Binomial Theorem Test

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Binomial Theorem Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

If the middle terms in the expansion of $$\left(x^2+\dfrac{1}{x}\right)^{2n}$$ is $$184756x^{10}$$, then what is the value of $$n$$ ?

A
$$10$$

B
$$8$$

C
$$5$$

D
$$4$$

Solution

Given,

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

$$2n$$ is even

$$r=\dfrac{2n}{2} =n$$

$$MT=^{2n}C_n(x^2)^{2n-n}\left ( \dfrac{1}{x} \right )^n$$

$$=^{2n}C_n(x^2)^n(x)^{-n}$$

$$=^{2n}C_n(x)^{2n-n}$$

$$=^{2n}C_nx^n$$

$$\therefore 184756x^{10}=^{2n}C_nx^n$$

$$\Rightarrow x^{10}=x^n$$

$$\therefore n=10$$
Question 2
1 / -0

If the sum of the coefficients in the expansions of $$(a^2 x^2 - 2ax + 1)^{51}$$ is zero, then $$a$$ is equal to

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$-2$$

E
$$2$$

Solution

$$(a^2 x^2 - 2ax + 1)^{51}$$
For sum of coefficients put $$x = 1$$
$$(a^2 - 2a + 1)^{51} = 0$$
$$\Rightarrow \{(a - 1)^2 \}^{51} = 0$$
$$\Rightarrow a = 1$$
Question 3
1 / -0

$$\dfrac{1}{9!} + \dfrac{1}{3! 7!} + \dfrac{1}{5! 5!} + \dfrac{1}{7! 3!} + \dfrac{1}{9!}$$ is equal to

A
$$\dfrac{2^9}{10!}$$

B
$$\dfrac{2^{10}}{8!}$$

C
$$\dfrac{2^{11}}{9!}$$

D
$$\dfrac{2^{10}}{7!}$$

E
$$\dfrac{2^{8}}{9!}$$

Solution

$$\dfrac{1}{9!} + \dfrac{1}{3! 7!} + \dfrac{1}{5! 7!} + \dfrac{1}{7! 3!} + \dfrac{1}{9!}$$

$$=\dfrac{1}{10!}\left[\dfrac{10}{9!1!} + \dfrac{10}{3! 7!} + \dfrac{10}{5! 7!} + \dfrac{10}{7! 3!} + \dfrac{10}{9!1!}\right]$$

$$=\dfrac{1}{10!}[^{10}C_1+\,^{10}C_3+\,^{10}C_5+\,^{10}C_7+\,^{10}C_9]$$
$$=\dfrac{1}{10!}.2^9=\dfrac{2^9}{10!}$$
Question 4
1 / -0

The coefficient of $$x^4$$ in the expansion of $$(1-2x)^5$$ is equal to

A
$$40$$

B
$$320$$

C
$$-320$$

D
$$80$$

Solution

General term of $$(1 - 2x)^5$$ is given by

$$T_{r+1} = \, ^5C_r(-2x)^r$$
$$= \, ^5C_r(-2)^r x^r$$

For coefficient of $$x^4$$, power of $$x = 4$$

$$\therefore r = 4$$
$$\therefore$$ Coefficient pf $$x^4 = \, ^5C_4(-2)^4$$

$$= 5\times 16 = 80$$
Question 5
1 / -0

Sum of last $$30$$ coefficents in the binomial expansion of $$(1 + x)^{59}$$ is

A
$$2^{29}$$

B
$$2^{59}$$

C
$$2^{58}$$

D
$$2^{59} - 2^{29}$$

E
$$2^{60}$$

Solution

We have, $$(1 + x)^{59}$$
Sum of last $$30$$ coefficient of the binomial expansion
$$=\ ^{59} C_{30} + \ ^{59} C_{31} + .... + \ ^{59}C_{59}$$
We know that,
$$^{59} C_0 + \ ^{59}C_1 + \ ^{59}C_2 + .... + \ ^{59} C_{59} = 2^{59}$$
$$\Rightarrow (^{59}C_0 +\ ^{59}C_{59}) + (^{59} C_1 +\ ^{59}C_{58}) + ..... + (^{59} C_{29} + \ ^{59}C_{30}) = 2^{59}$$
$$\Rightarrow 2 (^{59} C_{59} + \ ^{59}C_{58} + .... + \ ^{59}C_{31} + \ ^{59}C_{30}) = 2^{59}$$ $$[\because \ ^nC_r = \ ^nC_{n - r}]$$
$$\Rightarrow \ ^{59} C_{30} +\ ^{59} C_{31} + .... \ ^{59} C_{39} = \dfrac{2^{59}}{2} = 2^{58}$$
$$\therefore $$ Sum of last $$30$$ coefficient of the binomial expansion $$(1 + x)^{59} $$ is $$2^{58}$$.
Question 6
1 / -0

$$(^7C_0 + \ ^7C_1) + (^7 C_2 + \ ^7C_3) + .... + (^7C_6 + \ ^7 C_7) = $$

A
$$2^8 - 2$$

B
$$2^7 - 1$$

C
$$2^7$$

D
$$2^8 - 1$$

E
$$2^7 - 2$$

Solution

$$(^7C_0+^7C_1)+(^7C_2+^7C_3)+------+(^7C_6+^7C_7)$$
$$=^7C_0+\,^7C_1+----+\,^7C_7$$
$$=2^7[\because C_0+C_1+C_2+----+C_n=2^n]$$
Question 7
1 / -0

The arithmetic mean of $$^nC_0 , \ ^nC_1, \ ^nC_2 ...., \ ^nC_n$$ is

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

B
$$\dfrac{2^n}{n}$$

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

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

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

Solution

$$\because (1 + x)^n = \ ^nC_0 + \ ^nC_1 x + \ ^nC_2 . x^2 + ... + \ ^nC_n . x^n$$
Take $$x = 1$$
$$(1 + 1)^n = \ ^nC_0 + \ ^nC_1 . (1) + \ ^nC_2 . (1)^2 + ... + \ ^nC_0 . (1)^n$$
$$2^n = \ ^nC_0 + \ ^nC_1 + \ ^n C_2 + ... + \ ^nC_n$$
Now arithmetic mean
$$\bar{X} = \dfrac{\displaystyle \sum_{i = 1}^n x_i}{N} $$, where $$N = (n + 1)$$
$$\Rightarrow \bar{X} = \dfrac{^nC_0 + \ ^nC_1 + \ ^n C_2 + .... + \ ^nC_n}{n + 1}$$
$$\Rightarrow \bar{X} = \dfrac{2^n}{n+ 1}$$
Question 8
1 / -0

The largest term in the expansion of $$(2 + 3x)^{25}$$ where x = 2 is its

A
13th term

B
19th term

C
20th term

D
26th term

Solution
Question 9
1 / -0

In the expansion of $$(1 + x)^{43}$$, the coefficients of the (2r+1)th and the (r + 2)th terms are equal, then the value of r, is

A
14

B
15

C
16

D
17

Solution
Question 10
1 / -0

The total number of terms in the expansion of $$(x + a)^{100} + (x - a)^{100}$$ after simplification,

A
50

B
51

C
154

D
202

Solution

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Binomial Theorem Test - 53

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Binomial Theorem Test - 53

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Question 1

$$10$$

$$8$$

$$5$$

$$4$$

Solution

Question 2

$$0$$

$$1$$

$$-1$$

$$-2$$

$$2$$

Solution

Question 3

$$\dfrac{2^9}{10!}$$

$$\dfrac{2^{10}}{8!}$$

$$\dfrac{2^{11}}{9!}$$

$$\dfrac{2^{10}}{7!}$$

$$\dfrac{2^{8}}{9!}$$

Solution

Question 4

$$40$$

$$320$$

$$-320$$

$$80$$

Solution

Question 5

$$2^{29}$$

$$2^{59}$$

$$2^{58}$$

$$2^{59} - 2^{29}$$

$$2^{60}$$

Solution

Question 6

$$2^8 - 2$$

$$2^7 - 1$$

$$2^7$$

$$2^8 - 1$$

$$2^7 - 2$$

Solution

Question 7

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

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

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

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

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

Solution

Question 8

13th term

19th term

20th term

26th term

Solution

Question 9

14

15

16

17

Solution

Question 10

50

51

154

202

Solution

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