If $$A = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right]$$ and $$I = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right],$$ then which one of the following holds for all $$n \ge 1,$$ by

Given $$A = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right]$$ $$\therefore$$ $$A \times A$$ = $${A^2}$$ = $$\left[ {\matrix{ 1 & 0 \cr 2 & 1 \cr } } \right]$$ and $${A^3}$$ = $${A^2} \times A$$ = $$\left[ {\matrix{ 1 & 0 \cr 3 & 1 \cr } } \right]$$ So we can say $${A^n}$$ = $$\left[ {\matrix{ 1 & 0 \cr n & 1 \cr } } \right]$$ Now $$nA - \left( {n - 1} \right){\rm I}$$ = $$\left[ {\matrix{ n & 0 \cr n & n \cr } } \right]$$ - $$\left[ {\matrix{ {n - 1} &amp

Binomial Theorem — Page 11 | JEE Mathematics

Q101

The sum of all rational terms in the expansion of

(2+\sqrt{3})^8

is :

A 16923

B 18817

C 3763

D 33845

Correct Answer

Option B

Solution

To find the sum of all rational terms in the expansion of $(2+\sqrt{3})^8$ , we consider the binomial expansion: $S = { }^8 C_0 (2)^8 + { }^8 C_1 (2)^7 (\sqrt{3}) + \ldots + { }^8 C_8 (\sqrt{3})^8$ We need to identify and sum only the rational terms.

In the binomial expansion, a term is rational if the exponent of $\sqrt{3}$ is even.

Thus, the rational terms are: $\begin{aligned} &{ }^8 C_0 (2)^8 + { }^8 C_2 (2)^6 (\sqrt{3})^2 + { }^8 C_4 (2)^4 (\sqrt{3})^4 + \\ &{ }^8 C_6 (2)^2 (\sqrt{3})^6 + { }^8 C_8 (\sqrt{3})^8 \end{aligned}$ Evaluating these terms: ${ }^8 C_0 (2)^8 = 256$ ${ }^8 C_2 (2)^6 (3) = { }^8 C_2 \times 64 \times 3 = 1344$ ${ }^8 C_4 (2)^4 (3)^2 = { }^8 C_4 \times 16 \times 9 = 3024$ ${ }^8 C_6 (2)^2 (3)^3 = { }^8 C_6 \times 4 \times 27 = 4032$ ${ }^8 C_8 (3)^4 = 1 \times 81 = 81$ Summing these rational terms gives: $256 + 1344 + 3024 + 4032 + 81 = 18817$ Therefore, the sum of all rational terms in the expansion is $18,817$ .

Q102

In the expansion of

\left(\sqrt[3]{2}+\dfrac{1}{\sqrt[3]{3}}\right)^n, n \in \mathrm{~N}

, if the ratio of

15^{\text{th }}

term from the beginning to the

15^{\text{th }}

term from the end is

\dfrac{1}{6}

, then the value of

{ }^n \mathrm{C}_3

is

A 4960

B 2300

C 1040

D 4060

Correct Answer

Option B

Solution

In the expansion of $(a+b)^n$ $15^{\text{th }}$ term from beginning: $T_{15}={ }^n C_{14} a^{n-14} b^{14}$ $15^{\text{th }}$ term from end: $T_{15}^{\prime}={ }^n C_{14} b^{n-14} a^{14}$

\begin{aligned} & \therefore \quad \frac{T_{15}}{T_{15}^{\prime}}=\frac{1}{6} \\ & \left(\frac{a}{b}\right)^{n-28}=\frac{1}{6} \\ & \left(6^{\frac{1}{3}}\right)^{n-28}=6^{-1} \\ & \Rightarrow \quad \frac{n-28}{3}=-1 \\ & n=25 \\ & \therefore \quad{ }^{25} C_3=2300 \end{aligned}

Q103

If

{a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + .......} } }

having

n

radical signs then by methods of mathematical induction which is true

A

{a_n} > 7\,\,\forall \,\,n \ge 1

B

{a_n} < 7\,\,\forall \,\,n \ge 1

C

{a_n} < 4\,\,\forall \,\,n \ge 1

D

{a_n} > 3\,\,\forall \,\,n \ge 1

Correct Answer

Option D

Solution

Given

{a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + .......} } }

$\therefore$

{a_n} = \sqrt {7 + {a_n}}

$\Rightarrow$

a_n^2 = 7 + {a_n}

$\Rightarrow$

a_n^2 - {a_n} - 7 = 0

\Rightarrow {a_n} = {{1 \pm \sqrt {1 - 4 \times 1 \times - 7} } \over 2}

\Rightarrow {a_n} = {{1 \pm \sqrt {29} } \over 2}

As

{a_n}

> 0, $\therefore$

{a_n} = {{1 + \sqrt {29} } \over 2}

= 3.19 So

{a_n} > 3\,\,\forall \,\,n \ge 1

Q104

Let

S(K)

= 1 + 3 + 5... + \left( {2K - 1} \right) = 3 + {K^2}.

Then which of the following is true

A Principle of mathematical induction can be used to prove the formula

B

S\left( K \right) \Rightarrow S\left( {K + 1} \right)

C

S\left( K \right) \ne S\left( {K + 1} \right)

D

S\left( 1 \right)

is correct

Correct Answer

Option B

Solution

Given

S(K)

= 1 + 3 + 5... + \left( {2K - 1} \right) = 3 + {K^2}

When k = 1, S(1): 1 = 3 + 1, L.H.S of S(k) $\ne$ R.H.S of S(k) So S(1) is not true.

As S(1) is not true so principle of mathematical induction can not be used.

S(K+1) = 1 + 3 + 5...

+ (2K - 1) + (2K + 1) = 3 + (k + 1)2 Now let S(k) is true $\therefore$ 1 + 3 + 5 +........(2k - 1) = 3 + k2 $\Rightarrow$ 1 + 3 + 5 +........(2k - 1) + (2k + 1) = 3 + k2 + 2k +1 = 3 + (k + 1)2 $\Rightarrow$ S(k + 1) is true.

$\therefore$ S(k) $\Rightarrow$ S(k + 1)

Q105

If

A = \left[ \begin{array}{ll}1 & 0 \\ 1 & 1 \end{array} \right]

and

I = \left[ \begin{array}{ll}1 & 0 \\ 0 & 1 \end{array} \right],

then which one of the following holds for all

n \ge 1,

by the principle of mathematical induction?

A

{A^n} = nA - \left( {n - 1} \right){\rm I}

B

{A^n} = {2^{n - 1}}A - \left( {n - 1} \right){\rm I}

C

{A^n} = nA + \left( {n - 1} \right){\rm I}

D

{A^n} = {2^{n - 1}}A + \left( {n - 1} \right){\rm I}

Correct Answer

Option A

Solution

Given

A = \left[ \begin{array}{ll}1 & 0 \\ 1 & 1 \end{array} \right]

$\therefore$

A \times A

=

{A^2}

=

\left[ \begin{array}{ll}1 & 0 \\ 2 & 1 \end{array} \right]

and

{A^3}

=

{A^2} \times A

=

\left[ \begin{array}{ll}1 & 0 \\ 3 & 1 \end{array} \right]

So we can say

{A^n}

=

\left[ \begin{array}{ll}1 & 0 \\ n & 1 \end{array} \right]

Now

nA - \left( {n - 1} \right){\rm I}

=

\left[ \begin{array}{ll}n & 0 \\ n & n \end{array} \right]

-

\left[ \begin{array}{ll}{n - 1} & 0 \\ 0 & {n - 1} \end{array} \right]

=

\left[ \begin{array}{ll}1 & 0 \\ n & 1 \end{array} \right]

=

{A^n}

$\therefore$

{A^n} = nA - \left( {n - 1} \right){\rm I}

Q106

The term independent of x in the expansion of

(1 - {x^2} + 3{x^3}){\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}},\,x \ne 0

is :

A

{7 \over {40}}

B

{33 \over {200}}

C

{39 \over {200}}

D

{11 \over {50}}

Correct Answer

Option B

Solution

General term of Binomial expansion

{\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}

is

{T_{r + 1}} = {}^{11}{C_r}\,.\,{\left( {{5 \over 2}{x^3}} \right)^{11 - r}}\,.\,{\left( { - {1 \over {5{x^2}}}} \right)^r}

= {}^{11}{C_r}\,.\,{\left( {{5 \over 2}} \right)^{11 - r}}\,.\,{\left( { - {1 \over 5}} \right)^r}\,.\,{x^{33 - 5r}}

In the term,

\left( {1 - {x^2} + 3{x^3}} \right){\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}

Term independent of x is when (1)

33 - 5r = 0

\Rightarrow r = {{33} \over 5}\,\, \notin

integer (2)

33 - 5r = -2

\Rightarrow 5r = 35

\Rightarrow r = 7\,\, \in

integer (3)

33 - 5r = - 3

\Rightarrow 5r = 36

\Rightarrow r = {{36} \over 5}\,\, \notin

integer $\therefore$ Only for r = 7 independent of x term possible. $\therefore$ Independent of x term

= - \left( {{}^{11}{C_7}{{\left( {{5 \over 2}} \right)}^4}\,.\,{{\left( { - {1 \over 5}} \right)}^7}} \right)

= - \left( {{{11\,.\,10\,.\,9\,.\,8} \over {4\,.\,3\,.\,2\,.\,1}}\,.\,{{{5^4}} \over {{2^4}}}\,.\, - {1 \over {{5^7}}}} \right)

= {{11\,.\,10\,.\,3} \over {{2^4}\,.\,{5^3}}}

= {{11\,.\,3} \over {{2^3}\,.\,{5^2}}}

= {{33} \over {200}}

Q107

If the coefficients of x7 in

{\left( {{x^2} + {1 \over {bx}}} \right)^{11}}

and x

-

7 in

{\left( {{x} - {1 \over {bx^2}}} \right)^{11}}

, b

\ne

0, are equal, then the value of b is equal to :

A 2

B

-

1

C 1

D

-

2

Correct Answer

Option C

Solution

Coefficient of x7 in

{\left( {{x^2} + {1 \over {bx}}} \right)^{11}}

: General Term =

{}^{11}{C_r}{({x^2})^{11 - r}}.{\left( {{1 \over {bx}}} \right)^r}

=

{}^{11}{C_r}{x^{22 - 3r}}.{1 \over {{b^r}}}

22 - 3r = 7

r = 5

$\therefore$ Required Term =

{}^{11}{C_5}.{1 \over {{b^5}}}.{x^7}

Coefficient of x $-$ 7 in

{\left( {x - {1 \over {b{x^2}}}} \right)^{11}}

: General Term =

{}^{11}{C_r}{(x)^{11 - r}}.{\left( { - {1 \over {b{x^2}}}} \right)^r}

=

{}^{11}{C_r}{x^{11 - 3r}}.{{{{( - 1)}^r}} \over {{b^r}}}

11 - 3r = - 7

$\therefore$

r = 6

$\therefore$ Required Term =

{}^{11}{C_6}.{1 \over {{b^6}}}{x^{ - 7}}

According to the question,

{}^{11}{C_5}.{1 \over {{b^5}}} = {}^{11}{C_6}.{1 \over {{b^6}}}

Since, b $\ne$ 0 $\therefore$ b = 1

Q108

If

\dfrac{1}{n+1}{ }^{n} \mathrm{C}_{n}+\dfrac{1}{n}{ }^{n} \mathrm{C}_{n-1}+\ldots+\dfrac{1}{2}{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{0}=\dfrac{1023}{10}

then

n

is equal to :

A 9

B 6

C 7

D 8

Correct Answer

Option A

Solution

\frac{1}{n+1}{ }^{n} \mathrm{C}_{n}+\frac{1}{n}{ }^{n} \mathrm{C}_{n-1}+\ldots+\frac{1}{2}{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{0}=\frac{1023}{10}

\begin{aligned} & \Rightarrow \sum_{r=0}^n \frac{1}{r+1}{ }^n C_r=\frac{1023}{10} \\\\ & \quad\left( \because{ }^{n+1} C_{r+1}=\frac{n+1}{r+1}{ }^n C_r\right) \\\\ & \Rightarrow \sum_{r=0}^n \frac{1}{n+1}{ }^{n+1} C_{r+1}=\frac{1023}{10} \\\\ & \Rightarrow \frac{1}{n+1}\left[{ }^{n+1} C_1+{ }^{n+1} C_2+\ldots+{ }^{n+1} C_{n+1}\right]=\frac{1023}{10} \\\\ & \Rightarrow \frac{2^{n+1}-1}{n+1}=\frac{1023}{10}=\frac{2^{10}-1}{10} \\\\ & \Rightarrow n+1=10 \\\\ & \Rightarrow n=9 \end{aligned}

Q109

The remainder left out when

{8^{2n}} - {\left( {62} \right)^{2n + 1}}

is divided by 9 is :

A 2

B 7

C 8

D 0

Correct Answer

Option A

Solution

{8^{2n}} - {\left( {62} \right)^{2n + 1}}

=

{\left( {{8^2}} \right)^n} - {\left( {62} \right)^{2n + 1}}

=

{\left( {1 + 63} \right)^n} - {\left( {1 - 63} \right)^{2n + 1}}

=

\left( {1 + n.63 + {}^n{C_2}{{.63}^2} + ......} \right)

+

\left( {1 + {}^{2n + 1}{C_1}.\left( { - 63} \right) + {}^{2n + 1}{C_2}.{{\left( { - 63} \right)}^2} + ......} \right)

= 2 + 63

\left[ {\left( {n + {}^n{C_2} + ....} \right) + \left( { - {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2}.63 + ......} \right)} \right]

= 63 $\times$ [Some integral value] + 2 63 $\times$ [Some integral value] + 2 by dividing with 9 we will get 2 as remainder as 63 is multiple of 9.

Q110

A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of

{\left( {{2^{1/3}} + {1 \over {2{{\left( 3 \right)}^{1/3}}}}} \right)^{10}}

is :

A 1 : 2(6)1/3

B 1 : 4(6)1/3

C 2(36)1/3 : 1

D 4(36)1/3 : 1

Correct Answer

Option D

Solution

{{{T_5}} \over {T_5^1}} = {{{}^{10}{C_4}{{\left( {{2^{1/3}}} \right)}^{10 - 4}}{{\left( {{1 \over {2{{\left( 3 \right)}^{1/3}}}}} \right)}^4}} \over {{}^{10}{C_4}{{\left( {{1 \over {2\left( {{3^{1/3}}} \right)}}} \right)}^{10 - 4}}{{\left( {{2^{1/3}}} \right)}^4}}} = 4.{\left( {36} \right)^{1/3}}