Binomial Theorem Test

Result

Binomial Theorem Test - 26

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

If the third term in the expansion of $$\displaystyle (\frac{1}{x}+x^{\log_{10}x})^{5} $$ is $$1,000$$, then $$x$$-equals

A
10$$\displaystyle ^{2}$$

B
10$$\displaystyle ^{3}$$

C
$$10$$

D
None of these

Solution

$$T_{2+1}$$
$$=T_{3}$$
$$=\:^{5}C_{2}x^{-3}x^{2log_{10}(x)}$$
$$=10^{3}$$
Therefore
$$x^{-3}x^{2log_{10}(x)}=100$$
$$x^{2log_{10}(x)-3}=100$$
Now x has to be in powers of 10.
Therefore
Let $$x=10^k$$
Above equation reduces to
$$10^{k(2k-3)}=10^{2}$$
$$2k^{2}-3k-2=0$$
$$(2k+1)(k-2)=0$$
$$k=2$$ and $$k=\frac{-1}{2}$$
Hence $$x=10^{2}$$ and $$x=10^{-\frac{1}{2}}$$
Question 2
1 / -0

In the expansion of $$\displaystyle (1+x)^{23}$$, if $$\displaystyle r^{th}$$ ,$$\displaystyle (r+1)^{th}$$ , and $$\displaystyle (r+2)^{th}$$ terms are in A.P., then value of $$\displaystyle ^{23}C_{r}$$ equals

A
$$\displaystyle ^{23}C_{8}$$

B
$$\displaystyle ^{23}C_{7}$$

C
$$\displaystyle ^{23}C_{9}$$

D
None of these

Solution

Hence

$$\:^{n}C_{r-1}$$ $$\:^{n}C_{r}$$ and $$\:^{n}C_{r+1}$$ are in A.P.

For the above terms to be, in A.P, they must follow

$$(n-2r)^{2}=n+2$$

Replacing n by 23, we get

$$529+4r^{2}-92r=23+2$$

$$4r^{2}-92r+504=0$$

$$r^{2}-23r+126=0$$

$$(r-14)(r-9)=0$$

$$r=14$$ and $$r=9$$

$$\:^{23}C_{r}$$$$=\:^{23}C_{9}$$
Question 3
1 / -0

The number of integral terms in the expansion of $$\displaystyle \left ( \sqrt{3}+\sqrt[5]{5} \right )^{256}$$ is

A
$$25$$

B
$$26$$

C
$$24$$

D
None of these

Solution

Total number of integral terms will be
$$\cfrac{256}{L.C.M(2,5)}+1$$
$$=\cfrac{256}{10}+1$$
$$=25.6+1$$
However $$25.6$$ is not an integer, hence take $$[25.6]=25$$ where $$[x]$$ is the greatest integer function.
Hence total number of integral term, will be
$$25+1$$
$$=26$$
Question 4
1 / -0

If $$\displaystyle C_{r}=^{n}C_{r}$$ and $$\displaystyle (C_{0}+C_{1})(C_{1}+C_{2})...(C_{n-1}+C_{n})=k$$ $$\displaystyle (C_{0} C_{1}C_{2}...C_{n})$$ then the value of $$ \displaystyle k$$ equals

A
$$\displaystyle \frac{(n+1)^{n+1}}{n!}$$

B
$$\displaystyle \frac{(n+1)^{n}}{n.n!}$$

C
$$\displaystyle \frac{(n)^{n}}{n!}$$

D
$$\displaystyle \frac{(n+1)^{n}}{n!} $$

Solution

$$\:^{n}C_{r+1}+\:^{n}C_{r}$$
$$=\:^{n+1}C_{r+1}$$
Hence simplifying the terms, we get
$$\:^{n+1}C_{1}.\:^{n+1}C_{2}....\:^{n+1}C_{n}$$
Now
$$\:^{n+1}C_{1}$$
$$=\dfrac {(n+1)!}{n!.1!}$$
$$=\dfrac {(n+1)}{n}\:^{n}C_{1}$$
Similarly
$$\:^{n+1}C_{2}$$
$$=\dfrac {(n+1)!}{(n-1)!.2!}$$
$$=\dfrac {(n+1)}{(n-1)}\:^{n}C_{2}$$
Hence substituting, in the above expression, we get
$$\dfrac {(n+1)^{n}}{n!}(\:^{n}C_{0}.\:^{n}C_{1}...\:^{n}C_{n})$$
Comparing coefficients, we get
$$K=\dfrac {(n+1)^{n}}{n!}$$
Question 5
1 / -0

If $$\displaystyle P$$ be the sum of odd term and $$\displaystyle Q$$ that of even terms in the expansion of $$\displaystyle (x+a)^{n}$$ , then the value of $$\displaystyle [(x+a)^{2n}-(x-a)^{2n}]$$ equals

A
$$\displaystyle PQ$$

B
$$\displaystyle 2PQ$$

C
$$4 $$$$\displaystyle PQ$$

D
None of these

Solution

$$P=\dfrac{(x+a)^{n}+(x-a)^{n}}{2}$$ ...(i)
$$Q=\dfrac{(x+a)^{n}-(x-a)^{n}}{2}$$ ...(ii)
Hence, $$2P.2Q$$
$$=4PQ$$
$$=[(x+a)^{n}+(x-a)^{n}][(x+a)^{n}-(x-a)^{n}]$$
$$=(x+a)^{2n}-(x-a)^{2n}$$
Question 6
1 / -0

$$^{ n+1 }{ { C }_{ 2 }^{ } }+2\left[ _{ }^{ 2 }{ { C }_{ 2 }^{ } }+^{ 3 }{ { C }_{ 2 }^{ } }+^{ 4 }{ { C }_{ 2 }^{ } }+...+^{ n }{ { C }_{ 2 }^{ } } \right] =$$

A
$$\displaystyle \frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } $$

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

C
$$\displaystyle \frac { n\left( n-1 \right) \left( 2n-1 \right) }{ 6 } $$

D
None of these

Solution

We have,
$$^{ n+1 }{ { C }_{ 2 } }+2\left[ ^{ 2 }{ { C }_{ 2 } }+^{ 3 }{ { C }_{ 2 } }+^{ 4 }{ { C }_{ 2 } }+...+^{ n }{ { C }_{ 2 } } \right] =^{ n+1 }{ { C }_{ 2 } }+2\left[ ^{ 3 }{ { C }_{ 3 } }+^{ 3 }{ { C }_{ 2 } }+^{ 4 }{ { C }_{ 2 } }+...+^{ n }{ { C }_{ 2 } } \right] \\ =^{ n+1 }{ { C }_{ 2 } }+2\left[ ^{ 4 }{ { C }_{ 3 } }+^{ 4 }{ { C }_{ 2 } }+...+^{ n }{ { C }_{ 2 } } \right] =^{ n+1 }{ { C }_{ 2 } }+2\left[ ^{ 5 }{ { C }_{ 3 } }+...+^{ n }{ { C }_{ 2 } } \right] \\ =^{ n+1 }{ { C }_{ 2 } }+2.^{ n+1 }{ { C }_{ 3 } }=^{ n+1 }{ { C }_{ 2 } }+^{ n+1 }{ { C }_{ 3 } }+^{ n+1 }{ { C }_{ 3 } }=^{ n+2 }{ { C }_{ 3 } }+^{ n+1 }{ { C }_{ 3 } }$$
$$\displaystyle =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 6 } +\frac { n\left( n+1 \right) \left( n-1 \right) }{ 6 } =\frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } $$
Question 7
1 / -0

Number of terms in the expansion of $$ \left ( 1-x \right )^{51}\left ( 1+x+x^{2} \right )^{50}$$ of

A
50

B
51

C
100

D
102

Solution

$$(1-x)^{51}(1+x+x^{2})^{50}$$
$$=(1-x)[(1-x)(1+x+x^{2})]^{50}$$
$$=(1-x)(1-x^{3})^{50}$$
$$=(1-x)[1-a_{1}x^{3}+a_{2}x^{6}-a_{3}x^{9}+....a_{50}x^{150}]$$
$$=(1-a_{1}x^{3}+a_{2}x^{6}-a_{3}x^{9}+....a_{50}x^{150})-(x-a_{1}x^{4}+a_{2}x^{7}-a_{3}x^{10}+....a_{50}x^{151})$$
$$=1-x-a_{1}x^{3}+a_{1}x^{4}+a_{2}x^{6}-a_{2}x^{7}+.....$$
Hence no terms gets cancelled.
Thus the total number of terms will be
$$51+51$$
$$=2(51)$$
$$=102$$
Question 8
1 / -0

If $${C_{r}}^{13}$$ denoted by $$C_{r}$$ then value of $$c_{1}+c_{5}+c_{7}+c_{9}+c_{11}$$ is equal to

A
$$2^{12}-287$$

B
$$2^{12}-165$$

C
$$2^{12}-C_{3}$$

D
$$2^{12}-C_{2}-C_{13}$$

Solution

Consider
$$(1+x)^{13}=\:^{13}C_{0}+\:^{13}C_{1}.x^{1}+\:^{13}C_{2}.x^{2}+\:^{13}C_{3}.x^{3}...+\:^{13}C_{13}.x^{13}$$
Now Let x=1.
$$2^{13}=\:^{13}C_{0}+\:^{13}C_{1}+\:^{13}C_{2}+\:^{13}C_{3}...+\:^{13}C_{13}$$
$$=2[\:^{13}C_{0}+\:^{13}C_{1}+\:^{13}C_{2}+\:^{13}C_{3}...+\:^{13}C_{6}]$$
Hence
$$2^{12}=a_{0}+a_{1}+...a_{6}$$ ...(i)
Similarly
$$(1-x)^{13}|_{x=-1}=a_{0}-a_{1}+a_{2}-a_{3}...a_{6}=0...(ii)$$
Adding i and ii, we get
$$2^{12}=2(a_{0}+a_{2}+a_{4}+a_{6})$$
$$=2(a_{0}+a_{11}+a_{9}+a_{7})$$
Hence
$$2^{11}=a_{0}+a_{11}+a_{9}+a_{7}$$ ...(a)
Subtracting ii from i, we get
$$2^{12}=2(a_{1}+a_{3}+a_{5})$$
Or
$$2^{11}=a_{1}+a_{3}+a_{5}$$ ...(b)
Adding a and b, we get
$$2^{12}=a_{1}+a_{5}+a_{7}+a_{9}+(a_{0}+a_{3})$$
$$2^{12}=a_{1}+a_{5}+a_{7}+a_{9}+(1+\:^{13}C_{3})$$
$$2^{12}=a_{1}+a_{5}+a_{7}+a_{9}+(287)$$
$$2^{12}-287=a_{1}+a_{5}+a_{7}+a_{9}$$.
Question 9
1 / -0

If coefficient of $$x^{100}$$ in $$1+\left ( 1+x \right )\left ( 1+x \right )^{2}+.....+\left ( 1+x \right )^{n}\left ( if\:n \geq 100\right )$$ is $$C_{101}^{201}$$ then the value of n equals

A
$$202$$

B
$$100$$

C
$$200$$

D
$$201$$

Solution

$${ ^{ n }{ C } }_{ r }+{ ^{ n }{ C } }_{ (r+1) }={ ^{ (n+1) }{ C } }_{ (r+1) }$$
coefficient of $${ x }^{ 100 }$$ is $$^{ 100 }{ { C }_{ 100 } }+^{ 101 }{ { C }_{ 100 } }+^{ 102 }{ { C }_{ 100 } }+........+^{ n }{ { C }_{ 100 } }$$.
Which is equal to $${ ^{ (n+1) }{ C } }_{ 101 }$$.
Therefore, $$n+1=201$$
Which implies $$ n=200$$
Question 10
1 / -0

If $$C_{0},C_{1},C_{2},.....C_{n},$$ are binomial coefficients,then $$\displaystyle\sum_{k= 0}^{n} C_{k}\:\sin kx \cos \left ( n-k \right )x$$ equals

A
$$2^{n} \sin nx$$

B
$$2^{n+1}\sin \left ( n+1 \right )x$$

C
$$2^{n-1}\sin nx$$

D
$$2^{n+1}\sin nx$$

Solution

$$S=\sum \:^{n}C_{k} sin(kx) cos(n-k)x$$
Now writing the terms in reverse, we get
$$S=\sum \:^{n}C_{n-k} sin(n-k)x cos(k)x$$
Hence
$$2S=\sum \:^{n}C_{k} [sin(kx) cos((n-k)x)+cos(kx)sin((n-k)x)]$$
$$=\sum \:^{n}C_{k} sin(nx)$$
Hence
$$2S=2^{n} sin(nx)$$
$$S=2^{n-1}sin(nx)$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 26

Result

Binomial Theorem Test - 26

Score

-

Rank

-

SHARING IS CARING

Question 1

10$$\displaystyle ^{2}$$

10$$\displaystyle ^{3}$$

$$10$$

None of these

Solution

Question 2

$$\displaystyle ^{23}C_{8}$$

$$\displaystyle ^{23}C_{7}$$

$$\displaystyle ^{23}C_{9}$$

None of these

Solution

Question 3

$$25$$

$$26$$

$$24$$

None of these

Solution

Question 4

$$\displaystyle \frac{(n+1)^{n+1}}{n!}$$

$$\displaystyle \frac{(n+1)^{n}}{n.n!}$$

$$\displaystyle \frac{(n)^{n}}{n!}$$

$$\displaystyle \frac{(n+1)^{n}}{n!} $$

Solution

Question 5

$$\displaystyle PQ$$

$$\displaystyle 2PQ$$

$$4 $$$$\displaystyle PQ$$

None of these

Solution

Question 6

$$\displaystyle \frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } $$

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

$$\displaystyle \frac { n\left( n-1 \right) \left( 2n-1 \right) }{ 6 } $$

None of these

Solution

Question 7

50

51

100

102

Solution

Question 8

$$2^{12}-287$$

$$2^{12}-165$$

$$2^{12}-C_{3}$$

$$2^{12}-C_{2}-C_{13}$$

Solution

Question 9

$$202$$

$$100$$

$$200$$

$$201$$

Solution

Question 10

$$2^{n} \sin nx$$

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

$$2^{n-1}\sin nx$$

$$2^{n+1}\sin nx$$

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.