Question 1: Using binomial theorem, write down the expansions of the following:

i) (2x+3y)^5         ii) (2x-3y)^4         iii) \Big( x - \frac{1}{x} \Big)^6         iv) (1-3x)^7

v) \Big( ax - \frac{b}{x} \Big)^6         vi) \Big( \sqrt{\frac{x}{a}} - \sqrt{ \frac{a}{x}}  \Big)^6         vii) ( \sqrt[3]{x} - \sqrt[3]{a} )^6        

viii) ( 1 + 2x - 3x^2)^5         ix) \Big( x+1- \frac{1}{x} \Big)^3         x) ( 1 - 2x + 3x^2)^3

Answer:

i) (2x+3y)^5

= ^5 C_0  (2x)^5 (3y)^0 +  ^5 C_1 (2x)^4 (3y)^1 + \ldots + ^5 C_0  (2x)^0 (3y)^5

= 32x^5 + 5 \times 16 x^4 \times 3y + 10 \times 8x^3 \times 9y^2 + 10 \times  4x^2 \times  27y^3 + 5 \times 3x \times 81y^2 + 243y^5

= 32x^5 + 240x^4y + 720x^3y^2 + 1080 x^2 y^3 + 810 xy^4 + 243 y^5

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ii) (2x-3y)^4

= ^4 C_0  (2x)^4 (3y)^0 - ^4 C_1  (2x)^3 (3y)^1 + ^4 C_2  (2x)^2 (3y)^2 - ^4 C_3  (2x)^1 (3y)^3 +^4 C_4  (2x)^0 (3y)^4

= 16x^4 - 4 \times  8x^3 \times  3y + 6 \times  4x^2 \times 9y^2 - 4 \times 2x \times 27y^3 + 81y^4

= 16x^4 - 96 x^3 y + 216 x^2y^2 - 216 xy^3 + 81y^4

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iii) \Big( x - \frac{1}{x} \Big)^6

= ^6 C_0  (x)^6 (  \frac{1}{x}  )^0 - ^6 C_1  (x)^5 (  \frac{1}{x}  )^1 + ^6 C_2  (x)^4 (  \frac{1}{x}  )^2 -  ^6 C_3  (x)^3 ( \frac{1}{x} )^3 + ^6 C_4  (x)^2 (\frac{1}{x}  )^4 \\ \\  { \hspace{5.0cm}  - ^6 C_5  (x)^1  (  \frac{1}{x}  )^5 + ^6 C_6  (x)^0 (  \frac{1}{x}  )^6}

= x^6 - 6x^5 \times  \frac{1}{x} + 15x^4 \times  \frac{1}{x^2} - 20x^3 \times  \frac{1}{x^3} + 15 x^2 \times  \frac{1}{x^4} - 6x \times  \frac{1}{x^5} + \frac{1}{x^6}

= x^6 - 6x^4 + 15x^2 - 20 + \frac{15}{x^2} - \frac{6}{x^4} + \frac{1}{x^6}

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iv) (1-3x)^7

= ^7 C_0  (3x)^0 - ^7 C_1  (3x)^1 + ^7 C_2  (3x)^2 - ^7 C_3  (3x)^3 +^7 C_4  (3x)^4 - ^7 C_5  (3x)^5 +^7 C_6  (3x)^6 \\ \\  { \hspace{12.0cm} - ^7 C_7  (3x)^7}

= 1 - 7 \times 3x + 21 \times  9x^2 - 35 \times  27 x^3 + 35 \times  81x^4 - 21 \times  24x^5 + 7 \times  729x^6 - 2187 x^7

= 1 - 21 x +189x^2 - 945x^3 + 2835x^4 - 5103x^5 + 5103x^6- 2187x^7 

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v) \Big( ax - \frac{b}{x} \Big)^6

= ^6 C_0  (ax)^6 (\frac{b}{x})^0 - ^6 C_1  (ax)^5 (\frac{b}{x})^1 + ^6 C_2  (ax)^4 (\frac{b}{x})^2 - ^6 C_3  (ax)^3 (\frac{b}{x})^3  \\ \\  { \hspace{5.0cm} + ^6 C_4  (ax)^2 (\frac{b}{x})^4 - ^6 C_5  (ax)^1 (\frac{b}{x})^5 +  ^6 C_6  (ax)^0 (\frac{b}{x})^6 }

= a^6x^6 - 6a^5x^5 ( \frac{b}{x}) + 15a^4x^4 ( \frac{b^2}{x^2}) - 20 a^3b^3 ( \frac{b^3}{x^3}) + 15a^2x^2 ( \frac{b^4}{x^4}) - 6ax ( \frac{b^5}{x^5}) + ( \frac{b^6}{x^6})

= a^6x^6 - 6a^5x^4b + 15a^4x^2b^2-20a^3b^3+15 (\frac{a^2b^4}{x^2} ) - 6 (\frac{ab^5}{x^4} ) + ( \frac{b^6}{x^6})

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vi) \Big( \sqrt{ \frac{x}{a}} - \sqrt{ \frac{a}{x}} \Big)^6

= ^6 C_0  (\sqrt{\frac{x}{a}})^6 (\sqrt{\frac{a}{x}})^0 - ^6 C_1  (\sqrt{\frac{x}{a}})^5 (\sqrt{\frac{a}{x}})^1 +^6 C_2  (\sqrt{\frac{x}{a}})^4 (\sqrt{\frac{a}{x}})^2 \\ \\  { \hspace{0.0cm} - ^6 C_3  (\sqrt{\frac{x}{a}})^3 (\sqrt{\frac{a}{x}})^3  +^6 C_4  (\sqrt{\frac{x}{a}})^2 (\sqrt{\frac{a}{x}})^4 - ^6 C_5  (\sqrt{\frac{x}{a}})^1 (\sqrt{\frac{a}{x}})^5 +  ^6 C_6  (\sqrt{\frac{x}{a}})^0 (\sqrt{\frac{a}{x}})^6}

= ( \frac{x^3}{a^3} ) - 6 ( \frac{x^2}{a^2} ) + 15 ( \frac{x}{a} ) - 20 + 15 ( \frac{a}{x} ) - 6 ( \frac{a^2}{x^2} ) + ( \frac{a^3}{x^3} )

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vii) ( \sqrt[3]{x} - \sqrt[3]{a} )^6

= ^6 C_0  (\sqrt[3]{x})^6 (\sqrt[3]{a})^0 -  ^6 C_1  (\sqrt[3]{x})^5 (\sqrt[3]{a})^1 + ^6 C_2  (\sqrt[3]{x})^4 (\sqrt[3]{a})^2 -  ^6 C_3  (\sqrt[3]{x})^3 (\sqrt[3]{a})^3 \\ \\  { \hspace{3.0cm} + ^6 C_4  (\sqrt[3]{x})^2 (\sqrt[3]{a})^4 -  ^6 C_5  (\sqrt[3]{x})^1 (\sqrt[3]{a})^5 + ^6 C_0  (\sqrt[3]{x})^0 (\sqrt[3]{a})^6 } 

= x^2 - 6 x^{\frac{5}{3}} a^{\frac{1}{3}} + 15 x^{\frac{4}{3}} a^{\frac{2}{3}} - 20xa + 15 x^{\frac{2}{3}} a^{\frac{4}{3}} - 6 x^{\frac{1}{3}} a^{\frac{5}{3}} + a^2

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viii) ( 1 + 2x - 3x^2)^5 = [ (1+2x) - 3x^2 ]^5

= ^5 C_0  (1+2x)^5 (3x^2)^0 - ^5 C_1  (1+2x)^5 (3x^2)^1 +^5 C_2  (1+2x)^4 (3x^2)^2 \\ \\  { \hspace{3.0cm} - ^5 C_3  (1+2x)^3 (3x^2)^3 +^5 C_4  (1+2x)^2 (3x^2)^4 - ^5 C_5  (1+2x)^1 (3x^2)^5}

= (1+2x)^5 - 5 (1+2x)^4 \times (3x^2) + 10 (1+2x)^3 \times 9x^4 \\ \\  { \hspace{1.0cm}- 10 (1+2x)^2 \times 27x^6 + 5 (1+2x) \times 81x^8 - 243 x^{10} }     … … … i)

Now we calculate the following:

(1+2x)^5 = ^5 C_0 (2x)^5 +^5 C_1 (2x)^4 +^5 C_2 (2x)^3 +^5 C_3 (2x)^2 +^5 C_4 (2x)^1 +^5 C_5 (2x)^0

= 32x^5 + 80x^4 + 80^3 + 40x^2 + 10x + 1

(1+2x)^4 = ^4 C_0 (2x)^4 +^4 C_1 (2x)^3 +^4 C_2 (2x)^2 +^4 C_3 (2x)^1 +^4 C_4 (2x)^0

= 16x^4 + 32^3 + 24x^2 + 8x + 1

(1+2x)^3 = ^3 C_0 (2x)^3 +^3 C_1 (2x)^2 +^3 C_2 (2x)^1 +^3 C_3 (2x)^0

=  8^3 + 12x^2 + 6x + 1

(1+2x)^2 = ^2 C_0 (2x)^2 +^2 C_1 (2x)^1 +^2 C_2 (2x)^0

=   4x^2 + 4x + 1 

Substituting in i) we get

= [32x^5 + 80x^4 + 80^3 + 40x^2 + 10x + 1] - 5 [16x^4 + 32^3 + 24x^2 + 8x + 1] \times (3x^2) \\ \\  { \hspace{0.0cm}+ 10 [8^3 + 12x^2 + 6x + 1] \times 9x^4 - 10 [4x^2 + 4x + 1] \times 27x^6 + 5 (1+2x) \times 81x^8 - 243 x^{10} }

= 1+10x+25x^2-40x^3-190x^4+92x^5+570x^6-360x^7-675x^8+810x^9-243x^{10}

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ix) \Big( x+1- \frac{1}{x} \Big)^3

= ^3 C_0 (x+1)^3 ( \frac{1}{x} )^0 - ^3 C_1 (x+1)^2 ( \frac{1}{x} )^1 + ^3 C_2 (x+1)^1 ( \frac{1}{x} )^2 - ^3 C_3 (x+1)^0 ( \frac{1}{x} )^3

= ( x+1)^3 - 3 \Big[ \frac{(x+1)^2}{x} \Big]+ 3 \Big[ \frac{(x+1)}{x^2} \Big] - \frac{1}{x^3}

= [^3 C_0 x^3 + ^3 C_1 x^2 + ^3 C_2 x^1 + ^3 C_3 x^0 ] - 3 \Big[  \frac{x^2 + 1 + 2x}{x} \Big] + 3 \Big[ \frac{(x+1)}{x^2} \Big] - \frac{1}{x^3}

= (x^3+3x^2+3x+1) - \Big( 3x + \frac{3}{x} + 6 \Big) + \Big( \frac{3}{x} + \frac{3}{x^2} \Big) - \frac{1}{x^3}

= x^3 + 3x^2 -5 + \frac{3}{x^2} - \frac{1}{x^3}

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x) ( 1 - 2x + 3x^2)^3

= ^3 C_0 (1-2x)^3 (3x^2)^0 + ^3 C_1 (1-2x)^2 (3x^2)^1 + ^3 C_2 (1-2x)^1 (3x^2)^2 \\ \\ {\hspace{10.0cm}+ ^3 C_0 (1-2x)^0 (3x^2)^1 }

= (1-2x)^3 + 9x^2 ( 1- 2x)^2 + 27x^4 (1-2x) + 27x^6

= 1 - 6x + 21 x^2 - 44x^3 - 54x^5 + 27x^6 

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Question 2: Evaluate the following:

i) ( \sqrt{x+1} + \sqrt{x-1})^6 +  ( \sqrt{x+1} - \sqrt{x-1})^6

ii) ( x + \sqrt{x^2-1})^6 + ( x - \sqrt{x^2-1})^6      iii) (1+2\sqrt{x})^5-(1-2\sqrt{x})^5

iv) (\sqrt{2}+1)^6 + (\sqrt{2}-1)^6       v) (3 + \sqrt{2})^5 - (3 - \sqrt{2})^5

vi) (2 + \sqrt{3})^7 + (2 - \sqrt{3})^7       vii) (\sqrt{3} + 1)^5 - (\sqrt{3} - 1)^5

viii) (0.99)^5 + ( 1.01)^5        ix) (\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6

x) \{a^2 + \sqrt{a^2-1} \}^4 + \{a^2 - \sqrt{a^2-1} \}^4

Answer:

i) ( \sqrt{x+1} + \sqrt{x-1})^6 +  ( \sqrt{x+1} - \sqrt{x-1})^6

= 2 [ ^6 C_0 (\sqrt{x+1})^6 (\sqrt{x-1})^0  + ^6 C_2 (\sqrt{x+1})^4 (\sqrt{x-1})^2 \\ \\ { \hspace{3.0cm} + ^6 C_4 (\sqrt{x+1})^2 (\sqrt{x-1})^4 + ^6 C_6 (\sqrt{x+1})^0 (\sqrt{x-1})^4 ] }

= 2 [ (x+1)^3 + 15(x+1)^2 (x-1) + 15(x+1)(x-1)^2 + ( x-1)^3]

= 2 [ x^3 + 1 + 3x + 3x^2 + 15 ( x^2 + 2x+1)(x-1) + 15(x+1)^2(x^2+1-2x) \\ \\ { \hspace{10.0cm}+ x^3 - 1 + 3x -3x^2 ] }

= 2 [ 2x^3 + 6x + 15x^3 - 15x^2 +30x^2 - 30x + 15x - 15 + 15x^3 + 15x^2 \\ \\ { \hspace{10.0cm}- 30x^2 -30x + 15x+15] }

= 2 [ 32x^3 - 24x]

= 16x[4x^3 - 3]

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ii) ( x + \sqrt{x^2-1})^6 + ( x - \sqrt{x^2-1})^6

= 2 [ ^6 C_0 x^6 (\sqrt{x^2-1})^0  + ^6 C_2 x^4 (\sqrt{x^2-1})^2 + ^6 C_4 x^2 (\sqrt{x^2-1})^4 + ^6 C_6 x^0 (\sqrt{x^2-1})^6 ]

= 2 [ x^6 + 15x^4 (x^2 -1)+ 15x^2 (x^2 -1)^2 + (x^2 -1)^3 ]

= 2 [ x^6 + 15x^6 - 15x^4 + 15x^2 (x^4 - 2x^2+1) + ( x^6 - 1 + 3x^2 - 3x^4) ]

= 2 [ x^6 + 15x^6 - 15x^4 + 15x^6-30x^4+15x^2+x^6  - 1 + 3x^2 - 3x^4) ]

= 64x^6 - 96x^4 + 36x^2 - 2

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iii) (1+2\sqrt{x})^5-(1-2\sqrt{x})^5

= 2 [ ^5 C_0 (2\sqrt{x})^0 + ^5 C_2 (2\sqrt{x})^2 + ^5 C_4 (2\sqrt{x})^4 ]

= 2 [ 1 + 10 \times 4x + 5 \times 16x^2]

= 2 ( 1 + 40x + 80x^2]

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iv) (\sqrt{2}+1)^6 + (\sqrt{2}-1)^6

= 2 [ ^6 C_0 (\sqrt{2})^6 + ^6 C_2 (\sqrt{2})^4 + ^6 C_4 (\sqrt{2})^2 + ^6 C_6 (\sqrt{2})^0 ]

= 2 [ 8 + 15 \times 4 + 15 \times 2 + 1]

= 2 \times 99

= 198

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v) (3 + \sqrt{2})^5 - (3 - \sqrt{2})^5

= 2 [ ^5 C_1 ( 3)^4 (\sqrt{2})^1 + ^5 C_3 ( 3)^2 (\sqrt{2})^3 + ^5 C_5 ( 3)^0 (\sqrt{2})^4 ]

= 2 [ 5 \times 81 \times \sqrt{2} + 10 \times 9 \times 2\sqrt{2} + 4 \sqrt{2} ]

= 2\sqrt{2} (405 + 180 + 4 )

= 1178 \sqrt{2}

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vi) (2 + \sqrt{3})^7 + (2 - \sqrt{3})^7

= 2 [ ^7 C_0 (2)^7(\sqrt{3})^0 + ^7 C_2 (2)^5(\sqrt{3})^2 + ^7 C_4 (2)^3(\sqrt{3})^4 + ^7 C_5 (2)^1(\sqrt{3})^6 ]

= 2 [ 128 + 21\times 32 \times 3 + 35 \times 8\times 9 + 7\times 2 \times 27 ]

=2 ( 128 + 2016 + 2520 + 378]

= 10084

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vii) (\sqrt{3} + 1)^5 - (\sqrt{3} - 1)^5

= 2 [ ^5 C_1 (\sqrt{3})^4 + ^5 C_3 (\sqrt{3})^2 + ^5 C_5 (\sqrt{3})^0 ]

= 2 [ 5 \times 9 + 10 \times 3 + 1 ]

= 2 \times 76

= 152

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viii) (0.99)^5 + ( 1.01)^5

= ( 1 - 0.01)^5 + ( 1+0.01)^5

= 2 [ ^5 C_0 (0.01)^0 + ^5 C_2 (0.01)^2 + ^5 C_4 (0.01)^4 ]

= 2 [ 1 + 10 \times 0.0001 + 5 \times 0.00000001 ]

= 2 \times 1.00100005

= 2.0020001

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ix) (\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6

= 2 [ ^6 C_1 (\sqrt{3})^5 (\sqrt{2})^1 + ^6 C_3 (\sqrt{3})^3 (\sqrt{2})^3 + ^6 C_5 (\sqrt{3})^1 (\sqrt{2})^5 ]

= 2 [ 6 \times 9 \sqrt{3} \times \sqrt{2} + 20 \times 3 \sqrt{3} \times 2\sqrt{2} + 6 \times  \sqrt{3} \times 4\sqrt{2}]

= 2 [ \sqrt{6}(54+120+24)]

= 396\sqrt{6}

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x) \{a^2 + \sqrt{a^2-1} \}^4 + \{a^2 - \sqrt{a^2-1} \}^4

= 2 [ ^4 C_0 (a^2)^4 (\sqrt{a^2-1})^0 + ^4 C_2 (a^2)^2 (\sqrt{a^2-1})^2 + ^4 C_4 (a^2)^0 (\sqrt{a^2-1})^4 ]

= 2[ a^8 + 6a^4(a^2-1) + (a^2-1)^2]

= 2[a^8+6a^6-6a^4+a^4+1-2a^2]

= 2a^2 + 12a^6 - 10a^4 - 4a^2 + 2

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Question 3: Find (a+b)^4 - ( a-b)^4 . Hence, evaluate (\sqrt{3}+\sqrt{2})^4 - ( \sqrt{3}-\sqrt{2})^4 .

Answer:

(a+b)^4 - ( a-b)^4 = 2 [^4 C_1 a^3 b^1 + ^4 C_3 a^1 b^3 ] = 2 [ 4a^3b+4ab^3] = 8ab(a^2+b^2)

Now we see, a = \sqrt{3} and b = \sqrt{2} 

In (\sqrt{3}+\sqrt{2})^4 - ( \sqrt{3}-\sqrt{2})^4

= 8 \sqrt{3} \sqrt{2} [ (\sqrt{3})^2 + (\sqrt{2})^2]

= 40 \sqrt{6}

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Question 4: Find (x+1)^6 + (x-1)^6 . Hence, evaluate ( \sqrt{2}+1)^6+(\sqrt{2}-1)^6

Answer:

(x+1)^6 + (x-1)^6 = 2[ ^6 C_0 x^6 +^6 C_2 x^4 +^6 C_4 x^2 +^6 C_6 x^0 ]

= 2 [ x^6 + 15x^4+15x^2+1]

We see, 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 [ 8 + 60 + 30 + 1]

= 198

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Question 5: Using binomial theorem evaluate each of the following:

i) (96)^3      ii) (102)^5      iii) (101)^4      iv)  (98)^5

Answer:

i) (96)^3 = (100-4)^3

= ^3 C_0 (100)^3 ( -4)^0 +^3 C_1 (100)^2 ( -4)^1 +^3 C_2 (100)^1 ( -4)^2 +^3 C_3 (100)^0 ( -4)^3 

= 100^3 - 3 \times 100^2  \times 4 + 3 \times 100 \times 4^2 - 4^3 

= 1000000 - 120000 + 4800-64 

= 1004800-120064 

= 884736 

ii) (102)^5 = ( 100+2)^5

= ^5 C_0 (100)^5 ( 2)^0 + ^5 C_1 (100)^4 ( 2)^1 + ^5 C_2 (100)^3 ( 2)^2 + ^5 C_3 (100)^2 ( 2)^3 \\ \\ { \hspace{6.0cm}+ ^5 C_4 (100)^1 ( 2)^4 + ^5 C_5 (100)^0 ( 2)^5}

= 100^5 + 5 \times 100^4 \times 2 + 10 \times 100^3 \times 2^2 + 10 \times 100^2 \times 2^3 + 5 \times 100 \times 2^4 + 2^5

= 10000000000 + 1000000000 + 40000000+ 800000+8000+32

= 11040808032

iii) (101)^4 = (100+1)^4

= ^4 C_0 (100)^4 + ^4 C_1 (100)^3 + ^4 C_2 (100)^2 + ^4 C_3 (100)^1 + ^4 C_4 (100)^0 

= 100^4 + 4 \times 100^3 + 6 \times 100^2 + 400 + 1 

= 100000000 + 4000000+60000+400+1 

= 104060401 

iv)  (98)^5 = ( 100 - 2)^5

= ^5 C_0 (100)^5 (-2)^0 + ^5 C_1 (100)^4 (-2)^1 + ^5 C_2 (100)^3 (-2)^2 + ^5 C_3 (100)^2 (-2)^3 + ^5 C_4 (100)^1 (-2)^4 + ^5 C_5 (100)^0 (-2)^5

= 100^5 - 10 \times 100^4 + 40 \times 100^3 - 80 \times 100^2  + 80 \times 100 - 32

= 10000000000 - 1000000000+40000000-800000+8000-32

= 10040008000-1000800032

= 9039207968 

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Question 6: Using binomial theorem prove that 2^{3n}-7n-1 is divisible by 49 , where n \in N

Answer:

2^{3n}-7n-1

= 8^n - 7n -1

= (1+7)^n - 7n - 1

= [ ^n C_0 7^0+ ^n C_1 7^1+ ^n C_2 7^2+ \ldots + ^n C_n 7^n ] - 7n - 1

= 1+7n + [ ^n C_2 7^2+ \ldots + ^n C_n 7^n ] - 7n - 1

= [ ^n C_2 7^2+ \ldots + ^n C_n 7^n ]

= 49 [ ^n C_2 + \ldots + ^n C_n 7^{n-2} ]

Hence 2^{3n}-7n-1 is divisible by 49 , where n \in N

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Question 7: Using binomial theorem prove that 3^{2n+2}-8n-9 is divisible by 64, n \in N

Answer:

3^{2n+2}-8n-9

= 9^{n+1} - 8n - 9

= (1+8)^{n+1} - 8n - 9

= [ ^{n+1} C_0 8^0 + ^{n+1} C_1 8^1 + ^{n+1} C_2 8^2 + \ldots + ^{n+1} C_{n+1} 8^{n+1} ] - 8n - 9

= 1 + 8(n+1) + [  ^{n+1} C_2 8^2 + \ldots + ^{n+1} C_{n+1} 8^{n+1} ] - 8n - 9

= [  ^{n+1} C_2 8^2 + \ldots + ^{n+1} C_{n+1} 8^{n+1} ]

= 64 [  ^{n+1} C_2 + \ldots + ^{n+1} C_{n+1} 8^{n-1} ]

Hence 3^{2n+2}-8n-9 is divisible by 64, n \in N

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Question 8: If n is a positive integer, prove that 3^{3n}-26n-1 is divisible by 676 .

Answer:

3^{3n}-26n-1

= 27^n - 26n - 1

= (1+26)^n - 26n - 1

= [ ^{n} C_0 (26)^0 + ^{n} C_1 (26)^1 + ^{n} C_2 (26)^2 + \ldots + ^{n} C_{n+1} (26)^{n} ]- 26n - 1

= 1+26n + [^{n} C_2 (26)^2 + \ldots + ^{n} C_{n+1} (26)^{n} ]- 26n - 1

= [^{n} C_2 (26)^2 + \ldots + ^{n} C_{n+1} (26)^{n} ]

= 676 [^{n} C_2  + \ldots + ^{n} C_{n+1} (26)^{n-2} ]

Hence 3^{3n}-26n-1 is divisible by 676 .

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Question 9: Using binomial theorem, indicate which is larger (1.1)^{10000} or 10000 .

Answer:

(1.1)^{10000}

= (1+0.1)^{10000}

=  ^{10000} C_0 (0.1)^0 + ^{10000} C_1 (0.1)^1 +^{10000} C_2 (0.1)^2 + \ldots + ^{10000} C_{10000} (0.1)^{10000}

= 1 + 10000(0.1) + \text{other positive terms}

= 10001 + \text{other positive terms}

Therefore (1.1)^{10000} is greater than 10000 .

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Question 10: Using binomial theorem, determine which is smaller (1.2)^{4000} or 800 .

Answer:

(1.2)^{4000}

= (1+0.2)^{4000}

= ^{4000} C_0 (0.2)^0 + ^{4000} C_1 (0.2)^1 +^{4000} C_2 (0.2)^2 + \ldots + ^{4000} C_{4000} (0.1)^{4000}

= 1 + 4000(0.2) + \text{other positive terms}

= 801  + \text{other positive terms}

Therefore 800 is the smaller term.

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Question 11: Find the value of (1.01)^{10}+(1-0.01)^{10} correct to 7 decimal places.

Answer:

(1.01)^{10}+(1-0.01)^{10}

= (1+ 0.01)^{10}+(1-0.01)^{10}

= 2 [ ^{10} C_0 (0.01)^0 + ^{10} C_2 (0.01)^2 +^{10} C_4 (0.01)^4 +^{10} C_6 (0.01)^6 +^{10} C_8 (0.01)^8 +^{10} C_{10} (0.01)^{10} ]

= 2 [ 1 + 45 ( 0.0001) + 210 ( 0.0000001) + \ldots ]

= 2.0090042 + \ldots

Hence (1.01)^{10}+(1-0.01)^{10} correct to 7 decimal places is 2.0090042

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Question 12: Show that 2^{4n+4}- 15n - 16 , where n \in N is divisible by 225 .

Answer:

2^{4n+4}- 15n - 16

= {16}^{n+1}- 15n - 16

= (1+15)^{n+1}- 15n - 16

= [^{n+1} C_0 (15)^0 +^{n+1} C_1 (15)^1 +^{n+1} C_2 (15)^2 + \ldots + ^{n+1} C_{n+1} (15)^{n+1}   ]- 15n - 16

= 1 + 15( n+1) + [^{n+1} C_2 (15)^2 + \ldots + ^{n+1} C_{n+1} (15)^{n+1}   ] - 15n - 16

= [^{n+1} C_2 (15)^2 + \ldots + ^{n+1} C_{n+1} (15)^{n+1}   ]

= 225 [^{n+1} C_2  + \ldots + ^{n+1} C_{n+1} (15)^{n-1}   ]

Hence 2^{4n+4}- 15n - 16 , where n \in N is divisible by 225 .