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MATHS 3 Marks Question

Binomial Theorem · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 15 questions.

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Question 23 Marks
Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of ${\left( {\sqrt[4] 2 + \frac{1}{{\sqrt [4]{3} }}} \right)^n}$ is $\sqrt 6 :1$.
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Question 53 Marks
If a and b are distinct integers, prove that a - b is a factor of an - bn, whenever n is a positive integer.
[Hint write an = (a – b + b)n and expand]
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Question 83 Marks
Expand the given expression ${\left( {x + \frac{1}{x}} \right)^6}$
Answer
Using binomial theorem for the expansion of ${\left( {x + \frac{1}{x}} \right)^6}$ we have
${\left( {x + \frac{1}{x}} \right)^6}$= ${ = ^6}{C_0}{(x)^6}{ + ^6}{C_1}{(x)^5}\left( {\frac{1}{x}} \right)$${ + ^6}{C_2}{(x)^4}{\left( {\frac{1}{x}} \right)^2}{ + ^6}{C_3}{(x)^3}{\left( {\frac{1}{x}} \right)^3}$
${ + ^6}{C_4}{(x)^2}{\left( {\frac{1}{x}} \right)^4}{ + ^6}{C_5}(x){\left( {\frac{1}{x}} \right)^5}$${ + ^6}{C_6}{\left( {\frac{1}{6}} \right)^6}$
$ = {x^6} + 6 \cdot {x^5} \cdot \frac{1}{x} + 15 \cdot 4{x^4} \cdot \frac{1}{{{x^2}}} + $$20 \cdot {x^3} \cdot \frac{1}{{{x^3}}} + 15 \cdot {x^2} \cdot \frac{1}{{{x^4}}} + 6 \cdot x \cdot \frac{1}{{{x^5}}} + \frac{1}{{{x^6}}}$
$ = {x^6} + 6{x^4} + 15{x^2} + 20 + \frac{{15}}{{{x^2}}} + \frac{6}{{{x^4}}} + \frac{1}{{{x^6}}}$
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Question 93 Marks
Expand the given expression ${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$
Answer
Using binomial theorem for the expansion of ${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$ we have
${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$${ = ^5}{C_0}{\left( {\frac{x}{3}} \right)^5}{ + ^5}{C_1}{\left( {\frac{x}{3}} \right)^4}\left( {\frac{1}{x}} \right)$${ + ^5}{C_2}{\left( {\frac{x}{3}} \right)^3}{\left( {\frac{1}{x}} \right)^2}{ + ^5}{C_3}{\left( {\frac{x}{3}} \right)^2}{\left( {\frac{1}{x}} \right)^3}$
${ + ^5}{C_4}\left( {\frac{x}{3}} \right){\left( {\frac{1}{x}} \right)^4}{ + ^5}{C_5}{\left( {\frac{1}{x}} \right)^5}$
$ = \frac{{{x^5}}}{{243}} + 5 \cdot \frac{{{x^4}}}{{81}} \cdot \frac{1}{x} + 10 \cdot \frac{{{x^3}}}{{27}} \cdot \frac{1}{{{x^2}}}$$ + 10 \cdot \frac{{{x^2}}}{9} \cdot \frac{1}{{{x^3}}} + 5 \cdot \frac{x}{3} \cdot \frac{1}{{{x^4}}} + \frac{1}{{{x^5}}}$
$ = \frac{{{x^5}}}{{243}} + \frac{5}{{81}}{x^3} + \frac{{10}}{{27}}x + \frac{{10}}{{9x}} + \frac{5}{{3{x^3}}} + \frac{1}{{{x^5}}}$
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Question 103 Marks
Expand the given expression (2x - 3)6
Answer
Using binomial theorem for the expansion of (2x - 3)6 we have
${(2x - 3)^6}{ = ^6}{C_0}{(2x)^6}{ + ^6}{C_1}{(2x)^5}( - 3)$${ + ^6}{C_2}{(2x)^4}{( - 3)^2}{ + ^6}{C_3}{(2x)^3}{( - 3)^3}$
${ + ^6}{C_4}{(2x)^2}{( - 3)^4}{ + ^6}{C_5}2{x^5}{( - 3)^2}$${ + ^6}{C_6}{( - 3)^6}$
= 64x6 + 6.32x5 (-3) + 15.16x4.9 + 20.8x3 (-27) + 15.4x2.81 + 6.2x (-243) + 729
= 64x6 - 576x5 + 2160x4 - 4320x3 + 4860x2 - 2916x + 729
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Question 113 Marks
Expand the given expression ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$
Answer
Using binomial theorem for the expansion of ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$ we have
${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$${ = ^5}{C_0}{\left( {\frac{2}{x}} \right)^5}{ + ^5}{C_1}{\left( {\frac{2}{x}} \right)^4}\left( {\frac{{ - x}}{2}} \right)$${ + ^5}{C_2}{\left( {\frac{2}{x}} \right)^3}{\left( {\frac{{ - x}}{2}} \right)^2}{ + ^5}{C_3}{\left( {\frac{2}{x}} \right)^2}{\left( {\frac{{ - x}}{2}} \right)^3}$
${ + ^5}{C_4}\left( {\frac{2}{x}} \right){\left( {\frac{{ - x}}{2}} \right)^4}{ + ^5}{C_5}{\left( {\frac{{ - x}}{2}} \right)^5}$
$ = \frac{{32}}{{{x^5}}} + 5 \cdot \frac{{16}}{{{x^4}}} \cdot \frac{{ - x}}{2} + 10 \cdot \frac{8}{{{x^3}}} \cdot \frac{{{x^2}}}{4}$$ + 10 \cdot \frac{4}{{{x^2}}} \cdot \frac{{ - {x^3}}}{8} + 5 \cdot \frac{2}{x} \cdot \frac{{{x^4}}}{{16}} + \frac{{ - {x^5}}}{{32}}$
$ = \frac{{32}}{{{x^5}}} - \frac{{40}}{{{x^3}}} + \frac{{20}}{x} - 5x + \frac{5}{8}{x^3} - \frac{{{x^5}}}{{32}}$
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Question 123 Marks
Show that 9n+1 - 8n - 9 is divisible by 64 whenever n is a positive integer.
Answer
We have 9n+1 = (1 + 8)n+1
${ = ^{n + 1}}{C_0}{ + ^{n + 1}}{C_1}(8){ + ^{n + 1}}{C_2}{(8)^2} + $$^{n + 1}{C_3}{(8)^3} + ...{ + ^{n + 1}}{C_{n + 1}}{(8)^{n + 1}}$
$ = 1 + (n + 1) \times 8{ + ^{n + 1}}{C_2}{(8)^2}$${ + ^{n + 1}}{C_3}{(8)^3} + ...{ + ^{n + 1}}{C_{n + 1}}{(8)^{n + 1}}$
$ = 1 + 8n + 8{ + ^{n + 1}}{C_2}{(8)^2}{ + ^{n + 1}}{C_3}{(8)^3} + $$....{ + ^{n + 1}}{C_{n + 1}}{(8)^{n + 1}}$

$\style{font-family:Tahoma}{\style{font-size:8px}{=\;9\;+8n\;+^{n+1}C_2\left(8\right)^2\;+^{n+1}C_3\left(8\right)^3\;+........\;+^{n+1}C_{n+1}(8)^{n+1}}}$
$ \style{font-family:Tahoma}{\style{font-size:8px}{\Rightarrow9^{n+1}-8n-9=^{n+1}C_2{(8)^2}+^{n+1}C_3{(8)^3}}}$$ + ...{ + ^{n + 1}}{C_{n + 1}}{(8)^{n + 1}}$
$ = 64{[^{n + 1}}{C_2}{ + ^{n + 1}}{C_3} \cdot 8$$ + ...{ + ^{n + 1}}{C_{n + 1}} \cdot {8^{n + 1}}]$
which show that 9n+1 - 8n-9 is divisible by 64 wherever n is a positive integer

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Question 133 Marks
Find (x + 1)6 + (x - 1)6. Hence or otherwise evaluate ${(\sqrt 2 + 1)^6} + {(\sqrt 2 - 1)^6}$
Answer
(x + 1)6 + (x - 1)6 =$ = {[^6}{C_0}{x^6}{ + ^6}{C_1}{x^5}{ + ^6}{C_2}{x^4}{ + ^6}{C_3}{x^3}{ + ^6}{C_4}{x^2}{ + ^6}{C_5}x{ + ^6}{C_6}]$
$ + {[^6}{C_0}{x^6}{ + ^6}{C_1}{x^5}( - 1){ + ^6}{C_2}{x^4}{( - 1)^2}{ + ^6}{C_3}{x^3}{( - 1)^3}$${ + ^6}{C_4}{x^2}{( - 1)^4}{ + ^6}{C_5}x{( - 1)^5}{ + ^6}{C_6}{( - 1)^6}]$
= [x6 + 6x5 + 15x4 + 20x3 + 15x$^2$+ 6x + 1] + [x6 - 6x5 + 15x4 - 20x3 + 15x2 - 6x + 1]
= 2x6 + 30x4 + 30x2 + 2
= 2(x6 + 15x4 + 15x2 + 1)
Putting $x = \sqrt 2 $
${(\sqrt 2 + 1)^6} + {(\sqrt 2 - 1)^6} = 2[{(\sqrt 2 )^6} + 15{(\sqrt 2 )^4} + 15{(\sqrt 2 )^2} + 1]$
$ = 2\left[ {8 + 15 \times 4 + 15 \times 2 + 1} \right]$
= 2 [8 + 60 + 30 + 1]
$ = 2 \times 99 = 198$
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Question 143 Marks
Find (a + b)4 - (a - b)4. Hence, evaluate ${(\sqrt 3 + \sqrt 2 )^4} - {(\sqrt 3 - \sqrt 2 )^4}$
Answer
(a + b)4 $ = {[^4}{C_0}{a^4}{ + ^4}{C_1}{a^3}b{ + ^4}{C_2}{a^2}{b^2}$${ + ^4}{C_3}a{b^3}{ + ^4}{C_4}{b^4}]$
$\style{font-family:Tahoma}{\style{font-size:8px}{and\;\left(a-b\right)^4{\;=\lbrack^4}C_0a^4-^4C_1a^3b+^4C_2a^2{b^2}}}$${ - ^4}{C_3}a{b^3}{ + ^4}{C_4}{b^4}]$
$$$\style{font-family:Tahoma}{\style{font-size:8px}{\begin{array}{l}{\therefore\;(a+b)^4\;-\left(a-b\right)^4\;=}2\left[{}^4C_1a^3b\;+^4C_3ab^3\right]\\=2\left[4a^3b\;+4ab^3\right]\;=\;8ab\left[a^2+b^2\right]\\\therefore\left(\sqrt3\;+\sqrt2\right)^4\;-\;\left(\sqrt3\;-\sqrt2\right)^{4\;}=\;8.\sqrt3.\sqrt2\left[\left(\sqrt3\right)^2+\left(\sqrt2\right)^2\right]\\=\;8.\sqrt3.\sqrt2\left[3+2\right]\;=\;40.\sqrt3.\sqrt2\;=\;40\sqrt6\end{array}}}$
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Question 153 Marks
Expand the given expression (1 - 2x)5
Answer
Using binomial theorem for the expansion of (1 - 2x)5 we have
${(1 - 2x)^5}{ = ^5}{C_0}{ + ^5}{C_1}( - 2x)+$$^5{C_2}{( - 2x)^2}{ + ^5}{C_3}{( - 2x)^3}{ + ^5}{C_4}{( - 2x)^4}$${ + ^5}{C_5}{( - 2x)^5}$
= 1 + 5 (-2x) + 10(-2x)2 + 10(-2x)3 + 5(-2x)4 + (-2x)5
= 1 - 10x + 40x2 - 80x3 + 80x4 - 32x5
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