Binomial Theorem Test

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Binomial Theorem Test - 76

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Weekly Quiz Competition

Question 1
1 / -0

In the expansion of $$\left( 1+ \right) ^{ 50 },$$ the sum of the coefficient of odd powers of x is _______.

A
226

B
249

C
250

D
251

Solution
Question 2
1 / -0

Find the middle term(s) in the expansion of :
$$\left(2ax-\dfrac{b}{x^{2}}\right)^{12}$$

A
$$\dfrac {59136\ a^6 b^6} {x^6}$$

B
$$\dfrac {59163\ a^5 b^5} {x^5}$$

C
$$\dfrac {59631\ a^7 b^7} {x^7}$$

D
None of these

Solution

Given that to find middle term in expansion of $$\left(20x-\dfrac{b}{x^{2}}\right)^{12}$$

$$n=12\Rightarrow$$ even

for even number middle term $$=\left(\dfrac{n}{2}+1\right)th$$

middle term $$=\left(\dfrac{12}{2}+1\right)th\Rightarrow 7$$

We know that $$T_{r+1}=\ ^{n}C_{r}(a)^{n-r}b^{r}$$

$$T_{7}=T_{6+1}=\ ^{12}C_{6}(29 x)^{12-6}\left(\dfrac{-b}{x^{2}}\right)^{6}$$

$$=\dfrac{12!}{6!6!}2^{6}a^{6}x^{6}\dfrac{b^{6}}{x^{12}}$$

$$=\dfrac{12\times 11\times 10\times 9\times 8\times 7\times 6!}{6\times 5\times 4\times 3\times 2\times 1\times 6!}2^{6}\dfrac{a^{6}b^{6}}{x^{6}}$$

$$=\dfrac{59136 a^{6}b^{6}}{x^{6}}$$
Question 3
1 / -0

The constant term in the expansion of $$(x^{2}-\frac{1}{x^{2}})^{16}$$ is

A
$$C^{16}_{8}$$

B
$$C^{16}_{7}$$

C
$$C^{16}_{9}$$

D
$$C^{16}_{10}$$

Solution
Question 4
1 / -0

Find the middle term(s) in the expansion of :
$$\left(2x-\dfrac{x^{2}}{4}\right)^{9}$$

A
$$\dfrac {63}{5}x^{13}, -\dfrac {62}{31}x^{15}$$

B
$$\dfrac {63}{4}x^{13}, -\dfrac {63}{32}x^{14}$$

C
$$\dfrac {61}{4}x^{11}, \dfrac {61}{33}x^{13}$$

D
None of these

Solution

Given expansion term is $$\left(2x-\dfrac{x^{2}}{4}\right)^{9}$$

$$n=9=$$ odd

middle term $$=\left(\dfrac{9+1}{2}\right)^{th}+\left(\dfrac{9+3}{2}\right)^{th}$$

$$=5^{th}$$ or $$6^{th}$$ term

$$T_{5}=T_{4+1}=\ ^{9}C_{4}(2x)^{5}\left(\dfrac{-x^{2}}{4}\right)^{4}$$

$$=\dfrac{9!}{4!5!}2^{5}\times x^{5}(-1)^{4}\times \dfrac{x^{8}}{4^{4}}$$

$$=\dfrac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}\times \dfrac{2^{5}.x^{13}.1}{2^{5}}$$

$$T_{5}=\dfrac{63}{4}x^{13}$$

$$T_{6}=T_{5+1}=\ ^{9}C_{5}(2x)^{4}\left(\dfrac{-x^{2}}{4}\right)^{5}$$

$$=\dfrac{9!}{4!5!}\times 2^{4}x^{4}(-1)^{5}\dfrac{x^{10}}{4^{5}}$$

$$=\dfrac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}2^{4}\dfrac{x^{14}}{2^{10}}$$

$$T_{6}=\dfrac{-63}{32}x^{14}$$
Question 5
1 / -0

The coefficient of the middle term in the binomial expansion in powers of x of $$(1+\alpha x)^{4}$$ and of $$(1-\alpha x)^{6}$$ is the same if $$\alpha $$ equals :

A
$$\frac{3}{5}$$

B
$$\frac{10}{3}$$

C
$$\frac{-3}{10}$$

D
$$\frac{-5}{3}$$

Solution

$$\textbf{Step 1: Calculating coefficient of individual term}$$
$$\text{The middle term of the binomial expansion }(1+\alpha x)^4$$$$\text{ will be the }(\dfrac42+1)=\text{ 3rd term}$$
$$\text{The middle term of the binomial expansion }(1-\alpha x)^6 \text{will be the }(\dfrac62+1)=\text{ 4th term}$$
$$\text{Coefficient of3rd term of }(1+\alpha x)^4=^4C_2\alpha^2$$
$$\text{Coefficient of 4th term of }(1-\alpha x)^6=-^6C_3\alpha^3$$
$$\textbf{Step 2: Calculating }\mathbf{\alpha}$$
$$\text{Equating both the coefficients}$$
$$\implies ^4C_2\alpha^2=-^6C_3\alpha^3$$
$$\implies 6\alpha^2=-20\alpha^3 $$
$$\implies \alpha=-\dfrac6{20}=-\dfrac3{10}$$
$$\mathbf{\text{Hence, The value of }\alpha =-\dfrac3{10}}$$
Question 6
1 / -0

Find the middle term in the expansion of
$$\left(\dfrac{2}{3}x-\dfrac{3}{2x}\right)^{20}$$

A
$$^{20}C_{10}x^{10}y^{10}$$

B
$$^{20}C_{11}x^{11}y^{11}$$

C
$$^{20}C_{9}x^{11}y^{10}$$

D
None of these

Solution

Given that, $$\left(\dfrac{2x}{3}-\dfrac{3y}{2}\right)^{20}$$

$$n=20\Rightarrow$$ even

For even values of $$n$$, middle term is $$\left(\dfrac{n}{2}+1\right)th$$

$$\Rightarrow$$ middle term $$=\left(\dfrac{20}{2}+1\right)th=11th$$

Now use the formula $$T_{r+1}=\ ^{n}C_{r}(a)^{n-r}(b)^{r}$$

Here, $$r=10, a=\dfrac{2x}{3},b=\dfrac{3y}{2}$$

$$T_{11}=T_{10+1}=\ ^{20}C_{10}\left(\dfrac{2x}{3}\right)^{20-10}\left(\dfrac{3y}{2}\right)^{10}$$

$$T_{11}=\ ^{20}C_{10}\left(\dfrac{2x}{3}\right)^{10}\left(\dfrac{3y}{x}\right)^{10}$$

$$T_{11}=\ ^{20}C_{10}x^{10}y^{10}$$
Question 7
1 / -0

Find the middle term in the expansion of :
$$\left(x^{2}-\dfrac{2}{x}\right)^{10}$$

A
$$-8604 \ x^7$$

B
$$-8064 \ x^5$$

C
$$-804 \ x^4$$

D
None of these

Solution

Given to find middle term in the expansion of $$\left(x^{2}-\dfrac{2}{x}\right)^{10}$$

$$n=10\Rightarrow$$ even

for even numbers middle term is $$\left(\dfrac{n}{2}+1\right)th$$

middle term $$=\left(\dfrac{10}{2}+1\right)th=6^{th}$$

We know that $$T_{r+1}=\ ^{n}C_{r}(a)^{n-r}(b)^{r}$$

$$T_{6}=T_{5+1}=\ ^{10}C_{5}(x^{2})^{5}\left(\dfrac{-2}{x}\right)^{5}$$

$$=\dfrac{-10!}{5!5!}x^{10}\dfrac{2^{5}}{x^{5}}$$

$$T_{6}=-8064 x^{5}$$
Question 8
1 / -0

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

A
$$\dfrac {189}{8}x^{15},- \dfrac {21}{16} x^{17}$$

B
$$\dfrac {189}{8}x^{17},- \dfrac {21}{16} x^{19}$$

C
$$\dfrac {189}{7}x^{15},- \dfrac {23}{13} x^{19}$$

D
None of these

Solution

Given that to find middle term in expansion of $$\left(3x-\dfrac{x^{3}}{6}\right)^{9}$$

$$n=9\Rightarrow$$ odd

For odd values, there are two middle terms.

First middle term $$=\dfrac{n+1}{2}^{th}$$ term : $$\Rightarrow \left(\dfrac{9+1}{2}\right)^{th}$$ term $$=5^{th}$$ term

Second middle term $$=\left(\dfrac{n+1}{2}+1\right)^{th}$$ term; $$\Rightarrow \left(\dfrac{9+1}{2}+1\right)^{th}$$ term $$=6^{th}$$ term

We know that $$T_{r+1}=\ ^{9}C_{r}(3x)^{r}\left(\dfrac{-x^{3}}{6}\right)^{r}$$

$$=\ ^{9}C_{r}(3^{r})x^{r}(-1) x^{3r}\left(\dfrac{1}{6}\right)^{r}$$

$$T_{r+1}=T_{5}\Rightarrow r+1=5\Rightarrow r=4$$

$$T_{4+1}=\ ^{9}C_{4}(3x)^{9-4}\left(\dfrac{-x^{3}}{6}\right)^{4}$$ $$T_{5+1}=\ ^{9}C_{5}(3x)^{9-5}\left(\dfrac{-x^{3}}{6}\right)^{5}$$

$$=\dfrac{9!}{5!4!}3^{5}x^{5}\dfrac{x^{12}}{6^{4}}$$ $$=\dfrac{9!}{4!5!}3^{4}x^{4}\dfrac{x^{15}}{6^{5}}$$

$$=126\dfrac{3^{5}x^{12}\times x^{5}}{64}$$ $$=\dfrac{-21}{16}x^{19}$$

$$=\dfrac{189}{7}x^{17}$$
Question 9
1 / -0

Find the middle term(s) in the expansion of :
$$\left(\dfrac{p}{x}+\dfrac{x}{p}\right)^{9}$$

A
$$\dfrac {126x}{p}$$,$$\dfrac {126p}{x}$$

B
$$\dfrac {126}{p}$$,$$126x$$

C
$$\dfrac {160x^2}{p}$$,$$\dfrac {162x}{p^2}$$

D
None of these

Solution

Given that to find middle term in expansion of $$\left(\dfrac{P}{x}+\dfrac{x}{P}\right)^{9}$$

$$n=9\Rightarrow $$ odd

For off values there are two middle terms

First middle term $$=\left(\dfrac{n+1}{2}\right)^{th}$$ term $$\Rightarrow \left(\dfrac{9+1}{2}\right)^{th}$$ term $$=5^{th}$$ term

second middle term $$=\left(\dfrac{n+1}{2}+1\right)^{th}$$ term $$=\left(\dfrac{9+1}{2}+1\right)^{th}$$ term $$=6^{th}$$ term

We know that $$T_{r+1}=\ ^{9}C_{r}\left(\dfrac{P}{x}\right)^{9-r}\left(\dfrac{x}{P}\right)^{r}$$

$$T_{r+1}=T_{5}\Rightarrow r+1=5\Rightarrow r=4$$

$$T_{4+1}=\ ^{9}C_{4}\left(\dfrac{P}{x}\right)^{5}\left(\dfrac{x}{P}\right)^{4}$$

$$=\dfrac{9!}{4!5!}\dfrac{P^{5}}{x^{5}}\Rightarrow 126 \dfrac{P}{x}$$

$$T_{r+1}=T_{0}\Rightarrow r+1=6\Rightarrow r=5$$

$$T_{5+1}=\ ^{9}C_{5}\left(\dfrac{P}{x}\right)^{4}\left(\dfrac{x}{P}\right)^{5}$$

$$=\dfrac{9!}{4!5!}\dfrac{P^{4}}{x^{4}}\dfrac{r^{5}}{P^{5}}$$

$$=126\dfrac{x}{P}$$
Question 10
1 / -0

Find the coefficient of $$x^{7}$$ in the expansion of $$\left(x-\dfrac{1}{x^{2}}\right)^{40}$$

A
$$-^{40}C_{11}$$

B
$$-^{40}C_{10}$$

C
$$-^{40}C_{12}$$

D
None of these

Solution

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Binomial Theorem Test - 76

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Binomial Theorem Test - 76

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SHARING IS CARING

Question 1

226

249

250

251

Solution

Question 2

$$\dfrac {59136\ a^6 b^6} {x^6}$$

$$\dfrac {59163\ a^5 b^5} {x^5}$$

$$\dfrac {59631\ a^7 b^7} {x^7}$$

None of these

Solution

Question 3

$$C^{16}_{8}$$

$$C^{16}_{7}$$

$$C^{16}_{9}$$

$$C^{16}_{10}$$

Solution

Question 4

$$\dfrac {63}{5}x^{13}, -\dfrac {62}{31}x^{15}$$

$$\dfrac {63}{4}x^{13}, -\dfrac {63}{32}x^{14}$$

$$\dfrac {61}{4}x^{11}, \dfrac {61}{33}x^{13}$$

None of these

Solution

Question 5

$$\frac{3}{5}$$

$$\frac{10}{3}$$

$$\frac{-3}{10}$$

$$\frac{-5}{3}$$

Solution

Question 6

$$^{20}C_{10}x^{10}y^{10}$$

$$^{20}C_{11}x^{11}y^{11}$$

$$^{20}C_{9}x^{11}y^{10}$$

None of these

Solution

Question 7

$$-8604 \ x^7$$

$$-8064 \ x^5$$

$$-804 \ x^4$$

None of these

Solution

Question 8

$$\dfrac {189}{8}x^{15},- \dfrac {21}{16} x^{17}$$

$$\dfrac {189}{8}x^{17},- \dfrac {21}{16} x^{19}$$

$$\dfrac {189}{7}x^{15},- \dfrac {23}{13} x^{19}$$

None of these

Solution

Question 9

$$\dfrac {126x}{p}$$,$$\dfrac {126p}{x}$$

$$\dfrac {126}{p}$$,$$126x$$

$$\dfrac {160x^2}{p}$$,$$\dfrac {162x}{p^2}$$

None of these

Solution

Question 10

$$-^{40}C_{11}$$

$$-^{40}C_{10}$$

$$-^{40}C_{12}$$

None of these

Solution

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