Binomial Theorem Test

Result

Binomial Theorem Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

Find the coefficient of $$x^{11}$$ in the expansion of $$\left(x^{3}-\dfrac{2}{x^{2}}\right)^{12}$$

A
$$-25344$$

B
$$-25250$$

C
$$-25000$$

D
$$-25750$$

Solution
Question 2
1 / -0

Coefficient of $$\alpha $$ in the expansion of $$(\alpha +p)^{m-1}+(\alpha +p)^{m-2}(\alpha +q)^{m-3}(\alpha +q)^{2}+....(\alpha +q)^{m-1}$$ where $$\alpha \neq -q$$ and $$p\neq q$$ is:

A
$$\frac{^{m}C_{1}(p^{1}-q^{1})}{p-q}$$

B
$$\frac{^{m}C_{1}(p^{m-1}-q^{m-1})}{p-q}$$

C
$$\frac{^{m}C_{1}(p^{1}+q^{1})}{p-q}$$

D
$$\frac{^{m}C_{1}(p^{m-1}+q^{m-1})}{p-q}$$

Solution
Question 3
1 / -0

If the $$r^{th}$$ and the $$(r+1)^{th}$$ terms in the expansion of $$(p+q)^{n}$$ are equal, then $$\dfrac{(n+1)q}{r(p+q)}$$ is

A
$$1/2$$

B
$$1/4$$

C
$$1$$

D
$$0$$

Solution
Question 4
1 / -0

Number of distinct terms in the expansion of $$(x+y-z)^{16}$$ is

A
$$816$$

B
$$152$$

C
$$153$$

D
$$136$$

Solution
Question 5
1 / -0

The number of terms in the expansion of $$(1+x)^{101}(1+x^{2}-x)^{100}$$ in power of $$x$$ is:

A
$$302$$

B
$$301$$

C
$$202$$

D
$$101$$
Question 6
1 / -0

In the expansion of $$(1+x)^{30}$$, the sum of the coefficients of odd powers of $$x$$, is

A
$$2^{30}$$

B
$$2^{31}$$

C
$$0$$

D
$$2^{29}$$

Solution
Question 7
1 / -0

The coefficient of $$x^{10}$$ in the expansion of $$(1+x)^{2}(1+x^{2})^{3}(1+x^{3})^{4}$$ is qual to:

A
$$52$$

B
$$56$$

C
$$50$$

D
$$44$$

Solution

We know,
$$(a+b)^2=a^2+b^2+2ab$$
$$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$
$$(a + b)^4 = a^4 +4a^3b + 6a^2b^2 + 4ab^3 + b^4$$

$$(1+x)^2(1+x^2)^3(1+x^3)^4$$
$$\Rightarrow$$ $$(1+x^2+2x)(1+3x^2+3x^4+x^6)(1+4x^3+6x^6+4x^9+x^{12})$$
$$\Rightarrow$$ $$(1+3x^2+3x^4+x^6+x^2+3x^4+3x^6+x^8+2x+6x^3+6x^5+2x^7)(1+4x^3+6x^6+4x^9+x^{12})$$
$$\Rightarrow$$ $$(1+2x+4x^2+6x^4+6x^5+4x^6+2x^7+x^8)(1+4x^3+6x^6+4x^9+x^{12})$$
Now, coefficient of $$x^{10}$$ $$=(4\times 2)+(6\times 6)+(2\times 4)$$
$$=8+36+8$$
$$=52$$
Question 8
1 / -0

The coefficient of $$x^{n}$$ in the expansion of $$\dfrac{1}{(1-x)(1-2x)(1-3x)}$$ is

A
$$\dfrac{1}{2}(2^{n+2}-3^{n+3}+1)$$

B
$$\dfrac{1}{2}(3^{n+2}-2^{n+3}+1)$$

C
$$\dfrac{1}{2}(2^{n+3}-3^{n+2}+1)$$

D
$$none\ of\ these$$

Solution

We have
$$\dfrac{1}{{\left( {1 - x} \right)\left( {1 - 2x} \right)\left( {1 - 3x} \right)}}$$
$$ = \dfrac{1}{{2\left( {1 - x} \right)}} - \dfrac{4}{{1 - 2x}} + \dfrac{9}{{2\left( {1 - 3x} \right)}}$$
$$ = \dfrac{1}{2}{\left( {1 - x} \right)^{ - 1}} - 4{\left( {1 - 2x} \right)^{ - 1}} + \dfrac{9}{2}{\left( {1 - 3x} \right)^{ - 1}}$$
$$ = \dfrac{1}{2}\left( {1 + x + {x^2} + ..... + {x^n}} \right) - 4\left( {1 + 2x + {{\left( {2x} \right)}^2} + ..... + {{\left( {2x} \right)}^n}} \right) + \dfrac{9}{2}\left( {1 + 3x + {{\left( {3x} \right)}^2} + .... + {{\left( {3x} \right)}^n}} \right)$$

Coefficient of $${x^n}$$, we get
$$= \dfrac{1}{2}\left[ {1 - {{8.2}^n} + {{9.3}^n}} \right]$$
$$ = \dfrac{1}{2}\left[ {1 - {2^{n + 3}} + {3^{n + 2}}} \right]$$
$$ = \dfrac{1}{2}\left[ {3^{n+2} - 2^{n + 3}+1} \right]$$

Hence, the option $$(B)$$ is correct answer.
Question 9
1 / -0

If the $$6^{th}$$ term in the expansion of $$\left(\dfrac {1}{x^{8/3}}+x^{2}\log ^{x}_{10}\right)^{8}$$ is $$5600$$, then equals

A
$$1$$

B
$$\log^{10}_{e}$$

C
$$10$$

D
$$No\ such\ x\ exists$$
Question 10
1 / -0

In the expansion of $$(1+2x+3x^{2})^{10},$$ coefficient of $$x^{4}$$ is not divisible by

A
$$12$$

B
$$7$$

C
$$11$$

D
$$5$$

Solution