Binomial Theorem Test

Result

Binomial Theorem Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

The coefficient of $$x^4$$ in the expansion of $$(1+x+x^2+x^3)^{11}$$ is

A
$$990$$

B
$$495$$

C
$$330$$

D
none of these

Solution

$$=\left(1+x+x^{2}+x^{3}\right)^{11}$$

$$=\left((1+x)\left(1+x^{2}\right)\right)^{11}$$

$$=(1+x)^{ 11}\left[1+^{11} c_{1} x^{2}+{ }^{11} c_{2} x^{4}+\cdots^{11} c_{11}\left(x^{2}\right)^{11}\right.]$$

$$=\left[1+{ }^{11} c_{1} x+{ }^{11} c_{2} x^{2}+\cdots\right]\left[1+{ }^{11} c_{1} x^{2}+{ }^{11} c_{2} x^{4}+\cdots\right]$$

$$={ }^{11} c_{0} \cdot{ }^{11} c_{2} \cdot x^{4}+{ }^{11} c_{2} \cdot{ }^{11} c_{1} \cdot x^{4}+{ }^{11} c_{4}{ }^{11} c_{0} x^{4}$$

$$=\quad 990 x^{4}$$

Option $$(A)$$
Question 2
1 / -0

If the last term in the binomial expansion of $$\left(2^{1/3}-\dfrac {1}{\sqrt {2}}\right)^{n}$$ is $$\left(\dfrac {1}{3^{5/3}}\right)^{\log_{3}8}$$, then the $$5^{th}$$ terms form the beginning is:

A
$$210$$

B
$$420$$

C
$$103$$

D
$$None\ of\ these$$

Solution
Question 3
1 / -0

The co-efficient of $$x^{k}$$ in expansion of $$1+\left(1+x\right)+\left(1+x\right)^{2}++\left(1+x\right)^{n}$$ is : $$\left(n>k\right)$$

A
$$^{n}C_{k}$$

B
$$^{n+1}C_{k}$$

C
$$^{n+1}C_{k+1}$$

D
$$None\ of\ these$$
Question 4
1 / -0

For $$x\in R, x\neq -1$$ if $$(1+x)^{2016}+x(1+x)^{2015}+x(1+x)^{2014}+.+x^{2016}=\sum _{ i=0 }^{ 2016 }{ { a }_{ i }{ x }^{ i } }$$, then $$a_{17}$$ is equal to

A
$$\dfrac{2017!}{ 17!2000!}$$

B
$$\dfrac{2016!}{ 17!1999!}$$

C
$$\dfrac{2017!}{2000!}$$

D
$$\dfrac{2016!}{ 16!}$$

Solution
Question 5
1 / -0

The coefficients of $$x^{10}$$ in the expansion of $$(1+x)^{15}+(1+x)^{16}+(1+x)^{17}+....+(1+x)^{30}$$ is

A
$$^{31}C_{10}-^{15}C_{10}$$

B
$$^{31}C_{11}-^{15}C_{11}$$

C
$$^{30}C_{10}-^{15}C_{10}$$

D
$$^{31}C_{10}-^{14}C_{11}$$

Solution
Question 6
1 / -0

The sum of the coefficient in the expansion of $$(a+2b+c)^{11}$$ is-

A
$$4^{11}$$

B
$$32$$

C
$$31$$

D
$$None\ of\ these$$

Solution
Question 7
1 / -0

Coefficient of $$x^{11}$$ in the extension of $$(1+x^{2})^{4}(1+x^{3})^{7}(1+x^{4})^{12}$$ is

A
1051

B
1106

C
1113

D
1120

Solution
Question 8
1 / -0

If the variable takes the values 0,1,2,....., n with frequericies proportional to the binomial coefficients $$C\left( n,0 \right) ,C\left( n,1 \right) ,C\left( n,2 \right) ...,C\left( n,n \right) $$ respectively, then the variance of the distribution is :-

A
n

B
$$\frac { \sqrt { n } }{ 2 } $$

C
$$\frac { n }{ 2 } $$

D
$$\frac { n }{ 4 } $$

Solution
Question 9
1 / -0

The coefficient of $$x^n$$ in the binomial expansion of $$(1-x)^{-2}$$, is

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

B
$$n+1$$

C
n

D
$$2n$$

Solution
Question 10
1 / -0

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

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

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

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

D
$$None\ of\ these$$

Solution

$$\text { We know that } \\$$

$$\frac{1}{1-a} \text { is sum of } \infty GP=1+a+a^{2}+a^{3} \ldots\infty \\$$

$$\text { Similary: }$$

$$\frac{1}{(1-x)(1-2 x)(1-3 x)}={\left(1+x+x^{2}+\ldots\infty\right)\left(1+(2 x)+(2 x)^{2}\ldots\infty\right)}\\ $$

$${\left(1+(3 x)+(3 x)^{2}+(3 x)^{3}\ldots\infty\right.)}$$

From this equation, we observe that

coefficient of $$x^{n}$$ is $$: \rightarrow$$

$$\frac{1}{2}\left[2^{n+2}-3^{n+3}+1\right]$$

Option (a)