Expand each of the expression in Exercise 1to5 in 8.1
Answers
Answer:
Expand the expression (1 – 2x)5
Solution:
By using Binomial theorem the expression (1-2x)5 can be expected as
(1 – 2x)5
= 5C0(1)5 – 5C1(1)4 (2x)1– 5C2(1)3(2x)2 – 5C3(1)2(2x)3 – 5C4(1)1(2x)4 – 5C5(2x)5
= 1 – 5(2x) + 10(4x2) – 10(8x3) + 5(16x4) – (32x5)
= 1 – 10x + 40x2 – 80x3 + 80x4 – 32x5
Expand the expression (2/x – x/2)5
Solution:
By using Binomial theorem the expression (2/x – x/2)5can be expected as
(2/x – x/2)5
= 5C0(2/x)5 – 5C1(2/x)4 (x/2)1– 5C2(2/x)3(x/2)2 – 5C3(2/x)2(x/2)3 – 5C4(2/x)1(x/2)4 – 5C5(x/2)5
= 32/x5 – 5(16/x4)(x/2) + 10(8/x3)(x2/4) – 10(4/x2)(x3/8)+5(2/x)(x4/16) – (x5/32)
= 32/x5 – 40/x3 + 20/x – 5x + 5/8 x3 – (x3/32)
Expand the expression (2x – 3)6
Solution:
By using Binomial theorem the expression (2x – 3)6 can be expected as
(2x – 3)6
= 5C0(2x)5 – 5C1(2x)4 (3)1– 5C2(2x)3(3)2 – 5C3(2x)2(3)3 – 5C4(2x)1(3)4 – 5C5(3)5
= 64x6 – 6(32x5)(3) + 15(16x4)(9) – 20(8x3)(27) + 15(4x2)(81) –6(2x)(243) +729
= 64x5 – 57x5 + 2160x4 – 4320x3 + 4860x2 – 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
= 5C0(x/3)5 – 5C1(x/3)4 (1/x)1– 5C2(x/3)3(1/x)2 – 5C3(x/3)2(1/x)3 – 5C4(x/3)1(1/x)4 – 5C5(1/x)5
= (x5/243) + 5(x4/81)(1/x) + 10(x3/27)(1/x2) + 5(x2/9)(1/x3)+5(x/3)(1/x4) + (1/x5)
= (x5/243) + (5x3/81) + (10x/27) + (10/9x) + (5/3x3) + (1/x5)
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
= 5C0x5 – 5C1 x4 (1/x)1– 5C2 x3(1/x)2 – 5C3x2(1/x)3 – 5C4x1(1/x)4 – 5C5(1/x)5
= x6 +6(x)5(1/x) + 15(x)4 (1/x2) + 20(x)3(1/x3) + 15(x)2(1/x4)+6(x)(1/x5) + (1/x6)
= x6 + 6x4 + 15x2 + 20 + (15/x2) + (6/x4) + (1/x6)
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
= 3C0 (100)3 – 3C1 (100)2 4 + 3C2 (100)(4)2 – 3C3 (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
= 5C0 (100)5 + 5C1 (100)4 2 + 5C2 (100)3(2)2 + 5C3 (100)2(2)3 + 5C4 (100)1(2)4 + + 5C5(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
= 4C0 (100)4 + 4C1 (100)3 (1) + 4C2 (100)2(1)2 + 4C3 (100)(1)3 + 4C4 (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
= 5C0 (100)5 – 5C1 (100)4 (1) + 5C2 (100)3(1)2 – 5C3 (100)2(1)3 + 5C4 (100)(1)4 – 5C5(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
= 10000C0 + 10000C1(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 = 4C0 a4 + 4C1 a3b + 4C2 a2b2 + 4C3 ab3 +4C4 b4
(a – b)4 = 4C0 a4 – 4C1 a3b + 4C2 a2b2 – 4C3 ab3 +4C4 b4
(a + b)4 – (a – b)4 = [4C0 a4 + 4C1 a3b + 4C2 a2b2 + 4C3 ab3 +4C4 b4] – [4C0 a4 – 4C1 a3b + 4C2 a2b2 – 4C3 ab3 +4C4 b4]
= 2(4C1 a3b + 4C3ab3)
= 2(4a2b + 4ab3)
= 8ab(a2 + b2)
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 = 6C0 x6 + 6C1 x5 + 6C2 x412 + 6C3 x3(1)3 +6C4 x2(1)4 + 6C5x(1)5 + 6C6 (1)6
(x – 1)6 = 6C0 x6 – 6C1 x5 + 6C2 x412 – 6C3 x3(1)3 +6C4 x2(1)4 – 6C5x(1)5 + 6C6 (1)6
Therefore, (x + 1)6 – (x – 1)6 ={6C0 x6 + 6C1 x5 + 6C2 x412 + 6C3 x3(1)3 +6C4 x2(1)4 + 6C5x(1)5 + 6C6 (1)6} – {6C0 x6 – 6C1 x5 + 6C2 x412 – 6C3 x3(1)3 +6C4 x2(1)4 – 6C5x(1)5 + 6C6 (1)6}
= 2 {6C0x6 + 6C2x4 +6C4x2 +6C6 }
= 2{x6 + 15x4 +15x2 + 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
Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer.
Solution:
In order to show that 9n+1 – 8n – 9 is divisible by 64, it has to be proved that 9n+1 – 8n – 9 = 64k, where k is some natural number