Binomial Theorem Test

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Binomial Theorem Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

The middle term in the expansion of $$\left (x - \dfrac {1}{x}\right )^{18}$$ is

A
$$^{18}C_{9}$$

B
$$-^{18}C_{9}$$

C
$$^{18}C_{10}$$

D
$$-^{18}C_{10}$$

Solution

The general term in the expansion $$\left (x - \dfrac {1}{x}\right )^{18}$$ is given by
$$T_{r + 1} = ^{18}C_{r} (x)^{18 - r} \left (-\dfrac {1}{x}\right )^{r}$$
Here, $$n = 18$$
$$\therefore$$ The middle term is $$T_{9 + 1}$$, where $$r = 9$$
$$\therefore T_{9 + 1} = ^{18}C_{9} (-1)^{9} x^{18 - 2r}$$
$$= -^{18}C_{9}x^{18 - 18} = -^{18}C_{9}$$
Question 2
1 / -0

If $${ \left( 1+x+{ x }^{ 2 }+{ x }^{ 3 } \right) }^{ 5 }=\displaystyle\sum _{ k=0 }^{ 15 }{ { a }_{ k }{ x }^{ k } } $$ then $$\displaystyle\sum _{ k=0 }^{ 7 }{ { a }_{ 2k } } $$ is equal to

A
$$128$$

B
$$256$$

C
$$512$$

D
$$1024$$

Solution

Given, $${ \left( 1+x+{ x }^{ 2 }+{ x }^{ 3 } \right) }^{ 5 }=\displaystyle\sum _{ k=0 }^{ 15 }{ { a }_{ k }{ x }^{ k } } $$
$$\Rightarrow { \left[ \left( 1+x \right) +x\left( 1+x \right) \right] }^{ 5 }=\displaystyle\sum _{ k=0 }^{ 15 }{ { a }_{ k }{ x }^{ k } } $$
$$\Rightarrow { \left( 1+x \right) }^{ 10 }={ a }_{ 0 }{ x }^{ 0 }+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+\dots +{ a }_{ 15 }{ x }^{ 15 }$$
$$\Rightarrow _{ }^{ 10 }{ { C }_{ 0 } }+_{ }^{ 10 }{ { C }_{ 1 } }x+_{ }^{ 10 }{ { C }_{ 2 } }{ x }^{ 2 }+\dots +_{ }^{ 10 }{ { C }_{ 10 } }{ x }^{ 10 }$$
$$={ a }_{ 0 }+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+{ a }_{ 3 }{ x }^{ 3 }+\dots +{ a }_{ 15 }{ x }^{ 15 }$$
On equating the coefficient of constant and even powers of $$x$$, we get
$${ a }_{ 0 }=_{ }^{ 10 }{ { C }_{ 0 } },\quad { a }_{ 2 }=_{ }^{ 10 }{ { C }_{ 2 } },$$
$${ a }_{ 4 }=_{ }^{ 10 }{ { C }_{ 4 } },\dots ,{ a }_{ 10 }=_{ }^{ 10 }{ { C }_{ 10 } },$$
$${ a }_{ 12 }={ a }_{ 14 }=0$$
$$\therefore \displaystyle\sum _{ k=0 }^{ 7 }{ { a }_{ 2 }k } =_{ }^{ 10 }{ { C }_{ 0 } }+_{ }^{ 10 }{ { C }_{ 2 } }+^{ 10 }{ { C }_{ 4 } }+^{ 10 }{ { C }_{ 6 } }+^{ 10 }{ { C }_{ 8 } }+^{ 10 }{ { C }_{ 10 } }+0+0$$
$$={ 2 }^{ 10-1 }=29$$
$$=512$$
Question 3
1 / -0

Find the sum of coefficients in the expansion of $$\left(1-\dfrac 2x+\dfrac {4}{x^2}\right)^n$$ given that the number of terms are $$28$$

A
$$27$$

B
$$81$$

C
$$243$$

D
$$729$$

Solution

$$\left(1-\dfrac 2x+\dfrac {4}{x^2}\right)^n$$ and number of terms $$=28$$

Number of terms $$=^{n+3-1}C_{3-1} = ^{n+2}C_2$$

$$\dfrac {(n+2)!}{2! (n!)} = \dfrac {(n+2)(n+1)}{2} = 28$$

$$n^2+3n-54=0$$

$$n^2-6n+9n-54=0$$

$$(n+3)(n-6)=0$$

$$n=6$$

Number of terms $$=\left(1-\dfrac 2x+\dfrac {4}{x^2}\right)^6$$

Sum of coefficients is known when $$x=1$$

No. of terms $$=(1-2+4)^6 = 3^6 = 729$$
Question 4
1 / -0

$$(x+1)^n=C_0+C_1x^1+C_2x^2....C_nx^n$$.
Find the value of $$C_0+C_1+C_2.....+C_n$$

A
$$n!$$

B
$$(n!)^2$$

C
$$2^n$$

D
$$n!-2^n$$

Solution

$${ \left( x+1 \right) }^{ n }={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }+...+{ C }_{ n }{ x }^{ n }\\ $$
Put $$x=1$$, $${ \left( 2 \right) }^{ n }={ C }_{ 0 }+{ C }_{ 1 }+{ C }_{ 2 }+...+{ C }_{ n }\\ $$
Hence, $$C$$ is correct.
Question 5
1 / -0

Let $$n \in N$$, such that $$(1+x+x^2)^n=a_0+a_1x+a_2x^2...a_{2n}x^{2n}$$ The value of $$a_r$$ when $$(0\leq r\leq 2n)$$

A
$$a_{2n-r}$$

B
$$a_{n-r}$$

C
$$a_{2n}$$

D
$$n.a_{2n-1}$$

Solution

We have
$${(1+x+x^2)}^n$$ $$=$$ $$\sum _{r=0 }^{2n }{ a_rx^r}$$
Replace $$x$$ by $$\dfrac{1}{x}$$
Therefore, $$\left (1+\dfrac{1}{x}+\dfrac{1}{x^2}\right)=$$ $$\sum _{r=0 }^{2n }{ a_rx^r}$$
$$\Rightarrow$$ $${(1+x+x^2)}^n$$ $$=$$ $$\sum _{r=0 }^{2n }{ a_r{x}^{2n-r}}$$
$$\Rightarrow$$ $$\sum _{r=0 }^{2n }{ a_rx^r}$$ $$=$$ $$\sum _{r=0 }^{2n }{ a_r{x}^{2n-r}}$$
So, equating both the sides, we get $$a_r$$ $$=$$ $$a_{2n-r}$$.
Question 6
1 / -0

Find the coefficient of $$x^n$$ in expansion of $$(1 + x) (1 - x)^n$$.

A
$$(n - 1)$$

B
$$(-1)^{n - 1} n$$

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

D
$$(-1)^n (1- n)$$
Question 7
1 / -0

The general term in $$(x + y)^n$$ is given by
$$T_{r+1}=$$

A
$$^{n}C_{r+1}$$

B
$$^nC_{n-r-1}$$

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

D
none

Solution

The general term $${ T }_{ r+1 }=^{ n }{ { C }_{ r }{ x }^{ n-r }{ y }^{ r } }$$
So, $$D$$ is correct.
Question 8
1 / -0

Which of the following is incorrect?

A
$$^nC_r = ^nC_{n-r} $$

B
$$^nC_r = ^{n-1}C_r + ^nC_{n-r}$$

C
$$^nC_r = ^{n-1}C_r + ^{n-1}C_{r-1}$$

D
$$r!\ ^nC_r = ^nP_r$$

Solution

As we know that
$$(1+x)^n=\displaystyle^nC_{0}+\displaystyle^nC_{1}x+\displaystyle^nC_{2}x^2+......+\displaystyle^nC_{n}x^n$$

In this expansion coefficient of the first and last term be same
$$\displaystyle^nC_{r}=\displaystyle^nC_{n-r}$$

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

$$\dfrac{n!}{r!(n-r)!}=\dfrac{n!}{(n-r)!(r)!}$$
Hence it is correct
$$\displaystyle^nC_{r}=\displaystyle^nC_{n-r}$$

$$\displaystyle^nC_{r}=\displaystyle^{n-1}C_{r}+\displaystyle^{n-1}C_{r-1}$$
Then by fromula $$\displaystyle^nC_{r}+\displaystyle^nC_{r-1}=\displaystyle^{n+1}C_{r}$$
$$\displaystyle^nC_{r}=\displaystyle^{n-1+1}C_{r}$$
$$\displaystyle^nC_{r}=\displaystyle^{n}C_{r}$$
It is also correct

By formula of permutation
$$\dfrac{\displaystyle^nP_{r}}{r!}=\displaystyle^{n}C_{r}$$

$$\displaystyle^nP_{r}=r!\displaystyle^{n}C_{r}$$
Question 9
1 / -0

If the middle term of $$\left(\dfrac 1x +x\sin x\right)^{10}$$ is equal to $$7\dfrac 78$$, then the principal value of $$x$$ is:

A
$$60^0$$

B
$$30^0$$

C
$$45^0$$

D
$$90^0$$

Solution

Solution:
Middle term of the expansion $$\left(\cfrac1x+x\sin x\right)^{10}$$ is $$T_6.$$
$$T_6={}^{10}C_5\times\left(\cfrac1x\right)^5\times(x\sin x)^5=7\cfrac78$$
or, $${}^{10}C_5\times \sin^5x=\cfrac{63}8$$
or, $$252\times \sin^5x=\cfrac{63}8$$
or, $$\sin^5x=\cfrac{1}{32}$$
or, $$\sin x=\cfrac 12$$
or, $$x=30^\circ$$
Hence, B is the correct option.
Question 10
1 / -0

If the value of
$$C_0+2 \cdot C_1 + 3 \cdot C_2+........+(n+1)\cdot C_n=576$$, then n is ______.

A
$$7$$

B
$$5$$

C
$$6$$

D
$$9$$

Solution

We have: $$(1 + x)^n = C_0 + C_1x + C_2x^2 + ... + C_nx^n$$
Multiplying by $$x$$ on both sides, we get $$x(1 + x)^n = C_0x + C_1x^2 + C_2x^3 + ... + C_nx^{n + 1}$$
Differentiating both sides by $$x$$, we get
$$xn(1 + x)^{n - 1} + (1 + x)^n = C_0 + 2C_1x + 3C_2x^2 + ... + (n + 1)C_nx^n$$
Substituting $$x = 1$$, we have
$$n2^{n - 1} + 2^n = C_0 + 2C_1 + 3C_2 + ... + (n + 1)C_n$$
Thus, the R.H.S. is what has been asked, and so $$n2^{n - 1} + 2^n = 576$$
i.e. $$2^{n - 1} (2 + n) = 576$$
If $$n = 5, n + 2$$ becomes $$7$$ and $$576$$ is not divisible by $$7$$.
Similarly, $$n$$ cannot be $$9$$.
When $$n = 7, 2^6 \times 9 = 64 \times 9 = 576$$
Ans is option $$A$$.

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Binomial Theorem Test - 37

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Binomial Theorem Test - 37

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Question 1

$$^{18}C_{9}$$

$$-^{18}C_{9}$$

$$^{18}C_{10}$$

$$-^{18}C_{10}$$

Solution

Question 2

$$128$$

$$256$$

$$512$$

$$1024$$

Solution

Question 3

$$27$$

$$81$$

$$243$$

$$729$$

Solution

Question 4

$$n!$$

$$(n!)^2$$

$$2^n$$

$$n!-2^n$$

Solution

Question 5

$$a_{2n-r}$$

$$a_{n-r}$$

$$a_{2n}$$

$$n.a_{2n-1}$$

Solution

Question 6

$$(n - 1)$$

$$(-1)^{n - 1} n$$

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

$$(-1)^n (1- n)$$

Question 7

$$^{n}C_{r+1}$$

$$^nC_{n-r-1}$$

$$^{n-1}C_r$$

none

Solution

Question 8

$$^nC_r = ^nC_{n-r} $$

$$^nC_r = ^{n-1}C_r + ^nC_{n-r}$$

$$^nC_r = ^{n-1}C_r + ^{n-1}C_{r-1}$$

$$r!\ ^nC_r = ^nP_r$$

Solution

Question 9

$$60^0$$

$$30^0$$

$$45^0$$

$$90^0$$

Solution

Question 10

$$7$$

$$5$$

$$6$$

$$9$$

Solution

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