Binomial Theorem Test

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Binomial Theorem Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

If $$x+y= 1$$ then $$\displaystyle\sum_{r= 0}^{n} r^{n}C_{r}x^{r}y^{n-r}$$ equals

A
$$1$$

B
$$n$$

C
$$nx$$

D
$$ny$$

Solution

Expanding, we get
$$\:^{n}C_{1}xy^{n-1}+2\:^{n}C_{2}x^{2}y^{n-2}+3\:^{n}C_{3}x^{3}y^{n-3}+...n\:^{n}C_{n}x^{n}$$
$$=x[\:^{n}C_{1}y^{n-1}+2\:^{n}C_{2}x^{1}y^{n-2}+3\:^{n}C_{3}x^{2}y^{n-3}+...n\:^{n}C_{n}x^{n-1}]$$
$$=x[\dfrac{d(y+x)^{n}}{dx}]$$ ...(keeping $$y$$ constant.)
$$=x(n(x+y)^{n-1})$$
Now $$x+y=1$$,
Hence we get
$$x(n)$$
$$=nx$$
Question 2
1 / -0

If $$A$$ is the sum of the odd terms and $$B$$ the sum of even terms in the expansion of $${ \left( x+a \right) }^{ n }$$, then $${ A }^{ 2 }-{ B }^{ 2 }=$$

A
$${ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ n }$$

B
$${ \left( { x }^{ 2 }-{ a }^{ 2 } \right) }^{ n }$$

C
$$2{ \left( { x }^{ 2 }-{ a }^{ 2 } \right) }^{ n }$$

D
None of these

Solution

We have,
$${ \left( x+a \right) }^{ n }=^{ n }{ { C }_{ 0 }^{ } }{ x }^{ n }+^{ n }{ { C }_{ 1 }^{ } }{ x }^{ n-1 }{ a }^{ 1 }+^{ n }{ { C }_{ 2 }^{ } }{ x }^{ n-2 }{ a }^{ 2 }+^{ n }{ { C }_{ 3 }^{ } }{ x }^{ n-3 }{ a }^{ 3 }+...+^{ n }{ { C }_{ n }^{ } }{ a }^{ n }\\ =\left( ^{ n }{ { C }_{ 0 }^{ } }{ x }^{ n }+^{ n }{ { C }_{ 2 }^{ }{ x }^{ n-2 }{ a }^{ 2 } }+... \right) +\left( ^{ n }{ { C }_{ 1 }^{ } }{ x }^{ n-1 }{ a }^{ 1 }+^{ n }{ { C }_{ 3 }^{ } }{ x }^{ n-3 }{ a }^{ 3 }+... \right) \\ =A+B\\ { \left( x-a \right) }^{ n }=^{ n }{ { C }_{ 0 }^{ } }{ x }^{ n }-^{ n }{ { C }_{ 1 }^{ } }{ x }^{ n-1 }{ a }^{ 1 }+^{ n }{ { C }_{ 2 }^{ } }{ x }^{ n-2 }{ a }^{ 2 }-^{ n }{ { C }_{ 3 }^{ } }{ x }^{ n-3 }{ a }^{ 3 }\\ =\left( ^{ n }{ { C }_{ 0 }^{ } }{ x }^{ n }+^{ n }{ { C }_{ 2 }^{ }{ x }^{ n-2 }{ a }^{ 2 } }+... \right) -\left( ^{ n }{ { C }_{ 1 }^{ } }{ x }^{ n-1 }{ a }^{ 1 }+^{ n }{ { C }_{ 3 }^{ } }{ x }^{ n-3 }{ a }^{ 3 }+... \right) \\ =A-B\\ \therefore { A }^{ 2 }-{ B }^{ 2 }=\left( A+B \right) \left( A-B \right) ={ \left( x+a \right) }^{ n }{ \left( x-a \right) }^{ n }={ \left( { x }^{ 2 }-{ a }^{ 2 } \right) }^{ n }$$
Question 3
1 / -0

If $$\displaystyle \left ( 1+x+x^{2} \right )^{n}=\sum_{r=0}^{2n} a_{r}x^{r}=a_{0}+a_{1}x+a_{2}x^{2}+...+a^{2n}x^{2n}$$ and
$$ \displaystyle P=a_{0}+a_{3}+a_{6}+...$$
$$ \displaystyle Q=a_{1}+a_{4}+a_{7}+...$$
$$ \displaystyle R=a_{2}+a_{5}+a_{8}+...$$
then the set of values of $$ P, Q, R $$ are respectively equals

A
$$\displaystyle (1 ,1, 1)$$

B
$$\displaystyle (3^{n},3^{n},3^{n})$$

C
$$\displaystyle (3^{n+1},3^{n+1},3^{n+1})$$

D
$$\displaystyle (3^{n-1},3^{n-1},3^{n-1})$$

Solution

Sum of all coefficients will be $$3^{n}$$
$$=3(3^{n-1})$$
Now
The sum of coefficients of $$a_{0}+a_{3}+a_{6}....$$
will be equal to $$a_{1}+a_{4}+a_{7}+....$$ which will be further equal to
$$a_{2}+a_{5}+a_{8}+...$$
$$=$$sum of all coefficients/3
$$=\dfrac{3(3^{n-1})}{3}=3^{n-1}$$
Question 4
1 / -0

The evaluated value of $$\displaystyle \sum_{i=0}^{n} \sum_{j=1}^{n}$$ $$\displaystyle ^{n}C_{j}$$ $$\displaystyle ^{j}C_{i},$$ $$\displaystyle i\leq j$$

A
$$\displaystyle 3^{n}+1$$

B
$$\displaystyle 3^{n}-1$$

C
$$\displaystyle 3^{n+1}+1$$

D
None of these

Solution

Expanding, we get
$$\:^{n}C_{1}(\:^{1}C_{0}+\:^{1}C_{1})+\:^{n}C_{2}(\:^{2}C_{0}+\:^{2}C_{1}+\:^{2}C_{2})+....\:^{n}C_{n}(\:^{n}C_{1}+\:^{n}C_{2}+...\:^{n}C_{n})$$
$$=\:^{n}C_{1}2^{1}+\:^{n}C_{2}2^{2}+...\:^{n}C_{n}2^{n}$$
$$=\:^{n}C_{0}2^{0}+\:^{n}C_{1}2^{1}+\:^{n}C_{2}2^{2}+...\:^{n}C_{n}2^{n}-[\:^{n}C_{0}2^{0}]$$
$$=(1+2)^{n}-1$$
$$=3^{n}-1$$
Question 5
1 / -0

If $$C_{0},C_{1},C_{2}....,C_{n}$$ are Binomial Coefficients, such that $$\displaystyle S_{n}=\sum_{r=0}^{n}\frac{1}{{C_{r}}^{n}}$$ and $$\displaystyle t_{n}= \sum_{r= 0}^{n}\frac{r}{{C_{r}}^{n}}$$ then $$\displaystyle \frac{t_{n}}{s_{n}}$$ equals

A
$$\displaystyle \frac{n}{2}$$

B
$$\displaystyle\frac{n\left ( n+1 \right )}{2}$$

C
$$\displaystyle\frac{n+1}{2}$$

D
None of these

Solution

$$\displaystyle S_{ n }=\sum _{ r=0 }^{ n }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } =\sum _{ r=0 }^{ \frac { n }{ 2+1 } }{ \frac { 1 }{ ^{ n }{ C }_{ r } } +\frac { 1 }{ ^{ n }{ { C }_{ n-r } } } } $$
$$\displaystyle \Rightarrow { S }_{ n }=\sum _{ r=0 }^{ \frac { n+1 }{ 2 } }{ \frac { 2 }{ ^{ n }{ C_{ r } } } =2 } \sum _{ r=0 }^{ \frac { n+1 }{ 2 } }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } $$
$$\displaystyle \therefore { t }_{ n }=\sum _{ r=0 }^{ n }{ \frac { r }{ ^{ n }{ C }_{ r } } } =\sum _{ r=0 }^{ \frac { n+1 }{ 2 } }{ \frac { r+\left( n-r \right) }{ ^{ n }{ { C }_{ r } } } } $$
$$\displaystyle \Rightarrow n\sum _{ r=0 }^{ n }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } =n\left( \frac { { S }_{ n } }{ 2 } \right) =\frac { n }{ 2 } { S }_{ n }$$

Thus $$\dfrac{t_n}{S_n} = \dfrac{n}{2}$$
Question 6
1 / -0

The sum of the coefficients of all odd exponets of $$\displaystyle x$$ in the product of $$\displaystyle (1-x +x^{2}-x^{3}+x^{4}+...-x^{49}+x^{50})\times(1+x+x^{2}+x^{3}+...+x^{50})$$ equals

A
$$1$$

B
$$0$$

C
$$-1$$

D
None of these

Solution

Coefficient of $$x^{1}$$ will be
$$1+(-1)$$
$$=0$$ ..(i)
Coefficient of $$x^{3}$$ will be
$$1+(-1)+1+(-1)$$
$$=0$$
Hence the coefficients of $$x^{n}$$ where n is odd is zero.
Question 7
1 / -0

If$$\displaystyle\left ( 1+x \right )^{n}= \sum_{r= 0}^{n}C_{r}x^{r}$$, then the value of $$C_{0}-C_{2}+C_{4}-C_{6}+C_{8}-C_{10}+...$$ equals

A
$$\displaystyle2^{\tfrac{n}{2}}\cos \frac{nx}{4}$$

B
$$C_{1}-C_{3}+C_{5}-C_{7}+....$$

C
$$C_{0}+C_{4}+C_{8}+C_{12}+....$$

D
$$\displaystyle2^{\tfrac{n}{2}}\sin \frac{nx}{4}$$

Solution

$$(1+x)^{n}=\:^{n}C_{0}+\:^{n}C_{1}x+\:^{n}C_{2}x^{2}+...\:^{n}C_{n}x^{n}$$
And
$$(1-x)^{n}=\:^{n}C_{0}-\:^{n}C_{1}x+\:^{n}C_{2}x^{2}+...(-1)^{n}\:^{n}C_{n}x^{n}$$
Adding, we get
$$(1+x)^{n}+(1-x)^{n}=2[\:^{n}C_{0}+\:^{n}C_{2}x^{2}+\:^{n}C_{4}x^{4}+\:^{n}C_{6}x^{6}+...]$$
Substituting, $$x=\sqrt{-1}=i$$ we get
$$(1+i)^{n}+(1-i)^{n}=2[\:^{n}C_{0}-\:^{n}C_{2}+\:^{n}C_{4}x^{4}-\:^{n}C_{6}x^{6}+...]$$
$$2^{\dfrac{n}{2}-1}[e^{\dfrac{in\pi}{4}}+e^{\dfrac{-in\pi}{4}}]=\:^{n}C_{0}-\:^{n}C_{2}+\:^{n}C_{4}x^{4}-\:^{n}C_{6}x^{6}+...$$
Hence
Solving the LHS, we get $$2^{\frac{n}{2}}\cos(\displaystyle \dfrac{n\pi}{4})$$
Question 8
1 / -0

If $$\left ( 1+x \right )^{n}= \sum_{r= 0}^{n}C_{r}x^{r}$$ then the value of $$3C_{1}+7C_{2}+11C^{3}+....+\left ( 4n-1 \right )C_{n}$$ is

A
$$\left ( 4n-1 \right )2^{n}$$

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

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

D
$$\left ( 4n-1 \right )2^{n}-1$$

Solution

Let $$n=1$$, we get
$$S_{1}=3\:^{1}C_{1}$$
$$=3$$
$$=(2(1)-1)2^{1}+1$$
Let $$n=2$$ , we get
$$S_{2}=3\:^{2}C_{1}+7\:^{2}C_{2}$$
$$=6+7$$
$$=13$$
$$=(2(2)-1)2^{2}+1$$
Let $$n=3$$, we get
$$S_{3}=3\:^{3}C_{1}+7\:^{3}C_{2}+11\:^{3}C_{3}$$
$$=9+21+11$$
$$=41$$
$$=(2(3)-1)2^{3}+1$$
Hence
$$S_{n}$$$$=(2(n)-1)2^{n}+1$$
Question 9
1 / -0

The Coefficient of $$x^{53}$$ in $$\sum_{m= 0}^{100}{C_{m}}^{100}\left ( x-3 \right )^{100-m}2^{m}$$ is

A
$$=-\:^{100}C_{53}$$

B
$$=-\:^{101}C_{53}$$

C
$$=\:^{101}C_{47}$$

D
$$=-\:^{100}C_{47}$$

Solution

The general term is
$$G$$
$$=\:^{100}C_{m}(x-3)^{100-m}2^{m}$$
This is the general term of
$$((x-3)+2)^{100}$$
$$=(x-1)^{100}$$
$$=(1-x)^{100}$$
Hence the coefficient of $$x^{53}$$ is
$$T_{54}$$
$$=(-1)^{53}\:^{100}C_{53}$$
$$=-\:^{100}C_{53}$$
Question 10
1 / -0

The number of terms with integral coefficient in the expansion of $$\left ( 17^{\dfrac 13} +35^{\dfrac 12}\right )^{300}$$ is

A
$$50$$

B
$$100$$

C
$$150$$

D
$$51$$

Solution

The number of rational terms will be
$$1+\dfrac{300}{L.C.M(3,2)}$$

$$=1+\dfrac{300}{6}$$

$$=1+50=51$$ rational terms.

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Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 27

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Binomial Theorem Test - 27

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SHARING IS CARING

Question 1

$$1$$

$$n$$

$$nx$$

$$ny$$

Solution

Question 2

$${ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ n }$$

$${ \left( { x }^{ 2 }-{ a }^{ 2 } \right) }^{ n }$$

$$2{ \left( { x }^{ 2 }-{ a }^{ 2 } \right) }^{ n }$$

None of these

Solution

Question 3

$$\displaystyle (1 ,1, 1)$$

$$\displaystyle (3^{n},3^{n},3^{n})$$

$$\displaystyle (3^{n+1},3^{n+1},3^{n+1})$$

$$\displaystyle (3^{n-1},3^{n-1},3^{n-1})$$

Solution

Question 4

$$\displaystyle 3^{n}+1$$

$$\displaystyle 3^{n}-1$$

$$\displaystyle 3^{n+1}+1$$

None of these

Solution

Question 5

$$\displaystyle \frac{n}{2}$$

$$\displaystyle\frac{n\left ( n+1 \right )}{2}$$

$$\displaystyle\frac{n+1}{2}$$

None of these

Solution

Question 6

$$1$$

$$0$$

$$-1$$

None of these

Solution

Question 7

$$\displaystyle2^{\tfrac{n}{2}}\cos \frac{nx}{4}$$

$$C_{1}-C_{3}+C_{5}-C_{7}+....$$

$$C_{0}+C_{4}+C_{8}+C_{12}+....$$

$$\displaystyle2^{\tfrac{n}{2}}\sin \frac{nx}{4}$$

Solution

Question 8

$$\left ( 4n-1 \right )2^{n}$$

$$\left ( 2n-1 \right )2^{n}$$

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

$$\left ( 4n-1 \right )2^{n}-1$$

Solution

Question 9

$$=-\:^{100}C_{53}$$

$$=-\:^{101}C_{53}$$

$$=\:^{101}C_{47}$$

$$=-\:^{100}C_{47}$$

Solution

Question 10

$$50$$

$$100$$

$$150$$

$$51$$

Solution

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