here $${ a }_{ 1 }$$ is the coefficient of $$x$$ and $${ a }_{ 2 }$$ is the coefficient of $${ x }^{ 2 }$$,
$$\Longrightarrow { a }_{ 1 }=n-2$$ and $$\Longrightarrow { a }_{ 2 }=-2n+{ C }_{ 2 }+5$$
Substituting these values in the given relation we get,
$$\Longrightarrow { \left( n-2 \right) }^{ 2 }=2\left( 5-2n+\frac { n\left( n-1 \right) }{ 2 } \right) \\ \Longrightarrow n=6\\ $$
Question 9
1 / -0

In the expansion of $${ \left( 3x-\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ 10 }$$, the $$5^{th}$$ term from the end is

A
$$\cfrac { 16486 }{ { x }^{ 8 } } $$

B
$$\cfrac { 17010 }{ { x }^{ 8 } } $$

C
$$\cfrac { 13486 }{ { x }^{ 8 } } $$

D
None of these

Solution

There are $$11$$ terms in the expansion of $${ \left( 3x-\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ 10 }$$
Therefore, $$5^{th}$$ term from the end.
$$=(10-5+2)^{th}$$ term from beginning
$$={ T }_{ 7 }={ _{ }^{ 10 }{ C } }_{ 6 }{ \left( 3x \right) }^{ 10-6 }{ \left( -\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ 6 }$$
$$=\cfrac { 10! }{ 6!4! } { 3 }^{ 4 }{ x }^{ 4 }\cfrac { 1 }{ { x }^{ 12 } }$$
$$ =\cfrac { 17010 }{ { x }^{ 8 } } $$
Question 10
1 / -0

The coefficient of $$x^{49}$$ in the product $$(x - 1) (x - 2) (x - 3) .... (x - 50)$$ is

A
$$-2250$$

B
$$-1275$$

C
$$1275$$

D
$$2250$$

E
$$-49$$

Solution

Coefficient of $$x^{49}$$ in product of $$(x-1)(x-2)(x-3)(x-4)........(x-50)$$
We know that,
$$(x-1)(x-2)(x-3)(x-4)........(x-n)=x^n-(1+2+3+....+n)x^{n-1}+.......$$
Coefficient of $$x^{49}=-(1+2+3+...+50)$$
$$=-\cfrac{50\times51}{2}$$
$$=-1275$$
Hence, B is the correct option.

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Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 40

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Binomial Theorem Test - 40

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SHARING IS CARING

Question 1

$$\dfrac {n!}{(r!)^{2}}$$

$$\dfrac {n!}{(r + 1)! (r - 1)!}$$

$$\dfrac {n!}{\left (\dfrac {n + r}{2}\right )! \left (\dfrac {n - r}{2}\right )!}$$

$$\dfrac {n!}{\left [\left (\dfrac {n}{2}\right )!\right ]^{2}}$$

Solution

Question 2

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

$$\dfrac{1}{2}\,^{20}C_{10}$$

$$0$$

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

Solution

Question 3

$$(X^2-1)^{1008}$$

$$(X+1)^{2016}$$

$$(X^2-1)^{2016}$$

$$(X^2+1)^{2016}$$

Solution

Question 4

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

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

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

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

Solution

Question 5

24

47

48

96

Solution

Question 6

$$a_{0} + a_{2} + ..... + a_{18} = a_{1} + a_{3} + ..... + a_{17}$$

$$a_{0} + a_{2} + ..... + a_{18}$$ is even

$$a_{0} + a_{2} + ..... + a_{18}$$ is divisible by $$9$$

$$a_{0} + a_{2} + ..... + a_{18}$$ is divisible by $$3$$ but not by $$9$$

Solution

Question 7

$$\dfrac { 5 }{ n-4 } $$

$$\dfrac { 1 }{ 5\left( n-4 \right) } $$

$$\dfrac{n-5}{6}$$

$$\dfrac{n-4}{5}$$

Solution

Question 8

6

2

5

3

Solution

Question 9

$$\cfrac { 16486 }{ { x }^{ 8 } } $$

$$\cfrac { 17010 }{ { x }^{ 8 } } $$

$$\cfrac { 13486 }{ { x }^{ 8 } } $$

None of these

Solution

Question 10

$$-2250$$

$$-1275$$

$$1275$$

$$2250$$

$$-49$$

Solution

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