Binomial Theorem Test

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Binomial Theorem Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

In the expansion of $$\left( 1+x \right) ^{ n }$$, The binomial coefficients of three consecutive terms are respectively 220, 495 and 795, the value

A
$$10$$

B
$$11$$

C
$$12$$

D
$$13$$

Solution

Suppose the three consecutive terms are $$T_{r-1},\,T_r$$ and $$T_{r+1}.$$

Coefficient of these terms are $$^nC_{r-2},\,^nC_{r-1}$$ and $$^{n}C_r$$ respectively.

These coefficient are equal to $$220,\,495$$ and $$792$$

$$\therefore$$ $$\dfrac{^nC_{r-2}}{^nC_{r-1}}=\dfrac{220}{495}$$

$$\Rightarrow$$ $$\dfrac{r-1}{n-r+2}=\dfrac{4}{9}$$

$$\Rightarrow$$ $$9r-9=4n-4r+8$$

$$\Rightarrow$$ $$4n+17=13r$$ ------ ( 1 )

Also,

$$\dfrac{^nC_r}{^nC_{r-1}}=\dfrac{792}{495}$$

$$\Rightarrow$$ $$\dfrac{n-r+1}{r}=\dfrac{8}{5}$$

$$\Rightarrow$$ $$5n-5r+5=8r$$

$$\Rightarrow$$ $$5n+5=13r$$

$$\Rightarrow$$ $$5n+5=4n+17$$ [ From ( 1 ) ]

$$\Rightarrow$$ $$n=12$$
Question 2
1 / -0

The term independent of x in the expansion of $$\displaystyle \left ( x -\dfrac{1}{4} \right )^4\left ( x + \dfrac{1}{x} \right )^3$$

A
19/16

B
0

C
14

D
none of these

Solution
Question 3
1 / -0

If in the expansion of $$(1+x)^{20}$$,the coefficients of rth and (r+4)th terms are equal,then r is equal to

A
7

B
8

C
9

D
10

Solution

From given, we have,

$$(1+x)^{20}$$

and $$r$$th term is equal to $$(r+4)$$th term

$$\Rightarrow a=1,b=x,n=20$$

$$T_{r+1}=^nC_ra^{n-r}b^r$$

$$\Rightarrow T_{r+1}=^{20}C_r1^{n-r}x^r$$

$$\therefore T_{r+1}=^{20}C_rx^r$$

put $$r=r-1$$

$$\therefore T_r=^{20}C_{r-1}x^{r-1}$$..........(1)

put $$r=r+3$$

$$\therefore T_{r+4}=^{20}C_{r+3}x^{r+3}$$.......(2)

now,

$$^{20}C_{r-1}=^{20}C_{r+3}$$

we have formula,

$$^nC_r=^nC_p\rightarrow r=n-p$$

$$\Rightarrow r-1=20-(r+3)$$

$$r-1=20-r-3$$

$$2r=18$$

$$\therefore r=9$$
Question 4
1 / -0

If $$(1+2x+3x^2 )^{10} = a_0+a_1x+a_2x^2+\ ...\,+a_{20}x^{20}$$, then $$a_1$$ equals

A
$$10$$

B
$$20$$

C
$$210$$

D
None of these

Solution

Given,
$$(1+2x+3x^2 )^{10} = a_0+a_1x+a_2x^2+\ ...\ +a_{20}x^{20}$$

Then, $$a_1$$ = Coefficient of $$x$$ in $$(1+2x+3x^2)^{10}$$

Now,
$$(1+2x+3x^2)^{10}=\{( 1+2x)+3x^2\}^{10}$$
$$=\ ^{10}C_0 (1+2x)^{10}+\ ^{10}C_1 (1+2x)^9(3x^2)+...$$.

$$\therefore$$ Coefficient of $$x$$ in $$(1+2x+3x^2)^{10}$$
$$=$$ Coefficient of $$x$$ in $$^{10}C_0(1+2x)^{10}$$
$$=\ ^{10}C_0\cdot 2\cdot ^{10}C_1 $$
$$=1\cdot 2\cdot 10=20$$
Question 5
1 / -0

Let $$2.^{20}C_0 + 5. ^{20} C_1 + 8. ^{20}C_2 + ..... + 62. ^{20}C_{20}$$. then sum of this series is

A
$$16. 2^{22}$$

B
$$8. 2^{20}$$

C
$$8. 2^{21}$$

D
$$16. 2^{21}$$

Solution

$$2. ^{20}C_0 + 5. ^{20} C_1 + 8.^{20} C_2 \, + .... + \, 62. ^{20} C_{20} = \displaystyle \sum_{r = 0}^{20} \, (3r + 2) . ^{20} C_r$$
$$= 3 \displaystyle \sum_{r = 0}^{20} r. \, ^{20} C_r + 2 \displaystyle \sum_{r = 0}^{20} \, ^{20}C_r = 3 \times 20 \displaystyle \sum_{r = 1}^{20} \, ^{19} C_{r + 1} + 2. (2^{20}) = 60 . 2^{19} + 2.2^{20} = 16.2^{21}$$
Question 6
1 / -0

Find the $$7^{th}$$ term from the end in the expansion of $$\left(2x^{2}-\dfrac{3}{2x}\right)^{8}$$

A
$$4033\ x^{9}$$

B
$$4023\ x^{11}$$

C
$$4032\ x^{10}$$

D
None of these

Solution

Given to find the $$7^{th}$$ term in expression of $$\left(2x^{2}-\dfrac{3}{2x}\right)^{8}$$ from end

$$r^{th}$$ term from end $$=(n-r+2)^{th}$$ term starting

$$n=8, r=7$$

$$7^{th}$$ from end $$=(8-7+2)^{th}=3^{rd}$$ from starting

$$T_{3}=T_{2+1}=\ ^{8}C_{2}(2x^{2})^{6} \left(\dfrac{-3}{2x}\right)^{2}$$

$$=\dfrac{8!}{2!6!}2^{6}.x^{12}(-1)^{2}\dfrac{3^{2}}{2^{x}x^{2}}$$

$$T_{3}=4032 x^{10}$$
Question 7
1 / -0

Find the coefficient of $$x^{10}$$ in the expansion of $$\left(2x^{2}-\dfrac{1}{x}\right)^{20}$$

A
$$^{20}C_{8}. 2^{8}$$

B
$$^{20}C_{10}. 2^{10}$$

C
$$^{20}C_{11}. 2^{11}$$

D
None of these

Solution

Given that to find the coefficient of $$x^{10}$$ in the term $$\left(2x^{2}-\dfrac{1}{x}\right)$$

$$T_{r+1}=\ ^{20}C_{r}\left.\dfrac{}{}\right| (2x)^{20-r}\left(\dfrac{-1}{x}\right)^{r}$$

$$=(-1)^{r}\ ^{20}C_{r} 2^{20-r}x^{40-2r-r}$$

$$=(-1)^{r}\ ^{20}C_{r}2^{20-r}x^{40-3r}$$

$$40-3r=10\Rightarrow 30=3r\Rightarrow r=10$$

$$T_{11}=T_{10+1}=(-1)^{10}\ ^{20}C_{10} 20^{20-10}x^{10}$$

$$=\ ^{20}C_{10} 2^{20-10}x^{10}$$

$$T_{11}=\ ^{20}C_{10}2^{10}x^{10}$$
Question 8
1 / -0

The greatest term in the expansion of $$(2x + 3y)^{11}$$ when x = 9 and y = 4 is :

A
(165)$$(10)^8$$$$(12)^3$$

B
(330) $$(18)^7 (12)^4$$

C
462$$(18)^8 (12)^5$$

D
None of these

Solution
Question 9
1 / -0

Find the coefficient of $$x^{-15}$$ in the expansion of $$\left(3x^{2}-\dfrac{a}{3x^{3}}\right)^{10}$$

A
$$-\dfrac {40}{21}a^6$$

B
$$-\dfrac {30}{23}a^8$$

C
$$-\dfrac {40}{27}a^7$$

D
None of these

Solution

Given that, to find coefficient of $$x^{-15}$$ in equation of $$\left (3x^2 -\dfrac {a}{3x^2}\right)^{10}$$

$$T_{r+1}=^{10}C_{r} (3x)^{10-r} \left (\dfrac {-a}{3x^3}\right)^{10}$$

$$=(-1) ^{10}C_{r}(3)^{10-r} (x^{20-2r}) (a^r) (x^{-3r}) \left (\dfrac {1}{3}\right)^r$$

$$=(-1)^r\ ^{10}C_{r} 3^{10-r-r} x^{20-2r-3r} a^r$$

$$T_{r+1}=(-1)^r\ ^{10}C_{r} 3^{10-2r} x^{20-5r} a^r$$

$$20-5r =-15\ \Rightarrow 35=5r \Rightarrow r=7$$

$$T_{8}=T_{7+1}=(-1)^7\ ^{10}C_{7} 3^{10-2(7)} x^{20-5(7)} a^7$$

$$=-^{10}C_{7} 3^{r} x^{-15} a^7 \Rightarrow -\dfrac {40}{27}a^7 x^{-15}$$

Coefficient of $$x^{-15}$$ is $$-\dfrac {40}{27}a^7$$
Question 10
1 / -0

If $$p$$ is a real number and if the middle term in the expansion of $$\left(\dfrac{p}{2}+2\right)^{8}$$ is $$1120$$, find $$p$$

A
$$p=\pm 1$$

B
$$p=\pm 3$$

C
$$p=\pm 5$$

D
$$p=\pm 2$$

Solution

Given expansion is $$\left (\dfrac {P}{2}+2\right)^8$$

Since index is $$n=8$$, there is only one middle term i.e., $$\left (\dfrac {8}{2}+1\right)^{th}$$

$$\Rightarrow \ 5^{th }$$. term

$$T_5 =T_{4+1}=^8C_{4} \left(\dfrac {P}{2}\right)^{8-4} 2^4$$

$$\Rightarrow \ 1120=^{8}C_{4} P^4 \Rightarrow \ 1120=\dfrac {8\times 7\times 6\times 5\times 4!}{4! 4\times 3\times 2\times 1} P^4$$

$$\Rightarrow \ 1120=7\times 2\times 5\times P^4 =\dfrac {1120}{70}$$

$$\Rightarrow \ p^4 =16 \ \Rightarrow P^2 =4 \Rightarrow P=\pm 2$$

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Binomial Theorem Test - 52

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Binomial Theorem Test - 52

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SHARING IS CARING

Question 1

$$10$$

$$11$$

$$12$$

$$13$$

Solution

Question 2

19/16

0

14

none of these

Solution

Question 3

7

8

9

10

Solution

Question 4

$$10$$

$$20$$

$$210$$

None of these

Solution

Question 5

$$16. 2^{22}$$

$$8. 2^{20}$$

$$8. 2^{21}$$

$$16. 2^{21}$$

Solution

Question 6

$$4033\ x^{9}$$

$$4023\ x^{11}$$

$$4032\ x^{10}$$

None of these

Solution

Question 7

$$^{20}C_{8}. 2^{8}$$

$$^{20}C_{10}. 2^{10}$$

$$^{20}C_{11}. 2^{11}$$

None of these

Solution

Question 8

(165)$$(10)^8$$$$(12)^3$$

(330) $$(18)^7 (12)^4$$

462$$(18)^8 (12)^5$$

None of these

Solution

Question 9

$$-\dfrac {40}{21}a^6$$

$$-\dfrac {30}{23}a^8$$

$$-\dfrac {40}{27}a^7$$

None of these

Solution

Question 10

$$p=\pm 1$$

$$p=\pm 3$$

$$p=\pm 5$$

$$p=\pm 2$$

Solution

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