Binomial Theorem Test

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Binomial Theorem Test - 17

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Question 1
1 / -0

In the expansion of $$(1+x)^{n}.(1+y)^{n}.(1+\mathrm{z})^{n}$$ the sum of the coefficients of the terms of degree $$r$$ is

A
$$(^{n}C_{r})^{3}$$

B
$$^{3n}C_{r}$$

C
$$3\times nC_{r}$$

D
$$nC_{3r}$$

Solution

The given expression contains $$3n$$ factors
Using combination to choose $$r$$ brackets out of $$3n$$ brackets for a term of degree $$r$$, we get
$$\:^{3n}C_r$$
Hence, answer is option B
Question 2
1 / -0

Assertion (A) : Number of the disimilar terms in the sum of expansion $$(x+a)^{102}+(x-a)^{102}$$ is $$206$$
Reason (R) : Number of terms in the expansion of $$(x+b)^{n}$$ is n + 1

A
Both A and R are individually true and R is the correct explanation of A.

B
Both A and R are individually true and R is not correct explanation of A.

C
A is true but R is false

D
A is false but R is true

Solution

$$(x+a)^{102}+(x-a)^{102}$$
Since a is constant let $$a=1$$
Therefore the above question can be re-written as
$$(1+x)^{102}+(1-x)^{102}$$
$$=[1+\:^{102}C_{1}x^1+\:^{102}C_{2}x^2...+\:^{102}C_{102}x^{102}]+[1-\:^{102}C_{1}x^1+\:^{102}C_{2}x^2-\:^{102}C_{3}x^3...+\:^{102}C_{102}x^{102}]$$
$$=2[1+\:^{102}C_{2}x^2+\:^{102}C_{2}x^4...+\:^{102}C_{102}x^{102}]$$ ...(i)
Hence number of dissimilar terms in Eq(i) will be $$\dfrac{102}{2}+1$$
$$=52$$ terms.
Hence A is false.
Reason is true.
Hence option D is the correct option.
Question 3
1 / -0

If the coefficients of $$r^{th}$$ term and $$(r+1)^{th}$$ term in the expansion of $$(1+x)^{20}$$ are in the ration 1 : 2, then $$r=$$

A
6

B
7

C
8

D
9

Solution

The ratio of the coefficient of rth term and (r+1)th term 1 : 2
$$\Rightarrow ^{20}C_{r-1} : ^{20}C_r = 1: 2$$
$$\Rightarrow \displaystyle \frac{r}{21-r} = \frac{1}{2} \Rightarrow 2r = 21-r$$
$$\Rightarrow r =7$$
Question 4
1 / -0

The coefficient of $$x^4$$ in $$\displaystyle \left ( \frac{x}{2} - \frac{3}{x^2} \right )^{10}$$ is

A
$$\displaystyle \frac{45}{64}$$

B
$$\displaystyle \frac{243}{128}$$

C
$$\displaystyle \frac{405}{256}$$

D
$$\displaystyle \frac{810}{512}$$

Solution

$$\displaystyle \frac{10 \times 1- 4}{1+2} = \frac{10-4}{3} = \frac{6}{3}=2$$
Coefficient of $$x^4$$ is $$^{10}C_2 \left ( \frac{1}{2} \right )^{10-2}(-3)^2$$
$$\displaystyle \frac{10 \times 9}{2} \cdot \frac{1}{2^8} \cdot 3^2 = \frac{10 \times 81}{2 \times 2^8}$$
$$\displaystyle \frac{5 \times 81}{256} = \frac{405}{256}$$
Question 5
1 / -0

The coefficient of the $$8$$th term in the expansion of $$(1+x)^{10}$$ is

A
$$120$$

B
$$7$$

C
$$^{10}C_8$$

D
$$210$$

Solution

$$(1+x)^{10}=$$ $$^{10}C_0+\,^{10}C_1x+\,^{10}C_2x^2+........+\,^{10}C_7x^7+\,^{10}C_8x^8+\,^{10}C_9x^9+\,^{10}C_{10}x^{10}$$
So here, first term is $$^{10}C_0$$ then $$8^{th}$$ term will be $$10C_7x^7.$$
$$\Rightarrow$$ Coefficient of the $$8^{th}$$ term $$=\,^{10}C_7$$

$$=\dfrac{10!}{7!3!}$$

$$=\dfrac{10\times 9\times 8\times 7!}{7!\times 3\times 2\times 1}$$

$$=120$$
Question 6
1 / -0

If $$T_r$$ denotes the rth term in the expansion of $$\displaystyle \left ( x+\frac{1}{y} \right)^{23}$$ then

A
$$T_{12}=T_{13}$$

B
$$x^2T_{13}=T_{12}$$

C
$$T_{12} = xy T_{13}$$

D
$$T_{12} + T_{12} = 25$$

Solution

$$T_{12} = ^{23} C_{11}x^{12} \displaystyle \left ( \frac{1}{y} \right)^{11} = ^{23}C_{11} \frac{x^{12}}{y^{11}}$$,
$$T_{13}=^{23}C_{12} x^{11} \displaystyle \left ( \frac{1}{y} \right)^{12} = ^{23}C_{11}\frac{x^{11}}{y^{12}}$$,
$$T_{12} = xy \displaystyle \left ( ^{23}C_{11} \frac{x^{11}}{y^{12}} \right ) = xy T_{13}$$
Question 7
1 / -0

If the coefficients of $$x^7$$ and $$x^8$$ in $$\displaystyle \left ( 2 + \frac{x}{3} \right )^n$$ are equal then n $$=$$

A
45

B
55

C
35

D
27

Solution

Coefficient of $$x^7 = coefficient of $$x^8
$$\Rightarrow \displaystyle \frac{2^n \cdot ^nC_7}{6^7} = 2^n \cdot \frac{^n C_8}{6^8}$$
$$\Rightarrow 6 (^nC_7)=^nC_8$$
$$\Rightarrow 6 \cdot \displaystyle \frac{n!}{(n-7)!7!}=\frac{n!}{(n-8!)8!}$$
$$\Rightarrow \displaystyle \frac{6}{n-7}=\frac{1}{8} \Rightarrow n-7 =48$$
$$\Rightarrow n = 55$$
Question 8
1 / -0

The coefficient of $$x^3$$ in $$\displaystyle \left ( \sqrt{x^5}+ \frac{3}{\sqrt{x^3}} \right )^5$$ is

A
0

B
120

C
420

D
540

Solution

$$\displaystyle r=\frac{6 \times \frac{5}{2}-3}{\frac{5}{2}+\frac{3}{2}} = \frac{15-3}{4} =3$$
$$\therefore $$ Coefficient of $$x^3$$ is $$^6C_3 3^3$$
$$\displaystyle =\frac{6 \times 5 \times 4}{3 \times 2 \times 1} \cdot 27$$
$$=5 \times 4 \times 27=540$$
Question 9
1 / -0

Number of irrational terms in the expansion of $$\left(5^{\dfrac 16} + 2^{\dfrac 18}\right)^{100}$$ is

A
$$96$$

B
$$97$$

C
$$98$$

D
$$99$$

Solution

$$T_{r+1}=\:^{100}C_{r}5^{(r-100)/6}2^{r/8}$$
Hence we get rational terms when
$$r=8k$$ where k is an integer and
$$\dfrac{8k-100}6$$ is an integer
$$r=16,40,64,88$$
Hence we get in total 4 rational terms.
However, total number of terms will be $$101$$
Hence total number of irrational terms is $$101-4$$
$$=97$$ terms.
Question 10
1 / -0

In the binomial expansion of $$(a-b)^n, n \geq 5$$, the sum of $$5^{th}$$ and $$6^{th}$$ terms is zero, then $$\dfrac ab$$ equals

A
$$\displaystyle \frac{5}{n-4}$$

B
$$\displaystyle \frac{6}{n-5}$$

C
$$\displaystyle \frac{n-5}{6}$$

D
$$\displaystyle \frac{n-4}{5}$$

Solution

Given $$T_5+T_6=0$$

$$\Rightarrow ^nC_4 a^{n-4} \cdot b^4 - ^nC_5 a^{n-5}b^5=0$$

$$\Rightarrow ^nC_4 a^{n-4}b^4=^nC_5 a^{n-5}b^5$$

$$\Rightarrow \displaystyle \frac{a^{n-4}b^4}{a^{n-5}b^5}=\frac{^nC_5}{^nC_4}$$

$$\Rightarrow \displaystyle \frac{a}{b} = \frac{n-5+1}{5} \Rightarrow \frac{a}{b} = \frac{n-4}{5}$$

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Binomial Theorem Test - 17

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Binomial Theorem Test - 17

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SHARING IS CARING

Question 1

$$(^{n}C_{r})^{3}$$

$$^{3n}C_{r}$$

$$3\times nC_{r}$$

$$nC_{3r}$$

Solution

Question 2

Both A and R are individually true and R is the correct explanation of A.

Both A and R are individually true and R is not correct explanation of A.

A is true but R is false

A is false but R is true

Solution

Question 3

6

7

8

9

Solution

Question 4

$$\displaystyle \frac{45}{64}$$

$$\displaystyle \frac{243}{128}$$

$$\displaystyle \frac{405}{256}$$

$$\displaystyle \frac{810}{512}$$

Solution

Question 5

$$120$$

$$7$$

$$^{10}C_8$$

$$210$$

Solution

Question 6

$$T_{12}=T_{13}$$

$$x^2T_{13}=T_{12}$$

$$T_{12} = xy T_{13}$$

$$T_{12} + T_{12} = 25$$

Solution

Question 7

45

55

35

27

Solution

Question 8

0

120

420

540

Solution

Question 9

$$96$$

$$97$$

$$98$$

$$99$$

Solution

Question 10

$$\displaystyle \frac{5}{n-4}$$

$$\displaystyle \frac{6}{n-5}$$

$$\displaystyle \frac{n-5}{6}$$

$$\displaystyle \frac{n-4}{5}$$

Solution

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