Binomial Theorem Test

Result

Binomial Theorem Test - 23

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

Let $$n$$ and $$k$$ be positive integers such that $$\displaystyle n\ge \frac { k\left( k+1 \right) }{ 2 } .$$ The number of solution $$\left( { x }_{ 1 },{ x }_{ 2 },..,{ x }_{ k } \right) \ge 1;{ x }_{ 2 }\ge 2,...,{ x }_{ k }\ge k$$ all integers satisfying $${ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+...+{ x }_{ k }=n$$ is

A
$$^{ m }{ { C }_{ k-1 } }$$

B
$$^{ m }{ { C }_{ k } }$$3

C
$$^{ m }{ { C }_{ k+1 } }$$

D
None of these

Solution

The number of solutions of $${ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+...{ x }_{ k }=n$$
$$=$$ coefficients of $${ t }^{ n }$$ in $$\left( t+{ t }^{ 2 }+{ t }^{ 3 }+.. \right) \left( { t }^{ 2 }+{ t }^{ 3 }+.. \right) ...\left( { t }^{ k }+{ t }^{ k+1 }+.. \right) $$
$$=$$ coefficient of $${ t }^{ n }$$ in $${ t }^{ 1+2+3+...k }{ \left( 1+t+{ t }^{ 2 }+{ t }^{ 3 }+.. \right) }^{ k }$$
But $$1+2+3+...k=k\displaystyle\frac { k+1 }{ 2 } =r$$
and $$1+t+{ t }^{ 2 }+{ t }^{ 3 }+..=\displaystyle\frac { 1 }{ 1-t } $$
Thus the required number of solutions
$$=$$ coefficients of $${ t }^{ n-r }$$ in $${ \left( 1-t \right) }^{ -k }$$
$$=$$ coefficients of $${ t }^{ n-r }$$ in$$1+_{ }^{ k }{ { C }_{ 1 }^{ }t+ }^{ k+1 }{ { C }_{ 2 }^{ }{ t }^{ 2 }+.... }$$
$$\Rightarrow \ \ _{ }^{ k+n-r-1 }{ { C }_{ n-r }^{ } }=^{ k+n-r-1 }{ { C }_{ k-1 }^{ } }=^{ m }{ { C }_{ k-1 }^{ } }$$
Question 2
1 / -0

The number of irrational terms in the expansion of $${ \left( { 2 }^{ \dfrac 15 }+{ 3 }^{ \dfrac {1}{10} } \right) }^{ 55 }$$ is

A
$$47$$

B
$$56$$

C
$$50$$

D
$$48$$

Solution

For the above question
$$T_{r+1}=\:^{55}C_{r}2^{11-\dfrac{r}{5}}3^{\dfrac{r}{10}}$$
Hence we will have rational terms at $$r=0,10,20,30,40,50$$ respectively.
Hence there will be $$6$$ rational terms.
The total number of terms will be
$$55+1$$
$$=56$$ terms.
Hence the number of irrational terms will be
$$56-6$$
$$=50$$ terms.
Question 3
1 / -0

Find the middle term in the expansion of $${ \left( \cfrac { x }{ a } -\cfrac { a }{ x } \right) }^{ 21 }$$

A
$${ _{ }^{ 20 }{ C } }_{ 10 }\cfrac { x }{ a } , { _{ }^{ 21 }{ C } }_{ 10 }\cfrac { a }{ x } $$

B
$${ _{ }^{ 20 }{ C } }_{ 9 }\cfrac { x }{ a } , { _{ }^{ 21 }{ C } }_{ 10 }\cfrac { a }{ x } $$

C
$${ _{ }^{ 21 }{ C } }_{ 10 }\cfrac { x }{ a } , -{ {}_{ }^{ 21 }{ C } }_{ 10 }\cfrac { a }{ x } $$

D
$${ _{ }^{ 21 }{ C } }_{ 9 }\cfrac { x }{ a } , { _{ }^{ 21 }{ C } }_{ 10 }\cfrac { a }{ x } $$

Solution

There are 22 terms in the expansion of $$(\displaystyle \frac{x}{a}-\displaystyle \frac{a}{x})^{21}$$.
So, the middle terms are $$T_{11}$$ and $$T_{12}$$
$$T_{11}=^{21}\textrm{C}_{10}(\displaystyle \frac{x}{a})^{11}(\displaystyle \frac{-a}{x})^{10}=^{21}\textrm{C}_{10}(\displaystyle \frac{x}{a})$$
and
$$T_{12}=^{21}\textrm{C}_{11}(\displaystyle \frac{x}{a})^{10}(\displaystyle \frac{-a}{x})^{11}=-^{21}\textrm{C}_{11}(\displaystyle \frac{a}{x})=-^{21}\textrm{C}_{10}(\displaystyle \frac{a}{x})$$
Question 4
1 / -0

In the expansion of the expression $${ \left( x+a \right) }^{ 15 }$$, if the eleventh term in the geometric mean of the eighth and twelfth terms, which term in the expression is the greatest?

A
$${T}_{6}$$

B
$${T}_{7}$$

C
$${T}_{8}$$

D
$${T}_{9}$$

Solution

From the given condition, one get
$$\displaystyle { \left( _{ }^{ 15 }{ { C }_{ 10 }{ x }^{ 2 }{ a }^{ 10 } } \right) }^{ 2 }=\left( _{ }^{ 15 }{ { C }_{ 7 }{ x }^{ 8 }{ a }^{ 7 } } \right) \left( _{ }^{ 15 }{ { C }_{ 11 }{ x }^{ 4 }{ a }^{ 11 } } \right) \Rightarrow \frac { a }{ x } =\sqrt { \frac { 75 }{ 77 } } $$
Now $$\displaystyle \left( x+a \right) ^{ 15 }={ x }^{ 15 }{ \left( 1+\frac { a }{ x } \right) }^{ 15 }$$
consider the expansion of $$\displaystyle { \left( 1+\frac { a }{ x } \right) }^{ 15 }$$
$$\displaystyle { T }_{ r }={ T }_{ r+1 }\Rightarrow \frac { r }{ \left( n-r+1 \right) \left( \frac { a }{ x } \right) } =1$$
$$\displaystyle \Rightarrow r=16\sqrt { \frac { 75 }{ 77 } } -r\sqrt { \frac { 75 }{ 77 } } \Rightarrow r=\frac { 16\sqrt { \frac { 75 }{ 77 } } }{ 1+\sqrt { \frac { 75 }{ 77 } } } =7$$
If $$r=7\Rightarrow { T }_{ 7 }<{ T }_{ 8 }$$ and $$r=8\Rightarrow { T }_{ 8 }>{ T }_{ 9 }$$
Hence $${ T }_{ 8 }$$ is the greater term
Question 5
1 / -0

Find the sum of the series $$\displaystyle\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }\left[ \cfrac { 1 }{ { 2 }^{ r } } +\cfrac { { 3 }^{ r } }{ { 2 }^{ 2r } } +\cfrac { { 7 }^{ r } }{ { 2 }^{ 3r } } +\cfrac { { 15 }^{ r } }{ { 2 }^{ 4r } } +...upto\: m\: terms \right] $$

A
$$\displaystyle\cfrac { \left( { 2 }^{ mn }+1 \right) }{ \left( { 2 }^{ n }+1 \right) \left( { 2 }^{ mn } \right) } $$

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

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

D
$$\displaystyle\cfrac { \left( { 2 }^{ mn }-1 \right) }{ \left( { 2 }^{ n }+1 \right) \left( { 2 }^{ mn } \right) } $$

Solution

$$\displaystyle \sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } }. { _{ }^{ n }{ C } }_{ r }\left[ \cfrac { 1 }{ { 2 }^{ r } } +\cfrac { { 3 }^{ r } }{ { 2 }^{ 2r } } +\cfrac { { 7 }^{ r } }{ { 2 }^{ 3r } } +\cfrac { { 15 }^{ r } }{ { 2 }^{ 4r } } +... \right] $$ upto m terms
$$\displaystyle=\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }{ \left( \frac { 1 }{ 2 } \right) }^{ r }+\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }{ \left( \frac { 3 }{ 4 } \right) }^{ r }+\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }\left( \right) { \left( \frac { 7 }{ 8 } \right) }^{ r }+...$$
$$\displaystyle={ \left( 1-\frac { 1 }{ 2 } \right) }^{ n }+{ \left( 1-\frac { 3 }{ 4 } \right) }^{ n }+{ \left( 1-\frac { 7 }{ 8 } \right) }^{ n }+...\quad \quad \left[ \because \sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }{ x }^{ r }={ \left( 1-x \right) }^{ r } \right] $$
$$\displaystyle={ \left( \frac { 1 }{ 2 } \right) }^{ n }+{ \left( \frac { 1 }{ 4 } \right) }^{ n }+{ \left( \frac { 1 }{ 8 } \right) }^{ n }+...={ \left( \frac { 1 }{ 2 } \right) }^{ n }\left[ \frac { 1-{ \left( \frac { 1 }{ { 2 }^{ n } } \right) }^{ m } }{ 1-{ \left( \frac { 1 }{ 2 } \right) }^{ n } } \right] $$
$$\displaystyle=\frac { { 2 }^{ mn }-1 }{ { 2 }^{ mn }\left( { 2 }^{ n }-1 \right) } $$
Question 6
1 / -0

Determine the value of $$x$$ in the expression $${ \left( x+{ x }^{ t } \right) }^{ 5 }$$, if the third term in the expression is 10,00,000 where $$t=\log _{ 10 }{ x } $$.

A
$$10\quad $$

B
$${ 10 }^{ -5/2 }$$

C
Both (A)and (B)

D
None of these

Solution

$$T_{3}$$ $$=\:^{5}C_{2}x^{3+2t}$$
$$=10^{6}$$
$$x^{3+2t}=10^{5}$$
$$\ln(x)(3+2t)=5\ln(10)$$
$$\log_{10}(x)(3+2t)=5$$
$$t(3+2t)=5$$
$$2t^{2}+3t-5=0$$
$$t=1$$ and $$t=\dfrac{-5}{2}$$
Hence, $$x=10$$ and $$x=10^{\frac{-5}{2}}$$.
Question 7
1 / -0

If $$\cfrac { { _{ }^{ n }{ C } }_{ r }+4{ _{ }^{ n }{ C } }_{ r+1 }+6{ _{ }^{ n }{ C } }_{ r+2 }+4{ _{ }^{ n }{ C } }_{ r+3 }+{ _{ }^{ n }{ C } }_{ r+4 } }{ { _{ }^{ n }{ C } }_{ r }+3{ _{ }^{ n }{ C } }_{ r+1 }+3{ _{ }^{ n }{ C } }_{ r+2 }+{ _{ }^{ n }{ C } }_{ r+3 } } =\cfrac { n+k }{ r+k } $$. Find the value of k

A
2

B
4

C
6

D
8

Solution

$$\displaystyle\cfrac { { _{ }^{ n }{ C } }_{ r }+4{ _{ }^{ n }{ C } }_{ r+1 }+6{ _{ }^{ n }{ C } }_{ r+2 }+4{ _{ }^{ n }{ C } }_{ r+3 }+{ _{ }^{ n }{ C } }_{ r+4 } }{ { _{ }^{ n }{ C } }_{ r }+3{ _{ }^{ n }{ C } }_{ r+1 }+3{ _{ }^{ n }{ C } }_{ r+2 }+{ _{ }^{ n }{ C } }_{ r+3 } }$$
$$\displaystyle=1+\cfrac { { _{ }^{ n }{ C } }_{ r+1 }+3{ _{ }^{ n }{ C } }_{ r+2 }+3{ _{ }^{ n }{ C } }_{ r+3 }+{ _{ }^{ n }{ C } }_{ r+4 } }{ { _{ }^{ n }{ C } }_{ r }+3{ _{ }^{ n }{ C } }_{ r+1 }+3{ _{ }^{ n }{ C } }_{ r+2 }+{ _{ }^{ n }{ C } }_{ r+3 } }$$
using $${ _{ }^{ n }{ C } }_{ r }+{ _{ }^{ n }{ C } }_{ r-1 }={ _{ }^{ n+1 }{ C } }_{ r }$$
$$\displaystyle=1+\cfrac { { _{ }^{ n+1 }{ C } }_{ r+2 }+2{ _{ }^{ n+1 }{ C } }_{ r+3 }+{ _{ }^{ n+1 }{ C } }_{ r+4 } }{ { _{ }^{ n+1 }{ C } }_{ r+1 }+2{ _{ }^{ n+1 }{ C } }_{ r+2 }+{ _{ }^{ n+1 }{ C } }_{ r+3 } }$$
$$\displaystyle=1+\cfrac { { _{ }^{ n+2 }{ C } }_{ r+3 }+{ _{ }^{ n+2 }{ C } }_{ r+4 } }{ { _{ }^{ n+2 }{ C } }_{ r+2 }+{ _{ }^{ n+2 }{ C } }_{ r+3 } }$$
$$\displaystyle=1+\cfrac { { _{ }^{ n+3 }{ C } }_{ r+4 } }{ { _{ }^{ n+3 }{ C } }_{ r+3 } }$$
$$=\displaystyle\frac{n+4}{r+4}$$
$$\therefore k=4$$
Hence, option B.
Question 8
1 / -0

The value of $$\cfrac { 1 }{ \left( n-1 \right) ! } +\cfrac { 1 }{ \left( n-3 \right) !3! } +\cfrac { 1 }{ \left( n-5 \right) !5! }+.... $$

A
$$\cfrac { { 2 }^{ n+1 } }{ n! } $$

B
$$\cfrac { { 2 }^{ n-1 } }{ n! } $$

C
$$\cfrac { { 2 }^{ n-1 } }{ n+1! } $$

D
$$\cfrac { { 2 }^{ n+1 } }{ n-1! } $$

Solution

$$\cfrac { 1 }{ \left( n-1 \right) ! } +\cfrac { 1 }{ \left( n-3 \right) !3! } +\cfrac { 1 }{ \left( n-5 \right) !5! }+.... $$
$$=\cfrac{1}{n!}\left(\cfrac { n! }{ \left( n-1 \right) ! 1! } +\cfrac { n! }{ \left( n-3 \right) !3! } +\cfrac { n! }{ \left( n-5 \right) !5! }+.... \right)$$
$$=\cfrac{1}{n!}( ^nC_1+^nC_3+^nC_5+.............)=\cfrac{2^{n-1}}{n!}$$
Question 9
1 / -0

If $${ \left( 1+x \right) }^{ n }={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }+..........+{ C }_{ n }{ x }^{ R }$$, then the sum
$${ C }_{ 0 }+({ C }_{ 0 }+{ C }_{ 1 })+({ C }_{ 0 }+{ C }_{ 1 }+{ C }_{ 2 })+.....+({ C }_{ 0 }+{ C }_{ 1 }+{ C }_{ 2 }+.....+{ C }_{ n-1 }$$ is

A
$$n.{ 2 }^{ n+1 }$$

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

C
$$(n-1).{ 2 }^{ n-1 }$$

D
$$(n+1).{ 2 }^{ n+1 }$$

Solution

Simplifying, we get
$$n\:^{n}C_{0}+(n-1)\:^{n}C_{1}+(n-2)\:^{n}C_{2}+...(n-(n-1))\:^{n}C_{n-1}+(n-n)\:^{n}C_{n}$$
$$=n[\:^{n}C_{0}+\:^{n}C_{1}+...\:^{n}C_{n}]-[1.\:^{n}C_{1}+2\:^{n}C_{2}+...n\:^{n}C_{n}]$$
$$=n.2^{n}-n2^{n-1}$$
$$=n2^{n-1}(2-1)$$
$$=n2^{n-1}$$.
Question 10
1 / -0

If $$n$$ is a positive integer and $${ C }_{ k }={ _{ }^{ n }{ C } }_{ k }$$, find the value of $$\sum _{ k=1 }^{ n }{ { k }^{ 3 }{ \left( \cfrac { { C }_{ k } }{ { C }_{ k-1 } } \right) }^{ 2 } } $$

A
$$\cfrac { n{ \left( n-1 \right) }^{ 2 }(n+2) }{ 12 } $$

B
$$\cfrac { n{ \left( n+1 \right) }^{ 2 }(n-2) }{ 12 } $$

C
$$\cfrac { n{ \left( n+1 \right) }^{ 2 }(n+2) }{ 12 } $$

D
$$\cfrac { n{ \left( n-1 \right) }^{ 2 }(n-2) }{ 12 } $$

Solution

$$\displaystyle \sum _{ k=1 }^{ n }{ { k }^{ 3 }{ \left( \cfrac { { C }_{ k } }{ { C }_{ k-1 } } \right) }^{ 2 } } =\sum _{ k=1 }^{ n }{ { k }^{ 3 }{ \left( \frac { n-k+1 }{ k } \right) }^{ 2 } } $$
$$=\sum _{ k=1 }^{ n }{ { k }{ \left( n-k+1 \right) }^{ 2 } } $$
$$={ \left( n+1 \right) }^{ 2 }\sum _{ k=1 }^{ n }{ { k } } +\sum _{ k=1 }^{ n }{ { k }^{ 3 } } -2\left( n+1 \right) \sum _{ k=1 }^{ n }{ { k }^{ 2 } } $$
$$ =\dfrac { n{ \left( n+1 \right) }^{ 3 } }{ 2 } +\dfrac { { n }^{ 2 }{ \left( n+1 \right) }^{ 2 } }{ 4 } -\dfrac { n{ \left( n+1 \right) }^{ 2 }\left( 2n+1 \right) }{ 3 } $$
$$=\cfrac { n{ \left( n+1 \right) }^{ 2 }(n+2) }{ 12 } $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 23

Result

Binomial Theorem Test - 23

Score

-

Rank

-

SHARING IS CARING

Question 1

$$^{ m }{ { C }_{ k-1 } }$$

$$^{ m }{ { C }_{ k } }$$3

$$^{ m }{ { C }_{ k+1 } }$$

None of these

Solution

Question 2

$$47$$

$$56$$

$$50$$

$$48$$

Solution

Question 3

$${ _{ }^{ 20 }{ C } }_{ 10 }\cfrac { x }{ a } , { _{ }^{ 21 }{ C } }_{ 10 }\cfrac { a }{ x } $$

$${ _{ }^{ 20 }{ C } }_{ 9 }\cfrac { x }{ a } , { _{ }^{ 21 }{ C } }_{ 10 }\cfrac { a }{ x } $$

$${ _{ }^{ 21 }{ C } }_{ 10 }\cfrac { x }{ a } , -{ {}_{ }^{ 21 }{ C } }_{ 10 }\cfrac { a }{ x } $$

$${ _{ }^{ 21 }{ C } }_{ 9 }\cfrac { x }{ a } , { _{ }^{ 21 }{ C } }_{ 10 }\cfrac { a }{ x } $$

Solution

Question 4

$${T}_{6}$$

$${T}_{7}$$

$${T}_{8}$$

$${T}_{9}$$

Solution

Question 5

$$\displaystyle\cfrac { \left( { 2 }^{ mn }+1 \right) }{ \left( { 2 }^{ n }+1 \right) \left( { 2 }^{ mn } \right) } $$

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

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

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

Solution

Question 6

$$10\quad $$

$${ 10 }^{ -5/2 }$$

Both (A)and (B)

None of these

Solution

Question 7

2

4

6

8

Solution

Question 8

$$\cfrac { { 2 }^{ n+1 } }{ n! } $$

$$\cfrac { { 2 }^{ n-1 } }{ n! } $$

$$\cfrac { { 2 }^{ n-1 } }{ n+1! } $$

$$\cfrac { { 2 }^{ n+1 } }{ n-1! } $$

Solution

Question 9

$$n.{ 2 }^{ n+1 }$$

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

$$(n-1).{ 2 }^{ n-1 }$$

$$(n+1).{ 2 }^{ n+1 }$$

Solution

Question 10

$$\cfrac { n{ \left( n-1 \right) }^{ 2 }(n+2) }{ 12 } $$

$$\cfrac { n{ \left( n+1 \right) }^{ 2 }(n-2) }{ 12 } $$

$$\cfrac { n{ \left( n+1 \right) }^{ 2 }(n+2) }{ 12 } $$

$$\cfrac { n{ \left( n-1 \right) }^{ 2 }(n-2) }{ 12 } $$

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.