Binomial Theorem Test

Result

Binomial Theorem Test - 32

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Number of terms free from radical sign in the expansion of $$\displaystyle \left ( 1+3^{1/3}+7^{1/2} \right )^{10}$$ is

A
4

B
5

C
6

D
8

Solution

$$(1+(3^{\frac{1}{3}}+7^{\frac{1}{2}}))^{10}$$
Let
$$(3^{\frac{1}{3}}+7^{\frac{1}{2}})=x$$
Then the above expression reduces to,
$$(1+x)^{10}$$
Now writing the general term, we get
$$T_{r+1}=\:^{10}C_{r}x^{r}$$
$$=\:^{10}C_{r}(3^{\frac{1}{3}}+7^{\frac{1}{2}})^{r}$$.
Hence we get one rational term for each
$$r=0,2,3,4,6,8,9,10$$
Hence in total 8 terms.
Question 2
1 / -0

If $${C}_{r}$$ stands for $$^{n}{C}_{r}$$, then the sum of the series $$\displaystyle\frac { 2\left( \frac { n }{ 2 } \right) !\left( \frac { n }{ 2 } \right) ! }{ n! } \left[ { C }_{ 0 }^{ 2 }-2{ C }_{ 1 }^{ 2 }+3{ C }_{ 2 }^{ 2 }-...+{ \left( -1 \right) }^{ n }\left( n+1 \right) { C }_{ n }^{ 2 } \right] $$ where $$n$$ is an even positive integer, is

A
$$0$$

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

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

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

Solution

$$\displaystyle{ C }_{ 0 }^{ 2 }-2{ C }_{ 1 }^{ 2 }+3{ C }_{ 2 }^{ 2 }-4{ C }_{ 3 }^{ 2 }+...+{ \left( -1 \right) }^{ n }\left( n+1 \right) { C }_{ n }^{ 2 }$$
$$\displaystyle=\left[ { C }_{ 0 }^{ 2 }-{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }-{ C }_{ 3 }^{ 2 }+...+{ \left( -1 \right) }^{ n }{ C }_{ n }^{ 2 } \right] -\left[ { C }_{ 1 }^{ 2 }-2{ C }_{ 2 }^{ 2 }+3{ C }_{ 3 }^{ 2 }-...+{ \left( -1 \right) }^{ n }n{ C }_{ n }^{ 2 } \right] $$
$$\displaystyle={ \left( -1 \right) }^{ \frac { n }{ 2 } }\frac { n! }{ \left( \frac { n }{ 2 } \right) !\left( \frac { n }{ 2 } \right) ! } -{ \left( -1 \right) }^{ \frac { n }{ 2 } -1 }\frac { n }{ 2 } .^{ n }{ { C }_{ \frac { n }{ 2 } } }$$
$$\displaystyle={ \left( -1 \right) }^{ \frac { n }{ 2 } }\frac { n! }{ \left( \frac { n }{ 2 } \right) !\left( \frac { n }{ 2 } \right) ! } \left( 1+\frac { n }{ 2 } \right) $$
$$\displaystyle\therefore 2\left( \frac { n }{ 2 } \right) !\left( \frac { n }{ 2 } \right) !\left[ { C }_{ 0 }^{ 2 }-2{ C }_{ 1 }^{ 2 }+3{ C }_{ 2 }^{ 2 }-...+{ \left( -1 \right) }^{ n }\left( n+2 \right) { C }_{ n }^{ 2 } \right] $$
$$\displaystyle={ \left( -1 \right) }^{ \frac { n }{ 2 } }\left( n+2 \right) $$
Question 3
1 / -0

$$(r + 1)^{th}$$ term in the expansion of $$(x + a)^n$$ will be

A
$$^nC_rx^na^{n-r}$$

B
$$^nC_rx^{n-r}a^{r}$$

C
$$^nC_rx^{n-r}a^{n}$$

D
$$^nC_rx^ra^{n-r}$$

Solution

$$\\ { (x+a) }^{ n }={ x }^{ n }+{ C }_{ 1 }^{ n }{ x }^{ n-1 }a+...+{ C }_{ r }^{ n }{ x }^{ n-r }{ a }^{ r }+...{ a }^{ n }\\ \quad the\quad (r+1)th\quad term\quad is\quad { C }_{ r }^{ n }{ x }^{ n-r }{ a }^{ r }.\quad Hence\quad option\quad (B)\\ $$
Question 4
1 / -0

If coefficients of $$x^n$$ in $$(1+x)^{101}(1-x+x^2)^{100}$$ is non-zero then $$n$$ cannot be of the form

A
$$3t+1$$

B
$$3t$$

C
$$3t+2$$

D
$$4t+1$$

Solution

Given expression is $$(1+x)^{101} (1-x+x^{2})^{100}$$ , Let it be $$y$$
$$y = (1+x)((1+x)(1-x+x^{2}))^{100}=(1+x)(1+x^{3})^{100}$$
We know that $$(1+x^{3})^{100}=1+ax^{3}+bx^{6}+cx^{9}+....$$
$$\Rightarrow y= (1+x)(1+x^{3})^{100}=1+x+ax^{3}+ax^{4}+bx^{6}+bx^{7}+....$$
We can observe that $$x^{3t+2} $$ terms are missing , which means the coefficients of $$x^{3t+2}$$ terms are $$0$$
Therefore for the coefficient of $$x^{n}$$ not to be zero , $$n$$ cannot be $$3t+2$$
Question 5
1 / -0

$$\displaystyle \binom{n}{o}+\binom{n}{1}+\binom{n}{2}+.........+\binom{n}{n}=$$

A
$$\displaystyle 2^{n-1}$$

B
$$\displaystyle ^{2n}C_{n}$$

C
$$\displaystyle 2^n$$

D
$$\displaystyle 2^{n+1}$$

Solution

$$(1+x)^{n}=\:^{n}C_{0}+\:^{n}C_{1}x+\:^{n}C_{2}x^{2}+...\:^{n}C_{n}x^{n}$$
Substituting x=1, we get
$$2^{n}=\:^{n}C_{0}+\:^{n}C_{1}+\:^{n}C_{2}+...\:^{n}C_{n}$$
Question 6
1 / -0

In the expansion of $$(1 + x)^n$$, the sum of coefficients of odd powers of $$x$$ is

A
$$2^n+1$$

B
$$2^n-1$$

C
$$2^n$$

D
$$2^{n-1}$$

Solution

$$\left( 1+x \right) ^{ n }={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }+...+{ C }_{ n }{ x }^{ n }$$
Putting $$x=1$$ and $$x=1$$ and subtracting, we get.
$${ 2 }^{ n }=2\left( { C }_{ 1 }+{ C }_{ 3 }+{ C }_{ 5 }+... \right) $$
$$\therefore { C }_{ 1 }+{ C }_{ 3 }+{ C }_{ 5 }+...={ 2 }^{ n-1 }$$
Or the sum of the coefficients of the odd power of $$x$$ is $${2}^{n-1}$$
Question 7
1 / -0

If $$(1+x)^n=C_0+C_1x+C_2x^2+.....+C_nx^2$$, then the value of $$C_0+C_2+C_4+ .....$$ is

A
$$2^{n-1}$$

B
$$2^n-1$$

C
$$2^n$$

D
$$2^{n-1}-1$$

Solution

$$(1+x)^{n}=\:^{n}C_{0}+\:^{n}C_{1}x+\:^{n}C_{2}x^{2}+...\:^{n}C_{n}x^{n}$$
Let $$x=1$$
$$2^{n}=\:^{n}C_{0}+\:^{n}C_{1}+\:^{n}C_{2}+...\:^{n}C_{n}$$ ...(i)
Let $$x=-1$$
$$0=\:^{n}C_{0}-\:^{n}C_{1}+\:^{n}C_{2}+...(-1)^{n}\:^{n}C_{n}$$ ...(ii)
Adding both gives us
$$2^{n}=2[\:^{n}C_{0}+\:^{n}C_{2}+\:^{n}C_{4}+...]$$
$$2^{n=1}=\:^{n}C_{0}+\:^{n}C_{2}+\:^{n}C_{4}+...$$
Hence the answer is $$2^{n-1}$$
Question 8
1 / -0

$$^{10}C_1+^{10}C_3+^{10}C_5+^{10}C_7+^{10}C_9=$$

A
$$2^9$$

B
$$2^{10}$$

C
$$2^{10}-1$$

D
None of these

Solution

For $$(X+A)^{ n }={ { C }_{ 0 }^{ n }X }^{ n }+\cdots{ C }_{ r }^{ n }{ X }^{ n-r }{ A }^{ r }+\cdots{ C }_{ n }^{ n }A^{ n }$$
Put $$X=1; A=-1$$
Hence, $$0= { C }_{ 0 }^{ n }-{ C }_{ 1 }^{ n }+{ C }_{ 2 }^{ n }-{ C }_{ 3 }^{ n }\cdots$$ and so on
$$\Rightarrow{ C }_{ 0 }^{ n }+{ C }_{ 2 }^{ n }{ +C }_{ 4 }^{ n }+\cdots={ C }_{ 1 }^{ n }+{ C }_{ 3 }^{ n }+{ C }_{ 5 }^{ n }+\cdots $$ ...(1)
Now put $$A=1$$ and we see that
$${ 2 }^{ n }={ C }_{ 0 }^{ n }+{ C }_{ 1 }^{ n }+{ C }_{ 2 }^{ n }{ +{ C }_{ 3 }^{ n }+C }_{ 4 }^{ n }+{ C }_{ 5 }^{ n }+\cdots$$ ...(2)
Hence, the sum of
$${ C }_{ 1 }^{ n }+{ C }_{ 3 }^{ n }+{ C }_{ 5 }^{ n }+\cdots={ 2 }^{ n-1 }$$ (From (1) and (2))
In the problem, $$n=10$$
Hence, answer is $${ 2 }^{ 9 }$$
Question 9
1 / -0

The value of $$3{\;}^nC_0-8{\;}^nC_1+13 {\;}^nC_2-18 {\;}^nC_3+.....+n$$ terms is

A
$$0$$

B
$$3^n$$

C
$$5^n$$

D
none of these

Solution

The general term of the series is
$$Tr = (1)^r (3 + 5r) ^nC_r, r = 0, 1, 2, ......, n$$
$$\therefore$$ Sum of the series is given by
$$\displaystyle S=\sum_{r=0}^n(-1)^r(3+5r)^nC_r$$
$$=3\displaystyle \sum_{r=0}^n(-1)^r {\;}^nC_r+5\displaystyle \sum_{r=0}^n(-1)^r r^nC_r$$
$$=3(\displaystyle \sum_{r=0}^n(-1)^r {\;}^nC_r)+5(\displaystyle \sum_{r=1}^n(-1)^r r.\frac {n}{r} ^{n-1}C_{r-1})$$
$$=3(\displaystyle \sum_{r=0}^n(-1)^r {\;}^nC_r)+5n (\displaystyle \sum_{r=1}^n(-1)^r {\;}^{n-1}C_{r-1})$$
$$=3(0) + 5n(0) = 0$$
Question 10
1 / -0

Find the middle terms in the expansion of $$\displaystyle \left ( 2x^{2}-\frac{1}{x} \right )^{7}$$

A
$$\displaystyle -560x^{5},\:280x^{2}$$

B
$$\displaystyle -280x^{5},\:560x^{2}$$

C
$$\displaystyle 560x^{5},\:-280x^{2}$$

D
$$\displaystyle 280x^{5},\:-560x^{2}$$

Solution

The middle terms will be $$T_{4}$$ and $$T_{5}$$
Now, $$T_{4}$$
$$=T_{3+1}$$
$$=-\:^{7}C_{3}.2^{4}.x^{8-3}$$
$$=-35.(16).x^{5}$$
$$=-560.x^{5}$$
$$T_{5}$$
$$=T_{4+1}$$
$$=\:^{7}C_{4}.2^{3}.x^{6-4}$$
$$=35.(8).x^{2}$$
$$=280.x^{2}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 32

Result

Binomial Theorem Test - 32

Score

-

Rank

-

SHARING IS CARING

Question 1

4

5

6

8

Solution

Question 2

$$0$$

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

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

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

Solution

Question 3

$$^nC_rx^na^{n-r}$$

$$^nC_rx^{n-r}a^{r}$$

$$^nC_rx^{n-r}a^{n}$$

$$^nC_rx^ra^{n-r}$$

Solution

Question 4

$$3t+1$$

$$3t$$

$$3t+2$$

$$4t+1$$

Solution

Question 5

$$\displaystyle 2^{n-1}$$

$$\displaystyle ^{2n}C_{n}$$

$$\displaystyle 2^n$$

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

Solution

Question 6

$$2^n+1$$

$$2^n-1$$

$$2^n$$

$$2^{n-1}$$

Solution

Question 7

$$2^{n-1}$$

$$2^n-1$$

$$2^n$$

$$2^{n-1}-1$$

Solution

Question 8

$$2^9$$

$$2^{10}$$

$$2^{10}-1$$

None of these

Solution

Question 9

$$0$$

$$3^n$$

$$5^n$$

none of these

Solution

Question 10

$$\displaystyle -560x^{5},\:280x^{2}$$

$$\displaystyle -280x^{5},\:560x^{2}$$

$$\displaystyle 560x^{5},\:-280x^{2}$$

$$\displaystyle 280x^{5},\:-560x^{2}$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.