Binomial Theorem Test

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Binomial Theorem Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

The total number of terms in the expansion of $$(x+y)^{50}+(x-y)^{50}$$ is

A
$$51$$

B
$$26$$

C
$$102$$

D
$$25$$

Solution

Consider given the expression,

$${{\left( x+y \right)}^{50}}+{{\left( x-y \right)}^{50}}$$

We know that,

$${{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{+}^{n}}{{C}_{1}}{{x}^{n-1}}y{{+}^{n}}{{C}_{2}}{{x}^{n-2}}{{y}^{2}}+{{......}^{n}}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}{{.....}^{n}}{{C}_{n}}{{y}^{n}}$$

Now,

$${{\left( x+y \right)}^{50}}+{{\left( x-y \right)}^{50}}=2{{[}^{50}}{{C}_{0}}{{x}^{50}}{{+}^{50}}{{C}_{2}}{{x}^{50-2}}y{{+}^{50}}{{C}_{4}}{{x}^{50-4}}{{y}^{4}}+{{......}^{50}}{{C}_{r}}{{x}^{50-r}}{{y}^{r}}{{.....}^{50}}{{C}_{50}}{{y}^{50}}$$

Hence, there are $$26$$ terms

Hence, this is the answer.
Question 2
1 / -0

$${ _{ }^{ 15 }{ C } }_{ 3 }+{ _{ }^{ 15 }{ C } }_{ 5 }+....+{ _{ }^{ 15 }{ C } }_{ 15 }$$ will be equal to

A
$${2}^{14}$$

B
$${2}^{14}-15$$

C
$${2}^{14}+15$$

D
$${2}^{14}-1$$

Solution

We know
$${ _{ }^{ 15 }{ C } }_{ 1 }+{ _{ }^{ 15 }{ C } }_{ 3 }+{ _{ }^{ 15 }{ C } }_{ 5 }+.....{ _{ }^{ 15 }{ C } }_{ 15 }={ 2 }^{ 15-1 }$$
$$\therefore { _{ }^{ 15 }{ C } }_{ 3 }+{ _{ }^{ 15 }{ C } }_{ 5 }+...+{ _{ }^{ 15 }{ C } }_{ 15 }={ 2 }^{ 14 }-15$$
Question 3
1 / -0

The number of terms that are integers in the binomial expansion of $$(\sqrt {7} + \sqrt [3]{5})^{35}$$ is

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

Solution

The general term in the given expansion $${ \left( \sqrt { 7 } +\sqrt [ 3 ]{ 5 } \right) }^{ 35 }$$ is $${}^{35} C_{ r }{ 7 }^{ \frac { 35-r }{ 2 } }\cdot { 5 }^{ \frac { r }{ 3 } }$$,
If $$r$$ is a multiple of $$3$$ and $$35-r$$ is a multiple of $$2$$ then the terms are integers,
$$\therefore r=3,9,15,21,27,33$$ which are six values.
Question 4
1 / -0

[AS 1] If $$A = \dfrac{1}{3} B \, and \, B = \dfrac{1}{2} C$$, then A : B : C = ..

A
1 : 3 : 6

B
2 : 3 : 6

C
3 : 2 : 6

D
3 : 1 : 2

Solution

$$A=\frac{B}{3}$$ ........(1)

$$B=\frac{C}{2}$$

$$\Rightarrow C=2B$$ .........(2)

From (1) and (2),

$$A:B:C=\frac{B}{3}:B:2B$$

$$=\frac{1}{3}:1:2$$

$$=1:3:6$$
Question 5
1 / -0

The expression $$^{n}\textrm{C}_{0}+4\ ^{n}\textrm{C}_{1}+4^{2}\ ^{n}\textrm{C}_{2}+..........+4^{n}\ ^{n}\textrm{C}_{n}$$, equals

A
$$2^{2n}$$

B
$$2^{3n}$$

C
$$5^{n}$$

D
None of these

Solution

We know that
$$\begin{array}{l} ^{ n }{ C_{ 0 } }+{ 4^{ n } }{ C_{ 1 } }+{ 4^{ 2 } }^{ n }{ C_{ 2 } }+.......+{ 4^{ n } }\, \, { \, ^{ n } }{ C_{ n } } \\ { \left( { 1+x } \right) ^{ n } }{ =^{ n } }{ C_{ 0 } }{ +^{ n } }{ C_{ 1 } }x{ +^{ n } }{ C_{ 2 } }{ x^{ 2 } }+.....{ +^{ n } }{ C_{ n } } \\ Putting\, x=4 \\ { 5^{ n } }{ =^{ n } }{ C_{ 0 } }+{ 4^{ n } }{ C_{ 1 } }+{ 4^{ 2 } }^{ n }{ C_{ 2 } }+.......+{ 4^{ n } }\, \, \, { \, ^{ n } }{ C_{ n } } \\ Hence,\, option\, C\, is\, the\, correct\, answer. \end{array}$$
Question 6
1 / -0

The $$3rd$$ term of $${\left( {3x - \dfrac{{{y^3}}}{6}} \right)^4}$$ is

A
$$^4{C_2}{\left( {3x} \right)^2}{\left( { - \dfrac{{{y^3}}}{6}} \right)^2}$$

B
$$^4{C_2}{\left( {3x} \right)^2}{\left( { - \dfrac{{{y^2}}}{6}} \right)^2}$$

C
$$^4{C_2}{\left( {3x} \right)^3}{\left( { - \dfrac{{{y^3}}}{6}} \right)^2}$$

D
$$^4{C_2}{\left( {3x} \right)^3}{\left( { - \dfrac{{{y^2}}}{6}} \right)^2}$$

Solution

As we know the binomial expansion-
$$(a+b)^n=^nC_0a^nb^0+^nC_1a^{n-1}b^1+.......+^nC_na^0b^n$$

Similarly,
$$\left(3x-\dfrac{y^3}{6}\right)^4=^4C_0(3x)^4\left(-\dfrac{y^3}{6}\right)^0+^4C_1(3x)^3\left(-\dfrac{y^3}{6}\right)^1+^4C_2(3x)^2(-\dfrac{y^3}{6})^2+............$$

So, the $$3^{rd}$$ term $$=^4C_2(3x)^2(-\dfrac{y^3}{6})^2$$
Question 7
1 / -0

The number of zeroes at the end of $$(101)^{11}-1$$ is

A
$$8$$

B
$$4$$

C
$$1100$$

D
$$2$$

Solution
Question 8
1 / -0

The middle terms in the expansion of $$(x^{2}-a^{2})^{5}$$ is

A
$$10x^{6}a^{4},\ -10x^{4}a^{6}$$

B
$$-10x^{6}a^{4},\ 10x^{4}a^{6}$$

C
$$10x^{6}a^{4},\ 10x^{4}a^{6}$$

D
$$-10x^{6}a^{4},\ -10x^{4}a^{6}$$

Solution

$$ (a\pm b)^{n} =^{n}C_{0}a^{n}b^{0}\pm ^{n}C_{1}a^{n-1}b^{1}+^{n}C_{2}a^{n-2}b^{2}\pm ....+^{n}C_{n}a^{0}b^{n}$$
$$ (x^{2}-a^{2})^{5} = ^{5}C_{0}x^{10}-^{5}C_{1}x^{8}a^{2}+^{5}C_{2}x^{6}a^{4}-^{5}C_{3}x^{4}a^{6}+^{5}C_{4}x^{2}a^{8}-^{5}C_{5}x^{0}a^{10}$$
$$ = x^{10} - 5x^{8}a^{2}+10x^{6}a^{4} - 10x^{4}a^{6} +5x^{2}a^{8} - x^{0}a^{10}$$
Middle term = $$ 10x^{6}a^{4},-10x^{4}a^{6}$$
Question 9
1 / -0

If sum of the coefficients in the expansion of $$(2+3cx+c^2x^2)^{12}$$ vanishes, then $$c$$ equals to

A
$$-1,2$$

B
$$1,2$$

C
$$1,-2$$

D
$$-1,-2$$

Solution

sum of the coefficients in the expansion of $$(2+3cx+c^2x^2)^{12}$$ is obtained by putting $$x=1$$ in the given expression.
Since the sum of the co-efficients vanishes then,
$$2+3c+c^2=0$$
or, $$(c+1)(c+2)=0$$
or, $$c=-1,-2$$.
Question 10
1 / -0

$$^{100}C_{0}-^{100}C_{2}+^{100}C_{4}+^{100}C_{8}-........+^{100}C_{100}=$$__

A
$$2^{-49}$$

B
$$-2$$

C
$$2^{50}$$

D
$$-2^{50}$$

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Binomial Theorem Test - 10

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Binomial Theorem Test - 10

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Question 1

$$51$$

$$26$$

$$102$$

$$25$$

Solution

Question 2

$${2}^{14}$$

$${2}^{14}-15$$

$${2}^{14}+15$$

$${2}^{14}-1$$

Solution

Question 3

$$4$$

$$5$$

$$6$$

$$7$$

Solution

Question 4

1 : 3 : 6

2 : 3 : 6

3 : 2 : 6

3 : 1 : 2

Solution

Question 5

$$2^{2n}$$

$$2^{3n}$$

$$5^{n}$$

None of these

Solution

Question 6

$$^4{C_2}{\left( {3x} \right)^2}{\left( { - \dfrac{{{y^3}}}{6}} \right)^2}$$

$$^4{C_2}{\left( {3x} \right)^2}{\left( { - \dfrac{{{y^2}}}{6}} \right)^2}$$

$$^4{C_2}{\left( {3x} \right)^3}{\left( { - \dfrac{{{y^3}}}{6}} \right)^2}$$

$$^4{C_2}{\left( {3x} \right)^3}{\left( { - \dfrac{{{y^2}}}{6}} \right)^2}$$

Solution

Question 7

$$8$$

$$4$$

$$1100$$

$$2$$

Solution

Question 8

$$10x^{6}a^{4},\ -10x^{4}a^{6}$$

$$-10x^{6}a^{4},\ 10x^{4}a^{6}$$

$$10x^{6}a^{4},\ 10x^{4}a^{6}$$

$$-10x^{6}a^{4},\ -10x^{4}a^{6}$$

Solution

Question 9

$$-1,2$$

$$1,2$$

$$1,-2$$

$$-1,-2$$

Solution

Question 10

$$2^{-49}$$

$$-2$$

$$2^{50}$$

$$-2^{50}$$

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