**Expand the expression (1 – 2x)**^{5}

Solution:

By using Binomial theorem the expression (1-2x)^{5} can be expected as

(1 – 2x)^{5}

= ^{5}C_{0}(1)^{5} – ^{5}C_{1}(1)^{4} (2x)^{1}– ^{5}C_{2}(1)^{3}(2x)^{2 }– ^{5}C_{3}(1)^{2}(2x)^{3 }– ^{5}C_{4}(1)^{1}(2x)^{4} – ^{5}C_{5}(2x)^{5}

= 1 – 5(2x) + 10(4x^{2}) – 10(8x^{3}) + 5(16x^{4}) – (32x^{5})

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

**Expand the expression (**^{2}/_{x}–^{x}/_{2})^{5}

Solution:

By using Binomial theorem the expression (^{2}/_{x} – ^{x}/_{2})^{5}can be expected as

(^{2}/_{x} – ^{x}/_{2})^{5}

= ^{5}C_{0}(^{2}/_{x})^{5} – ^{5}C_{1}(^{2}/_{x})^{4} (^{x}/_{2})^{1}– ^{5}C_{2}(^{2}/_{x})^{3}(^{x}/_{2})^{2 }– ^{5}C_{3}(^{2}/_{x})^{2}(^{x}/_{2})^{3 }– ^{5}C_{4}(^{2}/_{x})^{1}(^{x}/_{2})^{4} – ^{5}C_{5}(^{x}/_{2})^{5}

= ^{32}/_{x}^{5} – 5(^{16}/_{x}^{4})(^{x}/_{2}) + 10(^{8}/_{x}^{3})(x^{2}/_{4}) – 10(^{4}/_{x}^{2})(x^{3}/_{8})+5(^{2}/_{x})(x^{4}/_{16}) – (x^{5}/_{32})

= ^{32}/_{x}^{5} – ^{40}/_{x}^{3} + ^{20}/_{x} – 5x + ^{5}/_{8 }x^{3} – (x^{3}/_{32})

**Expand the expression (2x – 3)**^{6}

Solution:

By using Binomial theorem the expression (2x – 3)^{6} can be expected as

(2x – 3)^{6}

= ^{5}C_{0}(2x)^{5} – ^{5}C_{1}(2x)^{4} (3)^{1}– ^{5}C_{2}(2x)^{3}(3)^{2 }– ^{5}C_{3}(2x)^{2}(3)^{3 }– ^{5}C_{4}(2x)^{1}(3)^{4} – ^{5}C_{5}(3)^{5}

= 64x^{6} – 6(32x^{5})(3) + 15(16x^{4})(9) – 20(8x^{3})(27) + 15(4x^{2})(81) –6(2x)(243) +729

= 64x^{5} – 57x^{5} + 2160x^{4} – 4320x^{3} + 4860x^{2} – 2916x + 729

**Expand the expression (**^{x}/_{3}+^{1}/_{x})^{5}

Solution:

By using Binomial theorem the expression (^{x}/_{3} + ^{1}/_{x})^{5}

can be expected as

(^{x}/_{3} + ^{1}/_{x})^{5}

= ^{5}C_{0}(^{x}/_{3})^{5} – ^{5}C_{1}(^{x}/_{3})^{4} (^{1}/_{x})^{1}– ^{5}C_{2}(^{x}/_{3})^{3}(^{1}/_{x})^{2 }– ^{5}C_{3}(^{x}/_{3})^{2}(^{1}/_{x})^{3 }– ^{5}C_{4}(^{x}/_{3})^{1}(^{1}/_{x})^{4} – ^{5}C_{5}(^{1}/_{x})^{5}

= (x^{5}/_{243}) + 5(x^{4}/_{81})(^{1}/_{x}) + 10(x^{3}/_{27})(^{1}/_{x}^{2}) + 5(x^{2}/_{9})(^{1}/_{x}^{3})+5(^{x}/_{3})(^{1}/_{x}^{4}) + (^{1}/_{x}^{5})

= (x^{5}/_{243}) + (5x^{3}/_{81}) + (^{10x}/_{27}) + (^{10}/_{9x}) + (^{5}/_{3x}^{3}) + (^{1}/_{x}^{5})

**Expand the expression (x +**^{1}/_{x})^{6}

Solution:

By using Binomial theorem the expression (x + ^{1}/_{x})^{6}

can be expected as

(x + ^{1}/_{x})^{6}

= ^{5}C_{0}x^{5} – ^{5}C_{1 }x^{4} (^{1}/_{x})^{1}– ^{5}C_{2 }x^{3}(^{1}/_{x})^{2 }– ^{5}C_{3}x^{2}(^{1}/_{x})^{3 }– ^{5}C_{4}x^{1}(^{1}/_{x})^{4} – ^{5}C_{5}(^{1}/_{x})^{5}

= x^{6} +6(x)^{5}(^{1}/_{x}) + 15(x)^{4} (^{1}/_{x}^{2}) + 20(x)^{3}(^{1}/_{x}^{3}) + 15(x)^{2}(^{1}/_{x}^{4})+6(x)(^{1}/_{x}^{5}) + (^{1}/_{x}^{6})

= x^{6} + 6x^{4} + 15x^{2} + 20 + (^{15}/_{x}^{2}) + (^{6}/_{x}^{4}) + (^{1}/_{x}^{6})

**Using binomial theorem, evaluate (96)**^{3}

Solution:

96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then binomial theorem can be applied.

(96)^{3} = (100 – 4)^{3}

= ^{3}C_{0} (100)^{3} – ^{3}C_{1} (100)^{2} 4 + ^{3}C_{2} (100)(4)^{2} – ^{3}C_{3} (4)^{3}

= (100)^{3} – 3(100)^{2} 4 +3(100)(4)^{2} – (4)^{3}

= 1000000 – 120000 + 4800 – 64

= 884736

**Using binomial theorem, evaluate (102)**^{5}

Solution:

102 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then binomial theorem can be applied.

It can be written that, 102 = 100 + 2

(102)^{5} = (100 + 2)^{5}

= ^{5}C_{0} (100)^{5} + ^{5}C_{1} (100)^{4} 2 + ^{5}C_{2} (100)^{3}(2)^{2} + ^{5}C_{3} (100)^{2}(2)^{3} + ^{5}C_{4} (100)^{1}(2)^{4} + + ^{5}C_{5}(2)^{5}

= (100)^{5} + 5(100)^{4} 2 +10 (100)^{3}(2)^{2} + 10(100)^{2}(2)^{3} + 5(100)(2)^{4} + (2)^{5}

= 11040808032

**Using binomial theorem, evaluate (101)**^{4}

Solution:

101 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then binomial theorem can be applied.

It can be written that, 101 = 100 + 1

(101)^{4} = (100 + 1)^{4}

= ^{4}C_{0} (100)^{4} + ^{4}C_{1} (100)^{3} (1) + ^{4}C_{2} (100)^{2}(1)^{2} + ^{4}C_{3} (100)(1)^{3} + ^{4}C_{4} (1)^{4}

= (100)^{4} +4(100)^{3} + 6(100)^{2} +4(100)+(1)^{4}

=104060401

**Using binomial theorem, evaluate (99)**^{5}

Solution:

99 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then binomial theorem can be applied.

It can be written that, 99 = 100 – 1

(99)^{5} = (100 – 1)^{5}

= ^{5}C_{0} (100)^{5} – ^{5}C_{1} (100)^{4} (1) + ^{5}C_{2} (100)^{3}(1)^{2} – ^{5}C_{3} (100)^{2}(1)^{3} + ^{5}C_{4} (100)(1)^{4} – ^{5}C_{5}(1)^{5}

= (100)^{5} – 5(100)^{4} + 10(100)^{3} – 10(100)^{2}+ 5(100) – 1

=9509900499

**Using binomial theorem , indicate which number is larger (1.1)**^{10000 }or 1000

Solution:

By spitting 1.1 and then applying binomial theorem, the first few terms of (1.1)^{10000 }can be obtained as

(1.1)^{10000 } = (1 + 0.1)^{10000}

= ^{10000}C_{0} + ^{10000}C_{1}(1.1) + (other positive temrs)

= 1 + 10000 x 1.1 + (other positive temrs)

>1000

Hence, (1.1)^{10000} > 1000

**Find (a + b)**^{4}– (a – b)^{4}Hence, evaluate (√3+ √2)^{4}– (√3 – √2)^{4}

Solution:

Using binomial theorem the expressions (a + b)^{4} and (a – b)^{4} can be expressed as

(a + b)^{4} = ^{4}C_{0} a^{4} + ^{4}C_{1} a^{3}b + ^{4}C_{2} a^{2}b^{2} + ^{4}C_{3} ab^{3} +^{4}C_{4} b^{4}

(a – b)^{4} = ^{4}C_{0} a^{4} – ^{4}C_{1} a^{3}b + ^{4}C_{2} a^{2}b^{2} – ^{4}C_{3} ab^{3} +^{4}C_{4} b^{4}

(a + b)^{4} – (a – b)^{4 }= [^{4}C_{0} a^{4} + ^{4}C_{1} a^{3}b + ^{4}C_{2} a^{2}b^{2} + ^{4}C_{3} ab^{3} +^{4}C_{4} b^{4}] – [^{4}C_{0} a^{4} – ^{4}C_{1} a^{3}b + ^{4}C_{2} a^{2}b^{2} – ^{4}C_{3} ab^{3} +^{4}C_{4} b^{4}]

= 2(^{4}C_{1 }a^{3}b + ^{4}C_{3}ab^{3})

= 2(4a^{2}b + 4ab^{3})

= 8ab(a^{2} + b^{2})

By putting a = √3 and b = √2, we obtain

(√3+√2)^{4} – (√3-√2)^{4} = 8(√3)( √2){(√3)^{2} + (√2)^{2}}

= 8(√6){3 + 2}

= 40√6

**Find (x+1)**^{6}+ (x – 1)^{6}. Hence or otherwise evaluate (√2 + 1)^{6}+ (√2 – 1)^{6}

Solution:

Using Binomial theorem the expression (√2 + 1)^{6} + (√2 – 1)^{6} can be expressed as

(x + 1)^{6} = ^{6}C_{0} x^{6} + ^{6}C_{1} x^{5} + ^{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}

(x – 1)^{6} = ^{6}C_{0} x^{6} – ^{6}C_{1} x^{5} + ^{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}

Therefore, (x + 1)^{6} – (x – 1)^{6} ={^{6}C_{0} x^{6} + ^{6}C_{1} x^{5} + ^{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}} – {^{6}C_{0} x^{6} – ^{6}C_{1} x^{5} + ^{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}}

= 2 {^{6}C_{0}x^{6} + ^{6}C_{2}x^{4} +^{6}C_{4}x^{2} +^{6}C_{6} }

= 2{x^{6} + 15x^{4} +15x^{2} + 1}

By putting x = (√2), we obtain

(√2 + 1)^{6} + (√2 – 1)^{6} = 2{(√2)^{6} + 15(√2)^{4} +15(√2)^{2} + 1}

= 2(8 + 15×4 + 15×2 + 1)

= 2(8 + 60 + 30 + 1)

= 2(99)

=198

- S
**how that 9**^{n+1}– 8n – 9 is divisible by 64, whenever n is a positive integer.

Solution:

In order to show that 9^{n+1} – 8n – 9 is divisible by 64, it has to be proved that 9^{n+1} – 8n – 9 = 64k, where k is some natural number

By binomial theorem

(1 + a)^{m} = ^{m}C_{0 } + ^{m}C_{1} a + ^{m}C_{2 }a^{2} + …+^{m}C_{m} a^{m}

For a = 8 and m = n + 1, we obtain

(1 + 8)^{n+1} = ^{n+1}C_{0} + ^{n+1}C_{1}(8)+ ^{n+1}C_{2}(8)^{2}+…+ ^{n+1}C_{n+1}(8)^{n+1}

9^{n+1} = 1 + (n+1)(8)+8^{2}[^{n+1}C_{2 }+ ^{n+1}C_{3 }8+ …+^{ n+1}C_{n+1}8^{n-1}]

9^{n+1}=9+8n+64[^{n+1}C_{2} + ^{n+1}C_{3} x 8 + …+^{n+1}C_{n+1}(8)^{n-1}]

9^{n+1} – 8n – 9 = 64k, where k = ^{n+1}C_{2} + ^{n+1}C_{3}x8 + …+^{n+1}C_{n+1}(8)^{n-1} is a natural number

Thus 9^{n+1} – 8n – 9 is divisible by 64, whenever n is a positive integer.

- Prove that