Binomial Theorem Test

Result

Binomial Theorem Test - 75

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Weekly Quiz Competition

Question 1
1 / -0

The coefficient of $$x^{m}$$ in $$(1 + x)^{p} + (1 + x)^{p + 1} + .... + (1 + x)^{n}, p\leq m\leq n$$ is

A
$$^{n + 1}C_{m + 1}$$

B
$$^{n - 1}C_{m - 1}$$

C
$$^{n}C_{m}$$

D
$$^{n}C_{m + 1}$$

Solution
Question 2
1 / -0

The sum of the binomial coefficients of $$\left[2x+\cfrac{1}{x}\right]^n$$ is equal to $$256$$. The constant term in the expansion is

A
1120

B
2110

C
1210

D
None of these

Solution
Question 3
1 / -0

The coefficient of $$x^{9}$$ in the expansion of (1+x)$$(1+x^{2})(1+x^{3})....(1+x^{100})$$ is____.

A
2

B
4

C
6

D
8

E
None of these.
Question 4
1 / -0

The first integral term in the expansion of $$(\sqrt{3}+\sqrt[3]{2})^{9}$$, is its

A
2nd term

B
3 rd term

C
4th term

D
5th term

Solution
Question 5
1 / -0

The sum of the coefficients in the expansion of $${ (1+x-{ 3x }^{ 2 }) }^{ 171 }$$ is

A
0

B
1

C
-1

D
None of these
Question 6
1 / -0

Co-efficient of $$x^{n-1}$$ in the expansion of, $$(x+3)^{n}+(x+3)^{n-1}(x+2)+(x+3)^{n-2}(x+2)^{2}+...+(x+2)^{n}$$ is:

A
$$^{n+1}C_{2}(3)$$

B
$$^{n-1}C_{2}(5)$$

C
$$^{n+1}C_{2}(5)$$

D
$$^{n}C_{2}(5)$$

Solution
Question 7
1 / -0

In the expansion of $$(x+y+z)^{10}$$

A
the coefficient of $$x^{2}y^{2}z^{3}$$ is 0

B
the coefficient of $$x^{8}yz$$ is 90

C
the number of terms is 66

D
none of these
Question 8
1 / -0

If in the expansion of $$\left( 2 ^ { x } + \dfrac { 1 } { 4 ^ { x } } \right) ^ { n } , T _ { 3 } = 7 T _ { 2 }$$ and sum of the binomial coefficients of second and third terms is $$36 ,$$ then the value of $$x$$ is -

A
$$- 1 / 3$$

B
$$- 1 / 2$$

C
$$1 / 3$$

D
$$1 / 2$$

Solution
Question 9
1 / -0

The first three terms in the expansion of $$(1 - ax)^{n}$$, where $$n$$ is positive integer are, $$1 - 4x + 7x^{2}$$, then the value of $$a$$

A
$$4$$

B
$$\dfrac {1}{4}$$

C
$$8$$

D
$$\dfrac {1}{2}$$
Question 10
1 / -0

The value of x for which the sixth term in the expansion of
$$\left [ 2^{\log_{2}{\sqrt{9^{x-1}+7}}}+\dfrac{1}{2^{\frac{1}{5}\log_{2}{(3^{x-1}+1)}}} \right ]^{7}$$ is equal to 84 is

A
4

B
1 or 2

C
0 or 1

D
3.

Solution

Here we have,
$$\left [ 2^{\log_{2}{\sqrt{9^{x-1}+7}}}+\dfrac{1}{2^{\frac{1}{5}\log_{2}{(x-1)}+1}} \right ]^{7}=\left[ \sqrt{9^{x-1}+7} +(3^{x-1}+1)^{(-1/5)}\right]^7$$

The sixth term in the expansion is given by,
$$T_6=\ ^7C_5 ( \sqrt{9^{x-1}+7})^2\cdot \left\{(3^{x-1}+1)^{(-1/5)}\right\}^5\\ \Rightarrow 84=21 \cdot(9^{x-1}+7)\cdot(3^{x-1}+1)^{-1}\\ \Rightarrow \dfrac{84}{21}=\dfrac{9^{x-1}+7}{3^{x-1}+1}\\\Rightarrow 4\cdot (3^{x-1}+1)=(3^{x-1})^2+7\\ \Rightarrow (3^{x-1})^2-4\times 3^{x-1}+3=0\\ \Rightarrow (3^{x-1}-1)(3^{x-1}-3)=0$$
$$ \Rightarrow 3^{x-1}=1$$ or $$3$$
$$\Rightarrow 3^{x-1}=3^0$$ or $$3^1$$
$$\Rightarrow x-1=0$$ or $$1$$
$$\Rightarrow x=1, 2$$

Hence, option (B) is correct.