Binomial Theorem Test

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Binomial Theorem Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

The $$7th$$ term in the expansion of $$\bigg(\dfrac{1}{2}+a\bigg)^8$$ is :

A
$$\,^8C_7\,\bigg(\dfrac{1}{2}\bigg)(a)^7$$

B
$$\,^8C_7\,\bigg(\dfrac{1}{2}\bigg)^7.a$$

C
$$\,^8C_6\,\bigg(\dfrac{1}{2}\bigg)^2(a)^6$$

D
$$\,^8C_6\,\bigg(\dfrac{1}{2}\bigg)^6(a)^2$$

Solution

$$7th$$ term in the expansion of $$\bigg(\dfrac{1}{2}+a\bigg)^8$$

$$=T_7=T_6+1$$

$$=\,^8C_6\,\bigg(\dfrac{1}{2}\bigg)^{8-6}(a)^6$$ $$(\because T_{r+1}=\,^nC_r\,a^{n-r}\,b^r)$$

$$=\,^8C_6\,\bigg(\dfrac{1}{2}\bigg)^2(a)^6$$
Question 2
1 / -0

The sum of the coefficients of even powers of $$x$$ in the expansion of
$$ (1+x+x^2+x^3)^5 $$ is

A
$$512$$

B
$$256$$

C
$$128$$

D
$$64$$

Solution

Let $$f(x)=(1+x+x^2+x^3)\:^5$$
$$=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3...$$ ...(i)
$$f(1)=a_{0}+a_{1}+a_{2}...$$ ...(ii)
$$f(-1)=a_{0}-a_{1}+a_{2}-a_{3}...$$ ...(iii)
Adding (ii) and (iii), we get
$$f(1)+f(-1)=2[a_{0}+a_{2}+a_{4}...]$$
$$(1+1+1+1)^{5}+(1-1+1-1)^{5}=2[a_{0}+a_{2}+a_{4}...]$$
$$4^{5}=2[a_{0}+a_{2}+a_{4}...]$$
$$2^{10}=2[a_{0}+a_{2}+a_{4}...]$$
$$2^9=a_{0}+a_{2}+a_{4}...$$
$$512=a_{0}+a_{2}+a_{4}...$$
Hence sum of the coefficients of even powers of $$x$$ is $$512$$.
Question 3
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Find the middle term of the expansion of $$\left ( 3x+\frac{1}{2x} \right )^7$$

A
$$^{7}C_{4}\dfrac{2^5 }{16 x^2}$$

B
$$^{7}C_{3}\dfrac{2^7 }{16 x}$$

C
$$^{7}C_{4}\dfrac{2^7 }{16 x}$$

D
$$^{-7}C_{3}\dfrac{2^7 }{16 x}$$

Question 4

1 / -0

Match the elements of List I with List II

List I	List II
A) lf $$\lambda$$ be the number of terms which are integers, in the expansion of $$(5^{\frac16}+7^{\frac19})^{1824}$$, then $$\lambda$$ is divisible by	P) 2
B) lf $$\lambda$$ be the number of terms which are rational in the expansion of $$(5^{\frac16}+2^{\frac18})^{100}$$, then $$\lambda$$ is divisible by	Q) 3
C) lf $$\lambda$$ be the number of terms which are irrational in the expansion of $$(3^{\frac14}+4^{\frac13})^{99}$$, then $$\lambda$$ is divisible by	R) 7
	S) 13
	T) 17

The correct option which matches all the elements correctly, is :

(A) - P,Q,T (B) - P (C) - R,S

(A) - P,T (B) - Q (C) - S,T

(A) - P,S,T (B) - S,Q (C) - Q

(A) - P,S (B) - P,A (C) - P,R

Solution

(i) $$(5^{\frac{1}{6}}+7^{\frac{1}{9}})^{1824}$$

$$T_{r+1}=^{1824}C_r (5)^{\frac{1824-r}{6}} 7^{\frac{r}{9}}$$

For integer terms, $$r$$ should be multiple of $$9$$.

For $$r=18,36,54,72,.....1818$$, terms comes as integer.

This is an A.P.

$$1818=18+(n-1)18$$

$$\Rightarrow n=101$$

Also, for $$r=0$$ , we would get an integer

So, total number of terms which gives integer values are $$101+1=102.$$

So, $$\lambda=102$$

So, $$\lambda$$ is divisible by $$2,3,17$$

(ii) $$(5^{\frac{1}{6}}+2^{\frac{1}{8}})^{1824}$$

$$T_{r+1}=^{100}C_r (5)^{\frac{100-r}{6}}2^{\frac{r}{8}}$$

For rational terms, $$r$$ should be multiple of $$8$$.

For $$r=16,40,64,88$$, terms comes as rational.

So, number of rational terms are $$4$$.

So, $$\lambda=4$$

which is divisble by $$2$$.

(iii) $$(3^{\frac{1}{4}}+4^{\frac{1}{3}})^{99}$$

$$T_{r+1}=^{99}C_r (3)^{\frac{99-r}{64}}4^{\frac{r}{3}}$$

For rational terms, $$r$$ should be multiple of $$3$$.

For $$r=3,15,27,.....97$$, terms comes as rational.

This is an AP

$$97=3+(n-1)12$$

$$\Rightarrow n=8$$

For $$r=99$$ also, there is a rational value

So, number of rational terms are $$8+1=9$$

Now, number of irrational terms = total number of terms -rational number of terms

$$=99+1-9=91$$

So, $$\lambda=91$$

which is divisble by $$7,13$$.

Hence, option A is the correct answer.

Question 5
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If $$S$$ be the sum of the coefficients in the expansion of $$(ax+by+-cz)^{n}$$ where $$a, b, c$$ are lengths of the sides of a triangle, then $$\lim_{n\rightarrow \infty }\dfrac{S}{(S^{1/n}+1)^{n}}$$ is

A
$$1$$

B
$$0$$

C
$$e^{\left(\dfrac{ab}{c}\right)}$$

D
$$e^{\displaystyle \left(\frac{a+b+c}{a+b+c-1}\right)}$$

Solution

$$S=(a+b-c)^{n}$$
$$\displaystyle \lim_{n\rightarrow \infty

}\dfrac{S}{\left ( S^{1/n}+1 \right )^{n}}$$ = $$\displaystyle

\lim_{n\rightarrow \infty }\left ( \dfrac{a+b-c}{a+b-c+1} \right )^{n}$$
$$\displaystyle =\lim_{n\rightarrow \infty } \left ( \dfrac{k}{k+1} \right )^{n} = L$$ (say)
Where $$k = a + b - c$$
Now In $$\Delta ABC a + b > c$$ $$\Rightarrow 0< \dfrac{k}{k+1}< 1 \therefore L = 0$$
Question 6
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The value of the expression $$\displaystyle \frac{C_1}{C_0}+2\frac{C_2}{C_1}+3\frac{C_3}{C_2}+\ldots\ldots\ldots +n\frac{C_n}{C_{n-1}}$$ is

A
$$ \dfrac{(n+1)(n+2)}{2} $$

B
$$ \dfrac{n(n+1)}{2} $$

C
$$ \dfrac{n(n-1)}{2} $$

D
$$ \dfrac{n(n+2)}{2} $$

Solution

Given:
$$\dfrac{C_1}{C_0} + 2\dfrac{C_2}{C_1}+3\dfrac{C_3}{C_2} +\cdots+n\dfrac{C_n}{C_{n-1}}$$

$$r^{th}$$ term of the above expression is
$$T_r = r\dfrac{C_r}{C_{r-1}} = r\dfrac{n-r+1}{r} = n-r+1$$

Now,
$$\displaystyle\sum_{r=1}^nT_r = \sum_{r=1}^n (n-r+1)$$

$$=n^2-\dfrac{n(n+1)}{2} + n$$

$$=\dfrac{n(n+1)}2$$
Question 7
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Arrange the values of $$n$$ in ascending order
A : If the term independent of $$x$$ in the expansion of $$\left(\displaystyle \sqrt{x}-\frac{n}{x^{2}}\right)^{10}$$ is $$405$$
B : If the fourth term in the expansion of $$\left(\displaystyle \frac{1}{n}+n^{\log_{n}10}\right)^{5}$$ is $$1000$$, ( $$ n< 10 $$)
C : In the binomial expansion of $$(1+x)^{n}$$ the coefficients of $$5^{\mathrm{t}\mathrm{h}},\ 6^{\mathrm{t}\mathrm{h}}$$ and $$7^{\mathrm{t}\mathrm{h}}$$ terms are in A.P.

A
$$A,B,C$$

B
$$B,A,C$$

C
$$A,C,B$$

D
$$C,A,B$$

Solution

A: Given: $$(\sqrt{x}-nx^{-2})^{10}$$

$$T_{r+1}=(-1)^{r}\:^{10} C_{r} x^{(5-{5r}/{2})} n^{r}$$ ...(1)

Term independent of $$x$$ implies
$$5-\dfrac{5r}{2}=0$$

Thus $$r=2$$
Substituting in equation (1), we get
$$T_{3}=\:^{10}C_{2}n^{2}=405$$
$$\Rightarrow 45n^{2}=405$$
$$\Rightarrow n^{2}=9$$
$$\Rightarrow n=3$$ ($$n$$ is $$\neq -3$$ as $$n$$ is a positive integer)

B: Given: $$(n^{-1}+n^{log_{n}10})^{5}$$
$$=(n^{-1}+10)^{5}$$
$$T_{r+1}=\:^{5}C_{r}n^{5-r}10^{r}$$ ...(2)
It is given that the third term is $$1000$$
Substituting in equation (2), we get
$$T_{4}=\:^{5}C_{2}n^{-2}10^{3}=1000$$
$$\Rightarrow 10(n^{-2}1000)=1000$$
$$\Rightarrow n^{-2}=10$$
$$\Rightarrow n=\dfrac{1}{\sqrt{10}}$$

C: For the terms to be in arithmetic progression
$$(n-2r)^{2}=n+2$$ ...(3)
Let us consider $$\:^{n}C_{r-1},\:^{n}C_{r},\:^{n}C_{r+1}$$ be in A.P.
Hence, $$T_{r},T_{r+1},T_{r+2}$$ are in A.P.
Here it is given that, the $$5^{th}, 6^{th}$$ and $$7^{th}$$ terms are in A.P.
Hence, $$r=5$$
Substituting in equation (3), we get
$$(n-10)^{2}=n+2$$
$$\Rightarrow n^{2}-21n+98=0$$
$$\Rightarrow(n-7)(n-14)=0$$
Hence, $$n=7, 14$$
Hence, the values of $$n$$ in ascending order is B,A,C.

Option B.
Question 8
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If the fourth term in the expansion of $$\displaystyle \left ( \sqrt{x^{\left ( \frac{1}{log x+1} \right )}}+x^{\frac{1}{12}} \right )^6$$ is equal to 200 and x > 1, then x is equal to

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

B
$$10$$

C
$$10^4$$

D
None of these

Solution

$${ \left( a+b \right) }^{ 6 } = { ^{ 6 }{ C } }_{ 0 }{ a }^{ 6 }+{ ^{ 6 }{ C } }_{ 1 }{ a }^{ 5 }b+{ ^{ 6 }{ C } }_{ 2 }{ a }^{ 4 }{ b }^{ 2 }+{ ^{ 6 }{ C } }_{ 3 }{ a }^{ 5 }{ b }^{ 3 }..........{ ^{ 6 }{ C } }_{ 6 }{ b }^{ 6 }$$
$$\hookrightarrow$$ 4th term
$${ t }_{ 4 }=200$$
$$\Rightarrow { ^{ 6 }{ C } }_{ 3 } { \left( \sqrt { { x }^{ { 1 }/{ ㏒x+1 } } } \right) }^{ 3 }{ \left( { x }^{ { 1 }/{ 12 } } \right) }^{ 3 } = 200$$
$$\Rightarrow20 \times \left[ { x }^{ { 3 }/{ 2(logx+1) } } \right] \left[ { x }^{ { 1 }/{ 4 } } \right] = 200$$
$$\Rightarrow{ x }^{ { \left[ { 3 }/{ 2(logx+1) }+{ 1 }/{ 4 } \right] } }=10$$
Taking log to base $$10$$ on both sides:
$$\left[ \dfrac { 3 }{ 2(logx+1) } +\dfrac { 1 }{ 4 } \right] { log }_{ 10 }x$$ = $$ { log }_{ 10 }10$$
$$\Rightarrow $$ $$\left[ \dfrac { 6+(1+logx) }{ 4(1+logx) } \right] logx = 1$$
$$\Rightarrow (7+logx) logx = 4 + 4 logx$$
$$\Rightarrow{ log }^{ 2 }x + 7 logx = 4 + 4 logx$$
$$\Rightarrow{ log }^{ 2 }x + 3 logx - 4 = 0$$
$$\Rightarrow(logx + 4 ) (logx - 1 ) = 0$$
$$\Rightarrow x = { 10 }^{ -4 }$$ or $$x = { 10 }^{ 1 }$$
Given that, $$x >1$$
$$\Rightarrow$$ $$ x = 10$$
Hence the answer is $$10.$$
Question 9
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The coefficient of $$x^r[0 \le r \le n-1]$$ in the expression of $$(x + 2)^{n-1} + (x+2)^{n-2} .(x+1) + (x+2)^{n-3}. (x+1)^2+...+(x+1)^{n-1}$$ is

A
$$^nC_r(2^r - 1)$$

B
$$^nC_r(2^{n-r} - 1)$$

C
$$^nC_r(2^r + 1)$$

D
$$^nC_r(2^{n-r} + 1)$$

Solution

Given expression is equal to $$\sum _{ p=1 }^{ n }{ { (x+3) }^{ (n-p) }\times { (x+2) }^{ (p-1) } }$$

Which is equal to $$ { (x+3) }^{ (n-1) }\times (\sum _{ p=1 }^{ n }{ { (\dfrac { (x+2) }{ (x+3) } ) }^{ (p-1) }) } = { (x+3) }^{ n }-{ (x+2) }^{ n }$$

So,coefficient of $${ x }^{ r } \space is \space^n{ C }_{ r }\times ({ 3 }^{ (n-r) }-{ 2 }^{ (n-r) })$$
Question 10
1 / -0

Find the sum of the series
$$3.{ _{ }^{ n }{ C } }_{ 0 }-8.{ _{ }^{ n }{ C } }_{ 1 }+13._{ }^{ n }{ { C }_{ 2 } }-18.{ _{ }^{ n }{ { C }_{ 3 } } }+\ldots+(n+1)\quad terms$$

A
$$0$$

B
$$-1$$

C
$$+1$$

D
None of these

Solution

Let $$n=2$$
Hence the above expression is reduced to
$$3(1)-8(2)+13(1)$$
$$=16-16=0$$
Let $$n=3$$
$$3(1)-8(3)+13(3)-18(1)$$
$$=42-42=0$$
Hence the sum of the series for $$n>1$$ is $$0$$

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Binomial Theorem Test - 55

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Binomial Theorem Test - 55

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Question 1

$$\,^8C_7\,\bigg(\dfrac{1}{2}\bigg)(a)^7$$

$$\,^8C_7\,\bigg(\dfrac{1}{2}\bigg)^7.a$$

$$\,^8C_6\,\bigg(\dfrac{1}{2}\bigg)^2(a)^6$$

$$\,^8C_6\,\bigg(\dfrac{1}{2}\bigg)^6(a)^2$$

Solution

Question 2

$$512$$

$$256$$

$$128$$

$$64$$

Solution

Question 3

$$^{7}C_{4}\dfrac{2^5 }{16 x^2}$$

$$^{7}C_{3}\dfrac{2^7 }{16 x}$$

$$^{7}C_{4}\dfrac{2^7 }{16 x}$$

$$^{-7}C_{3}\dfrac{2^7 }{16 x}$$

Question 4

(A) - P,Q,T (B) - P (C) - R,S

(A) - P,T (B) - Q (C) - S,T

(A) - P,S,T (B) - S,Q (C) - Q

(A) - P,S (B) - P,A (C) - P,R

Solution

Question 5

$$1$$

$$0$$

$$e^{\left(\dfrac{ab}{c}\right)}$$

$$e^{\displaystyle \left(\frac{a+b+c}{a+b+c-1}\right)}$$

Solution

Question 6

$$ \dfrac{(n+1)(n+2)}{2} $$

$$ \dfrac{n(n+1)}{2} $$

$$ \dfrac{n(n-1)}{2} $$

$$ \dfrac{n(n+2)}{2} $$

Solution

Question 7

$$A,B,C$$

$$B,A,C$$

$$A,C,B$$

$$C,A,B$$

Solution

Question 8

$$10^{\sqrt{2}}$$

$$10$$

$$10^4$$

None of these

Solution

Question 9

$$^nC_r(2^r - 1)$$

$$^nC_r(2^{n-r} - 1)$$

$$^nC_r(2^r + 1)$$

$$^nC_r(2^{n-r} + 1)$$

Solution

Question 10

$$0$$

$$-1$$

$$+1$$

None of these

Solution

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