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CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 15 - MCQExams.com
CBSE
Class 11 Engineering Maths
Binomial Theorem
Quiz 15
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:
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$$^{n+1}C_{2}(3)$$
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$$^{n-1}C_{2}(5)$$
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$$^{n+1}C_{2}(5)$$
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$$^{n}C_{2}(5)$$
In the expansion of $$(x+y+z)^{10}$$
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the coefficient of $$x^{2}y^{2}z^{3}$$ is 0
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the coefficient of $$x^{8}yz$$ is 90
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the number of terms is 66
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none of these
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 -
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$$- 1 / 3$$
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$$- 1 / 2$$
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$$1 / 3$$
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$$1 / 2$$
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 $$a$$
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$$4$$
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$$\dfrac {1}{4}$$
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$$8$$
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$$\dfrac {1}{2}$$
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
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4
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1 or 2
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0 or 1
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3.
Explanation
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.
In the expansion of $$\left( 1+ \right) ^{ 50 },$$ the sum of the coefficient of odd powers of x is _______.
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0%
226
0%
249
0%
250
0%
251
Find the middle term(s) in the expansion of :
$$\left(2ax-\dfrac{b}{x^{2}}\right)^{12}$$
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$$\dfrac {59136\ a^6 b^6} {x^6}$$
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$$\dfrac {59163\ a^5 b^5} {x^5}$$
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$$\dfrac {59631\ a^7 b^7} {x^7}$$
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None of these
Explanation
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 $$T_{r+1}=\ ^{n}C_{r}(a)^{n-r}b^{r}$$
$$T_{7}=T_{6+1}=\ ^{12}C_{6}(29 x)^{12-6}\left(\dfrac{-b}{x^{2}}\right)^{6}$$
$$=\dfrac{12!}{6!6!}2^{6}a^{6}x^{6}\dfrac{b^{6}}{x^{12}}$$
$$=\dfrac{12\times 11\times 10\times 9\times 8\times 7\times 6!}{6\times 5\times 4\times 3\times 2\times 1\times 6!}2^{6}\dfrac{a^{6}b^{6}}{x^{6}}$$
$$=\dfrac{59136 a^{6}b^{6}}{x^{6}}$$
The constant term in the expansion of $$(x^{2}-\frac{1}{x^{2}})^{16}$$ is
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$$C^{16}_{8}$$
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$$C^{16}_{7}$$
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$$C^{16}_{9}$$
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$$C^{16}_{10}$$
Find the middle term(s) in the expansion of :
$$\left(2x-\dfrac{x^{2}}{4}\right)^{9}$$
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$$\dfrac {63}{5}x^{13}, -\dfrac {62}{31}x^{15}$$
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$$\dfrac {63}{4}x^{13}, -\dfrac {63}{32}x^{14}$$
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$$\dfrac {61}{4}x^{11}, \dfrac {61}{33}x^{13}$$
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None of these
Explanation
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
$$T_{5}=T_{4+1}=\ ^{9}C_{4}(2x)^{5}\left(\dfrac{-x^{2}}{4}\right)^{4}$$
$$=\dfrac{9!}{4!5!}2^{5}\times x^{5}(-1)^{4}\times \dfrac{x^{8}}{4^{4}}$$
$$=\dfrac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}\times \dfrac{2^{5}.x^{13}.1}{2^{5}}$$
$$T_{5}=\dfrac{63}{4}x^{13}$$
$$T_{6}=T_{5+1}=\ ^{9}C_{5}(2x)^{4}\left(\dfrac{-x^{2}}{4}\right)^{5}$$
$$=\dfrac{9!}{4!5!}\times 2^{4}x^{4}(-1)^{5}\dfrac{x^{10}}{4^{5}}$$
$$=\dfrac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}2^{4}\dfrac{x^{14}}{2^{10}}$$
$$T_{6}=\dfrac{-63}{32}x^{14}$$
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 :
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$$\frac{3}{5}$$
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$$\frac{10}{3}$$
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$$\frac{-3}{10}$$
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$$\frac{-5}{3}$$
Explanation
$$\textbf{Step 1: Calculating coefficient of individual term}$$
$$\text{The middle term of the binomial expansion }(1+\alpha x)^4$$
$$\text{ will be the }(\dfrac42+1)=\text{ 3rd term}$$
$$\text{The middle term of the binomial expansion }(1-\alpha x)^6 \text{will be the }(\dfrac62+1)=\text{ 4th term}$$
$$\text{Coefficient of3rd term of }(1+\alpha x)^4=^4C_2\alpha^2$$
$$\text{Coefficient of 4th term of }(1-\alpha x)^6=-^6C_3\alpha^3$$
$$\textbf{Step 2: Calculating }\mathbf{\alpha}$$
$$\text{Equating both the coefficients}$$
$$\implies ^4C_2\alpha^2=-^6C_3\alpha^3$$
$$\implies 6\alpha^2=-20\alpha^3 $$
$$\implies \alpha=-\dfrac6{20}=-\dfrac3{10}$$
$$\mathbf{\text{Hence, The value of }\alpha =-\dfrac3{10}}$$
Find the coefficient of $$x^{7}$$ in the expansion of $$\left(x-\dfrac{1}{x^{2}}\right)^{40}$$
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$$-^{40}C_{11}$$
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$$-^{40}C_{10}$$
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$$-^{40}C_{12}$$
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None of these
Find the middle term in the expansion of
$$\left(\dfrac{2}{3}x-\dfrac{3}{2x}\right)^{20}$$
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$$^{20}C_{10}x^{10}y^{10}$$
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$$^{20}C_{11}x^{11}y^{11}$$
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$$^{20}C_{9}x^{11}y^{10}$$
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None of these
Explanation
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, $$r=10, a=\dfrac{2x}{3},b=\dfrac{3y}{2}$$
$$T_{11}=T_{10+1}=\ ^{20}C_{10}\left(\dfrac{2x}{3}\right)^{20-10}\left(\dfrac{3y}{2}\right)^{10}$$
$$T_{11}=\ ^{20}C_{10}\left(\dfrac{2x}{3}\right)^{10}\left(\dfrac{3y}{x}\right)^{10}$$
$$T_{11}=\ ^{20}C_{10}x^{10}y^{10}$$
Find the middle term(s) in the expansion of :
$$\left(\dfrac{p}{x}+\dfrac{x}{p}\right)^{9}$$
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$$\dfrac {126x}{p}$$,
$$\dfrac {126p}{x}$$
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$$\dfrac {126}{p}$$,
$$126x$$
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$$\dfrac {160x^2}{p}$$,
$$\dfrac {162x}{p^2}$$
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None of these
Explanation
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 $$T_{r+1}=\ ^{9}C_{r}\left(\dfrac{P}{x}\right)^{9-r}\left(\dfrac{x}{P}\right)^{r}$$
$$T_{r+1}=T_{5}\Rightarrow r+1=5\Rightarrow r=4$$
$$T_{4+1}=\ ^{9}C_{4}\left(\dfrac{P}{x}\right)^{5}\left(\dfrac{x}{P}\right)^{4}$$
$$=\dfrac{9!}{4!5!}\dfrac{P^{5}}{x^{5}}\Rightarrow 126 \dfrac{P}{x}$$
$$T_{r+1}=T_{0}\Rightarrow r+1=6\Rightarrow r=5$$
$$T_{5+1}=\ ^{9}C_{5}\left(\dfrac{P}{x}\right)^{4}\left(\dfrac{x}{P}\right)^{5}$$
$$=\dfrac{9!}{4!5!}\dfrac{P^{4}}{x^{4}}\dfrac{r^{5}}{P^{5}}$$
$$=126\dfrac{x}{P}$$
Find the middle term(s) in the expansion of :
$$\left(3x-\dfrac{x^{3}}{6}\right)^{9}$$
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$$\dfrac {189}{8}x^{15},- \dfrac {21}{16} x^{17}$$
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$$\dfrac {189}{8}x^{17},- \dfrac {21}{16} x^{19}$$
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$$\dfrac {189}{7}x^{15},- \dfrac {23}{13} x^{19}$$
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None of these
Explanation
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 $$T_{r+1}=\ ^{9}C_{r}(3x)^{r}\left(\dfrac{-x^{3}}{6}\right)^{r}$$
$$=\ ^{9}C_{r}(3^{r})x^{r}(-1) x^{3r}\left(\dfrac{1}{6}\right)^{r}$$
$$T_{r+1}=T_{5}\Rightarrow r+1=5\Rightarrow r=4$$
$$T_{4+1}=\ ^{9}C_{4}(3x)^{9-4}\left(\dfrac{-x^{3}}{6}\right)^{4}$$ $$T_{5+1}=\ ^{9}C_{5}(3x)^{9-5}\left(\dfrac{-x^{3}}{6}\right)^{5}$$
$$=\dfrac{9!}{5!4!}3^{5}x^{5}\dfrac{x^{12}}{6^{4}}$$ $$=\dfrac{9!}{4!5!}3^{4}x^{4}\dfrac{x^{15}}{6^{5}}$$
$$=126\dfrac{3^{5}x^{12}\times x^{5}}{64}$$ $$=\dfrac{-21}{16}x^{19}$$
$$=\dfrac{189}{7}x^{17}$$
Find the middle term in the expansion of :
$$\left(x^{2}-\dfrac{2}{x}\right)^{10}$$
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$$-8604 \ x^7$$
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$$-8064 \ x^5$$
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$$-804 \ x^4$$
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None of these
Explanation
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 $$T_{r+1}=\ ^{n}C_{r}(a)^{n-r}(b)^{r}$$
$$T_{6}=T_{5+1}=\ ^{10}C_{5}(x^{2})^{5}\left(\dfrac{-2}{x}\right)^{5}$$
$$=\dfrac{-10!}{5!5!}x^{10}\dfrac{2^{5}}{x^{5}}$$
$$T_{6}=-8064 x^{5}$$
In the expansion of $$(1 + ax + bx^2)(1-3x)^{15}$$, if coefficient of $$x^2$$ is $$0$$ then order pair $$(a, b)$$ is equal to
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$$(28, 325)$$
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$$(18, 315)$$
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$$(28, 315)$$
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$$(18, 325)$$
Explanation
Coefficient of $$x^2$$ in $$(1 + ax + bx^2 )(1-3x)^{15}$$
$$=\ ^{15}C_2(-3)^2+a. \ ^{15}C_1(-3)+ b. \ ^{15}C_0 = 0$$
$$\dfrac{15.14}{2}. 9 - 3.a.15+b = 0$$
$$15 \times 63 - 45a + b = 0$$ ....(1)
Coefficient of $$x^3$$ in $$(1 + ax + bx^2) (1 - 3x)^{15}$$
$$\ ^{15}C_3(-3)^3 + a . \ ^{15}C_2(-3)^2 + b.\ ^{15}C_1(-3) = 0$$
$$= \dfrac{15.14.13}{3\times 2} . 3^2 - a.3. \dfrac{15.14}{2} + 15.b = 0$$
$$7.13.3 - 21a + b = 0$$ ...(2)
by using (1) - (2)
$$672 - 24a = 0 \Rightarrow a = 28$$
Hence $$b = 315$$
The number of terms in the expansion of $$ \left\{ ( 2x +3y)^9 + ( 2x - 3y)^9 \right\} $$is
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10
0%
8
0%
4
0%
5
Explanation
.
$$\textbf{Step 1: Find number of terms of expansion}$$
$$\text{Given equation is -}$$
$$ \left\{ ( 2x +3y)^9 + ( 2x - 3y)^9 \right\}$$
$$\text{We know that for expansion of (x + y)$^n$ + (x – y)$^n$ $\\$$\Rightarrow$ If n is odd then, number of terms = $\dfrac{n+1}{2} $}$$
$$\text{Comparing with standard equation,n=9}$$
$$\therefore \text{Numer of terms=}$$
$$\dfrac{9+1}{2}$$
$$=\dfrac{10}{2}$$
$$=5 $$
$$\textbf{Therefore,the expansion of}$$
$$\mathbf{ \left\{ ( 2x +3y)^9 + ( 2x - 3y)^9 \right\}}$$
$$\textbf{has 5 terms.}$$
The number of terms in the expansion of $$ \left\{( x+a)^{16} + (x -a)^{16} \right\} $$ is
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0%
7
0%
8
0%
9
0%
17
Explanation
$$\textbf{Step 1: Find number of terms of expansion}$$
$$\text{Given equation is -}$$
$$ ({x} + {16})^{16} +( {x} - {16})^{16} $$
$$\text{We know that for expansion of (x + y)$^n$ + (x – y)$^n -$}$$
$$\text{Comparing with standard equation,n=16}$$
$$\Rightarrow \text{If n is even then, Number of terms = $\dfrac{n}{2} $+ 1}$$
$$\therefore \text{Numer of terms=}$$
$$\dfrac{16}{2} +1$$
$$=8+1$$
$$=9 $$
$$\textbf{Therefore,the expansion of $\mathbf{ ({x} + {a})^{16} +( {x} - {a})^{16} }$ has 9 terms.}$$
The numbers of terms in the expansion of $$ ( 3x +y)^{10} $$ is
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9
0%
11
0%
10
0%
8
Explanation
$$\textbf{Step 1: Expand the Expression to find number of terms}$$
$$\text{Given-Equation is } (3x+y)^{10}$$
$$\text{Binomial Theorem is Given by-}$$
$$(b+a)^n=b^n+C_{1}^n(b)^{n-1}a+C_{2}^n(b)^{n-2}a^2+C_{3}^n(b)^{n-3}a^3+..............+C_{n}^n(a)^n$$
$$\text{Comparing with standard formula, n=10, a=3x and b=y}$$
$$(3x+y)^{10}=(3x)^{10}+C_{1}^{10}(3x)^{10-1}(y)+C_{2}^{10}(3x)^{10-2}(y)^2+C_{3}^{10}(2x)^{10-3}(3y)^3+.....+C_{10}^{10}(y)^{10}$$
$$\text{From expansion, total number of terms=n+1}$$
$$\therefore \text{Total number of terms= 10+1}$$
$$=11$$
$$\textbf{Therefore, total number of terms are 11}$$
$$ \{ C_{ 0 }+3C_{ 1 }+5C_{ 2 }+...+(2n+1)C_{ n }\} =? $$
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$$ (n+1)2^{ n } $$
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$$ (n+1)2^{ n-1 } $$
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$$ (n-1)(n+2) $$
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none of these
Explanation
$$\textbf{Step 1: Find the value of given expression}$$
$$\text{Let X=} C_{ 0 }+3C_{ 1 }+5C_{ 2 }+...+(2n+1)C_{ n } $$
$$\Rightarrow X=(C_{ 0 }+C_{ 1 }+C_{ 2 }+...+C_{ n })+2(C_{ 1 }+2C_{ 2 }+3C_{ 3 }+...+nC_{ n }) $$
$$\text{In binomial expansion,sum of coefficient= } 2^n$$
$$\therefore (C_{ 0 }+C_{ 1 }+C_{ 2 }+...+C_{ n }) = 2^n$$ $$(1)$$
$$\Rightarrow X=(C_{ 0 }+C_{ 1 }+C_{ 2 }+...+C_{ n })+2(C_{ 1 }+2C_{ 2 }+3C_{ 3 }+...+nC_{ n }) $$
$$=2^n +2\times\bigg(n+2\dfrac{n(n-1)}{2}+3\dfrac{n(n-1)(n-2)}{3\times 2}+........+n.1\bigg)$$
$$\textbf{[From equation 1]}$$
$$=2^n +n\bigg(1+(n-1)+\dfrac{(n-1)(n-2)}{ 2}+........+1\bigg)$$
$$=2^n+2n[ ^{n-1}C_{ 0 }+^{n-1}C_{ 1 }+3C_{ 2 }+......+^{n-1}C_{ n-1 }] $$
$$=2^{ n }+(2n).2^{ n-1 }$$
$$[\because \mathbf{ C_{ 0 }+C_{ 1 }+C_{ 2 }+...+C_{ n-1 } = 2^{n-1}}]$$
$$=2^{ n }+(n).2^{ n-1+1 }$$
$$=2^{ n }+(n).2^{ n }$$
$$=(n+1).2^{ n } $$
$$\textbf{Hence, value of given expression is }$$
$$\mathbf{(n+1).2^{ n } }$$
The sum of coefficients of the two middle terms in the expansion of $$(1 + x)^{2n - 1}$$ is equal to
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$$^{2n -1}C_n$$
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$$^{2n -1}C_{n+1}$$
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$$^{2n}C_{n-1}$$
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$$^{2n}C_n$$
The cofficient of $$ x^{32} $$in the expansion of $$ ( x^4 - \frac {1}{x^3})^{15} $$is
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0%
273
0%
546
0%
1365
0%
1092
Explanation
$$\textbf{Step 1: Find value of general term for expansion}$$
$$\text{Given equation is}$$
$$\bigg ( x^4 -\dfrac {1}{x^3}\bigg)^{15} $$
$$\text{General term of expansion $ (p+q)^n$ is given by-}$$
$$T_n= T_{r +1 }=^nC_r p^{ (n-r)}.q^r$$
$$\text{Comparing given equation with standard form, p=x$^4$and q=-1/x$^3$ and n=15}$$
$$\Rightarrow T_n=T_{r+1}$$
$$ = ^{15}C_r (x^4)^{(15-r)} .\bigg (\dfrac {-1}{x^3}\bigg)^r $$
$$ = (-1)^r. ^{15}C_r. x^{(60-4r)} .\bigg (\dfrac {1}{x^3}\bigg)^r $$
$$=(-1)^r . ^{15}C_r x^{(60 -4r -3r)} $$
$$=(-1)^r . ^{15}C_r x^{(60 -7r)} $$
$$\textbf{Step 2: Find value of coefficient of term x}^{32}$$
$$\text{To obtain term of x}^{32} $$
$$\Rightarrow 60-7r=32 $$
$$\Rightarrow $$ $$7r=28$$
$$\therefore $$ $$r=4 $$
$$\text{For r=4,5th term is given by-}$$
$$ T_5=T_{4+1} = (-1)^4.^{15}C_4 x^{(60-7 \times 4)} = ^{15}C_4 x^{32} .$$
$$ \therefore $$ $$\text{Coefficient of } x^{32} $$$$=^{15}C_4 $$
$$= \dfrac { 15 \times 14 \times 13 \times 12 }{ 4 \times 3 \times 2\times 1} $$
$$ = 1365 $$
$$\textbf{Therefore,value of coefficient of term x$^{32}$ is 1365}$$
The number of terms in the expansion of $$\displaystyle{(\sqrt{3} + ^4 \sqrt{5})^{124}}$$ which are integers, is equal to
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0%
nil
0%
30
0%
31
0%
32
The coefficient of $$ x^3 $$ in the expansion of $$ (1+2x)^6(1-x)^7 $$ is
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0%
43
0%
-43
0%
63
0%
-63
Explanation
$$\textbf{Step 1: Expand the Expression }$$
$$\text{Binomial Theorem is Given by-}$$
$$(1+p)^n=1+^nC_{1}(p)^1+^nC_{2}(p)^2+^nC_{3}(p)^3+..............+^nC_{n}(p)^n$$
$$\text{For expansion of} (1+2x)^6:$$
$$\text{By Comparing with Formula p=2x and n=6}$$
$$(1+2x)^6 =\{ 1+^{ 6 }C_{ 1 }.2x+^{ 6 }C_{ 2 }.(2x)^{ 2 }+^{ 6 }C_{ 3 }.(2x)^{ 3 }+...\} $$
$$\text{For expansion of} (1-x)^7 :$$
$$\text{By Comparing with Formula p=x and n=7}$$
$$(1-x)^7 = \{ 1-^{ 7 }C_{ 1 }.x+^{ 7 }C_{ 2 }.x^{ 2 }-^{ 7 }C_{ 3 }.x^{ 3 }+..........\} $$
$$\textbf{Step 2: Find the coefficient of } \mathbf{ x^3}$$
$$ \text{Let P}= (1+2x)^6(1-x)^7 $$
$$\Rightarrow P =\{ 1+^{ 6 }C_{ 1 }.2x+^{ 6 }C_{ 2 }.(2x)^{ 2 }+^{ 6 }C_{ 3 }.(2x)^{ 3 }+...\} \times \{ 1-^{ 7 }C_{ 1 }.x+^{ 7 }C_{ 2 }.x^{ 2 }-^{ 7 }C_{ 3 }.x^{ 3 }+..\} $$
$$\therefore \text{Coefficient of }x^3 = -^{7}C_3 + ^{7}C_2 \times 2 ^{6}C_1 + 4\times ^{6}C_2 \times (- ^{7}C_1 )+8\times ^{7}C_3 $$
$$=1\times(-35)+(12\times21)+60 \times (-7) + (160 \times 1)$$
$$ = (-35 +252-420+160)$$
$$=-43$$
$$\textbf{Therefore, coefficient of } \mathbf{ x^3 } \textbf{is equal to -43}$$
The number of terms in the expansion of $$ ( 1 + \sqrt [3]{2}x)^9 + ( 1 - \sqrt [3]{2}x )^9 $$ is
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0%
5
0%
7
0%
9
0%
10
Explanation
$$\textbf{Step 1: Find number of terms of expansion}$$
$$\text{Given equation is -}$$
$$( 1 + \sqrt [3]{2x})^9 + ( 1 + \sqrt [3]{2x} )^9 $$
$$\text{We know that for expansion of (x + y)$^n$ + (x – y)$^n$}$$
$$\text{Comparing with standard equation,n=9}$$
$$\Rightarrow \text{ If n is even then, Number of terms = }$$ $$\dfrac{n+1}{2} $$
$$\therefore \text{Numer of terms=}$$
$$\dfrac{9+1}{2} $$
$$=\dfrac{10}{2}$$
$$=5 $$
$$\textbf{Therefore,the expansion of}$$
$$ \mathbf{ ( 1 + \sqrt [3]{2x})^9 + ( 1 + \sqrt [3]{2x} )^9 }$$
$$\textbf{has 5 terms.}$$
If p and q are positive integers, then the coefficients of $$ x^p $$ and $$ x^q $$ in the expansion of $$ ( 1+ x)^{p+q} $$ are
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0%
euqal
0%
equal with opposite signs
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reciprocal to each other
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none of these
Explanation
$$\textbf{Step 1: Find value of (p+1)th and (q+1)th term from expansion}$$
$$\text{General term of expansion $ (1+x)^n$ is given by-}$$
$$T_n= T_{r +1 }=^nC_r x^{ (n-r)}$$
$$\text{Comparing given equation with standard form, n=p+q}$$
$$\Rightarrow$$
$$ T_{p+1} = ^{(p+q)}C_p . x^{p+1-1} $$
$$ = ^{(p+q)}C_p . x^{p} $$
$$\therefore \text{Coefficient of x$^{p}$ is $ = ^{(p+q)}C_p $}$$
$$(1)$$
$$\Rightarrow T_{q+1} = ^{(p+q)}C_q. x^{q+1-1} $$
$$ = ^{(p+q)}C_p . x^{q} $$
$$\therefore \text{Coefficient of x$^{q}$ is $ = ^{(p+q)}C_q $}$$
$$(2)$$
$$\Rightarrow ^{(p+q)}C_p = ^{(p+q)}C_{(p+q)-p}$$
$$ \mathbf{[^{n}C_r = ^{n}C_{n-r}]}$$
$$=^{(p+q)}C_q $$
$$\therefore \text{Coefficient of x$^p$ and x$^q$ are same}$$
$$\textbf{[From equation (1) and (2)]}$$
$$\textbf{Therefore, coefficient of both terms are equal.}$$
The total number of terms in the expansion of $$ (a + y)^{100 } + ( a -y)^{100 } $$ is
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0%
50
0%
51
0%
101
0%
102
Explanation
$$\textbf{Step 1: Find number of terms of expansion}$$
$$\text{Given equation is -}$$
$$ ({a} + {y})^{100} +( {a} - {y})^{100} $$
$$\text{We know that for expansion of (x + y)$^n$ + (x – y)$^n -$ $\\$$\Rightarrow$ If n is even then, Number of terms = $\dfrac{n}{2} $+ 1}$$
$$\text{Comparing with standard equation,n=100}$$
$$\therefore \text{Numer of terms=}$$
$$\dfrac{100}{2} +1$$
$$=50+1$$
$$=51 $$
$$\textbf{Therefore,the expansion of $\mathbf{ ({a} + {y})^{100} +( {a} - {y})^{100} }$ has 51 terms.}$$
If the coefficients of the second, third and fourth terms in the expansion of $$ (1+x)^{2n} $$ are in AP , then
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$$ n^2-3n+8=0 $$
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$$ 2n^2 - 7n +5 = 0 $$
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$$ 2n^2-9n+7 =0 $$
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none of these
Explanation
$$\textbf{Step 1: Expand the Expression }$$
$$\text{Binomial Theorem is Given by-}$$
$$(1+x)^p=1+C_{1}^p(x)^1+C_{2}^p(x)^2+C_{3}^p(x)^3+..............+C_{p}^p(x)^p$$
$$\text{By Comparing with Formula p=2n}$$
$$\Rightarrow (1+x)^{-2n}$$
$$=C_{1}^{2n}(x)^1+C_{2}^{2n}(x)^2+C_{3}^{2n}(x)^3+..............+C_{2n}^{2n}(x)^{2n}$$
$$\textbf{Step 2: Find value of 2nd, 3rd and 4th term from expansion}$$
$$\text{General term of expansion$(1+x)^{2n}$ is given by-}$$
$$T_n= T_{r +1 }=^{2n}C_r x^{ r}$$
$$ \Rightarrow T_2=T_{1+1}=^{2n}C_1x^1 $$
$$\therefore \text{Coefficient of}$$
$$T_2 =^{2n}C_1x$$
$$=2n$$
$$\Rightarrow T_3=T_{2+1}=^{2n}C_2x^2 $$
$$\therefore \text{Coefficient of}$$
$$T_3$$
$$=^{2n}C_2x^2 $$
$$=\dfrac { 2n(2n-1) }{ 2 }$$
$$ \Rightarrow T_4=T_{3+1} =^{2n}C_3x^3 $$
$$\therefore \text{Coefficient of}$$
$$T_4$$
$$=^{2n}C_2x^3$$
$$=\dfrac { 2n(2n-1)(2n-2) }{ 6 }$$
$$\textbf{Step 3: Find condition for three terms to be in AP}$$
$$\text{Given-Coefficient of $T_2,T_3$ and $T_4$ are in AP }$$
$$\text{Condition for three terms a,b and c to be in AP- 2b=a+c }$$
$$\therefore 2 T_3=T_2 + T_4 $$
$$2\bigg(\dfrac { 2n(2n-1) }{ 2 }\bigg) =2n+\dfrac { 2n(2n-1)(2n-2) }{ 6 } $$
$$\Rightarrow$$
$$3(2n-1)=3+2n^2-3n+1$$
$$\Rightarrow$$
$$6n-3=4+2n^2-3n$$
$$\therefore $$
$$ 2n^2 - 9n+7=0 $$.
$$\textbf{Therefore,option c is the correct answer}$$
The term independent of x in the expansion of $$ (x -\frac {1}{x})^{12} $$ is
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0%
924
0%
462
0%
231
0%
693
Explanation
$$\textbf{Step 1: Find value of general term}$$
$$\text{Given equation is}$$
$$\bigg ( x -\dfrac {1}{x}\bigg)^{12} $$
$$\text{General term of expansion $ (p+q)^n$ is given by-}$$
$$T_n= T_{r +1 }=^nC_r p^{ (n-r)}.q^r$$
$$\text{Comparing given equation with standard form, p=x and q=-1/x and n=12}$$
$$\Rightarrow T_n=T_{r+1} = ^{12}C_r. (x)^{(12-r)} .\bigg( \dfrac {-1}{x}\bigg)^r $$
$$= (-1)^r. ^{12}C_r. (x)^{(12-r)} .x^{-r} $$
$$= (-1)^r. ^{12}C_r. (x)^{(12-r-r)} $$
$$= (-1)^r. ^{12}C_r. (x)^{(12-2r)} $$
$$\textbf{Step 2: Find value of coefficient of term independent of x}$$
$$\text{To obtain term independent of x-:Coefficient of x will be 0} $$
$$\Rightarrow 12-2r=0 $$
$$\Rightarrow $$ $$2r=12 $$
$$\therefore $$ $$r=6 $$
$$\text{For r=6,7th term is given by-}$$
$$ T_7=T_{6+1} = (-1)^6.^{12}C_6 x^{(12-2x6)} = ^{12}C_6 x^0 .$$
$$ \therefore $$ $$\text{Coefficient of } x^0 $$$$=^{12}C_6 $$
$$= \dfrac { 12\times11\times10 \times 9 \times 8 \times 7}{6\times5 \times 4 \times 3 \times 2 \times 1} $$
$$ = 924 $$
$$\textbf{Therefore,value of coefficient of term independent of x will be 924}$$
The sum of possible values of $$x$$ is
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$$1$$
0%
$$3$$
0%
$$4$$
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$$None\ of \these$$
The middle term in the expansion of $$(x/2+2)^{8}$$ is $$1120$$; then $$x\in R$$ is equal to
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$$-2$$
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$$3$$
0%
$$-3$$
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$$2$$
If the coefficients of $$r^{th}$$ and $$(r+1)^{th}$$ terms in the expansion of $$(3+7x)^{29}$$ are equal, then $$r$$ equals
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0%
$$15$$
0%
$$21$$
0%
$$14$$
0%
none of these
Explanation
Given expansion is
$$(3+7x)^{29}$$
General term
$$T_{r+1}=^{29}C_r.3^{29-r}.(7x)^{r}$$
For $$r^{th}$$ term
Let put $$r=r-1$$
$$\implies T_{r-1+1}=^{29}C_{r-1}3^{29-r+1}.(7x)^{r-1}~~~~-(1)$$
AS per question coefficient are equal
$$\implies ^{29}C_r.3^{29-r}.(7)^{r}=^{29}C_{r-1}.3^{30-r}.(7)^{r-1}$$
On solving the above expression we get
$$\dfrac{29-r+1}{r}\times7=3$$
$$\implies 210-7r=3r\\\implies 10r=210\\\implies r=21$$
The middle term in the expansion of $$(a x)^n$$
is
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0%
$$56a^3 x^5$$
0%
$$-56a^3 x^5$$
0%
$$70a^4 x^4$$
0%
$$-70a^4 x^4$$
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