Binomial Theorem Test

Result

Binomial Theorem Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

$$\text{5}^{th}$$ term from the end in the expansion of $${\left( {\frac{{{x^3}}}{2} - \frac{2}{{{x^2}}}} \right)^{12}}$$ is :

A
$$7920 {x^{ - 4}}$$

B
$$7920 {x^{ 5}}$$

C
$$7920 {x^{ 4}}$$

D
$$-7920{x^{ -4}}$$

Solution
Question 2
1 / -0

If the $$rth$$ term in the expansion of $$\left(\dfrac{x}{3}-\dfrac{2}{x^{2}}\right)^{10}$$ contains $$x^{4}$$ then $$r$$ is equal to

A
$$2$$

B
$$1$$

C
$$3$$

D
$$5$$

Solution
Question 3
1 / -0

The sum of the binomial coefficients in the expansion of $${ \left( { x }^{ -3/4 }+a{ x }^{ 5/4 } \right) }^{ n }$$ lies between $$200$$ and $$400$$ and the term independent of $$x$$ equals $$448$$. The value of $$a$$ is

A
$$1$$

B
$$2$$

C
$$1/2$$

D
for no value of $$a$$

Solution

$$\left(x^{-\dfrac{3}{4}}+ax^{5/4}\right)^n$$ $$={^{n}C_0}\left(x^{-3/4}\right)^n+{^{n}C_1}(x^{-3/4})^{n-1}(ax^{5/4})^1.......+{^{n}C_n}(ax^{5/4})^n$$

Put $$x=1$$

$$(1+a)^n={^{n}C_0}+{^{n}C_1}a+{^{n}C_2}a^2+......+{^{n}C_n}a^n$$

Put $$a=1$$

$$2^n={^{n}C_0}+{^{n}C_1}+{^{n}C_2}+........+{^{n}C_n}$$

$$200 < 2^n < 400$$

$$[\because 2^8=256]$$

$$n=8$$

$$(x^{-3/4}+ax^{5/4})^8$$

$$T_{r+1}={^8C_r}(x^{-3/4})^{8-r}(ax^{5/4})^r$$

$$\Rightarrow T_{3+1}={^8C_3}a^3=448$$

$$T_4=^8C_3a^3=448$$

$$\Rightarrow \dfrac{8\times 7\times 6}{3\times 2}=a^3=448\Rightarrow a^3=8$$

$$a=2$$

$$\begin{bmatrix} \because \dfrac{-3}{4}(8-r)+\dfrac{5r}{4}=0\\ -6+\dfrac{3r}{4}+\dfrac{5r}{4}=0\\ -6+2r=0\\ r=3\end{bmatrix}$$.
Question 4
1 / -0

The sum of the series
$$\dfrac {2\left(\dfrac {n}{2}\right)!\left(\dfrac {n}{2}\right)!}{(n!)}[C^{2}_{0}-2C^{2}_{1}+3C^{2}_{2}...+(-1)^{n}(n+1)C^{2}_{n}]$$
where $$n$$ is an even positive integer, is equal to

A
$$0$$

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

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

D
$$(-1)^{n}n$$

Solution
Question 5
1 / -0

Sum of coefficients of $${ x }^{ 2r }$$, $$r= 1,2,3,$$....... in $$(1+x{ ) }^{ n }$$ is

A
$$({ 2 }^{ n-1 }-1)$$

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

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

D
$$({ 2 }^{ n-2 }-1)$$

Solution
Question 6
1 / -0

Coefficient of $$x^{n}$$ in expansion of $$\dfrac{(1+2x)^{2}}{(1-x)^{3}}$$ is

A
$$2n$$

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

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

D
$$None\ of\ these$$
Question 7
1 / -0

$$\begin{array} { l } { \text { Assertion } ( \mathrm { A } ) : \text { The expansion of } ( 1 + x ) ^ { n } = } \\ { C _ { 0 } + C _ { 1 } x + C _ { 2 } x ^ { 2 } + \ldots + C _ { n } x ^ { n } } \\ { \text { Reason (R): If } x = - 1 , \text { then the above expansion is } } \\ { \text { zero } } \end{array}$$

A
$$\begin{array} { l } { \text { Both A and R are true and } \mathrm { R } \text { is the correct } } \\ { \text { explanation of } \mathrm { A } } \end{array}$$

B
$$\begin{array} { l } { \text { Both A and R are true and R is not the } } \\ { \text { correct explanation of } \mathrm { A } } \end{array}$$

C
$$A$$ is true, but $$R$$ is false

D
$$A$$ is false, but $$R$$ is true

Solution

We Know that Assertion is nothing but the standard binomial expansion of

$$(1+x)^n$$$$=C_{0}+C_{1}X+C_{2}X^2+ \dots +C_{n}x^n$$

So if $$x=-1$$ then

$$(1+(-1))^n=(1-1)^n=0$$

$$Ans:$$ $$Opt:[B]$$
Question 8
1 / -0

The coefficient of $$x^{99}$$ in $$(x+1)(x+3)(x+5).....(x+199)$$ is

A
$$1+2+3+...+99$$

B
$$1+3+5+...+199$$

C
$$1.3.5............199$$

D
$$None\ of\ these$$

Solution

$$\text x^{99} \text { in }(x+1)(x+3)(x+5) \ldots(x+199) \\$$

$$\text { Multiplying }(x+1)(x+3) \\$$
$$\text { will give } x^{2}+(1+3) x+(1)(3) \\$$
$$\text { so coefficient } x \text { is }(1+3) \\$$
$$\Rightarrow \operatorname{2n}(x+1)(x+3)(x+5)$$

So coefficient of $$x^{2}=(1+3+5)$$

So coefficient of $$x^{n}$$ in $$(n+1)$$ terms will be sum of coefficients

So we can say it will be sum of roots of equation

So coefficient of $$x^{99}$$ will be roots in 100 terms

$$1+3+5+\ldots+199$$
Question 9
1 / -0

The middle term in the expansion of $${\left(3x-\dfrac{{x}^{3}}{6}\right)}^{9}$$ is

A
$$\dfrac{21}{16}{x}^{19}$$

B
$$\dfrac{-21}{16}{x}^{19}$$

C
$$\dfrac{21}{16}{x}^{-19}$$

D
$$\dfrac{-21}{16}{x}^{-19}$$

Solution
Question 10
1 / -0

The sum of the co-efficient of all odd degree terms in the expansion of $$\left( x + \sqrt { x ^ { 3 } - 1 } \right) ^ { 5 } + \left( x - \sqrt { x ^ { 3 } - 1 } \right)$$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$- 1$$

Solution