( √5 - √2 ) ^5 expand using binomial theorem
Answers
Answer:
Question 1:
Expand the expression (1– 2x)5
Answer:
By using Binomial Theorem, the expression (1– 2x)5 can be expanded as
(1– 2x)5 = 5C0 (1)5 - 5C1 (1)4 (2x) + 5C2 (1)3 (2x)2 - 5C4 (1)1 (2x)4 + 5C5 (2x)5
= 1 – 5(2x) + 10(4x2) – 10(8x3) + 5(16x4) – 32x5
= 1 – 10x + 40x2 – 80x3 + 80x4 – 32x5
Question 2:
Expand the expression (2/x – x/2)5
Answer:
By using Binomial Theorem, the expression (2/x – x/2)5 can be expanded as
(2/x– x/2)5 = 5C0 (2/x)5 - 5C1 (2/x)4 (x/2) + 5C2 (2/x)3 (x/2)2 - 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 + 5x3/8 - x5/32
Question 3:
Expand the expression (2x – 3)6
Answer:
By using Binomial Theorem, the expression can be expanded as
(2x – 3)6 = 6C0 (2x)6 - 6C1 (2x)5 (3) + 6C2 (2x)4 (3)2 - 6C3 (2x)3 (3)2 + 6C4 (2x)2 (3)4 - 6C5 (2x)1 (3)5 +
6C6 (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
Question 4:
Expand the expression (x/3 + 1/x)5
Answer:
By using Binomial Theorem, the expression (2/x – x/2)5 can be expanded as
(x/3 + 1/x)5 = 5C0 (x/3)5 + 5C1 (x/3)4 (1/x) + 5C2 (x/3)3 (1/x)2 + 5C4 (x/3)1 (1/x)4 + 5C5 (1/x)5
= x5/343 + 5(x4/81)(1/x) + 10(x3/27)(1/x2) + 10(x2/9)(1/x3) + 5(x/3)(1/x4) + 1/x5
= x5/343 + 5x3/81 + 10x/27 +10/9x + 5/3x3 + 1/x5
Question 5:
Expand the expression (2/x – x/2)5
Answer:
By using Binomial Theorem, the expression (x + 1/x)6 can be expanded as
(x + 1/x)6 = 6C0 (x)6 + 6C1 (x)5 (1/x) + 6C2 (x)4 (1/x)2 + 6C3 (x)3 (1/x)4 + 6C4 (x)2 (1/x)4 + 6C5 (x) (1/x)5
+ 6C6 (1/x)6
= x6 + 6(x)5(1/x) + 14(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
Question 6:
Using Binomial Theorem, evaluate (96)3
Answer:
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.
It can be written that, 96 = 100 – 4
Now, (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
Question 7:
Using Binomial Theorem, evaluate (102)5
Answer:
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
Now, (102)5 = (100 + 2)5
= 5C0 (100)5 + 5C1 (100)4 (2) + 5C2 (100)3(2)2 + 5C3 (100)2 (2)3 + 5C4 (100)(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
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032
Question 8:
Using Binomial Theorem, evaluate (101)4
Answer:
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
Now, (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
= 100000000 + 4000000 + 60000 + 400 + 1
= 104060401
Question 9:
Using Binomial Theorem, evaluate (99)5
Answer:
99 can be written 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
Now, (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
= 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1
= 9509900499
Question 10:
Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.
Answer:
By splitting 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 terms
= 1 + 10000 * 1.1 + other positive terms
= 1 + 11000 + other positive terms > 1000
Hence, (1.1)10000 > 1000
Question 11:
Find (a + b)4 – (a – b)4. Hence, evaluate. (√3 + √2)4 – (√3 – √2)4
Answer:
Using Binomial Theorem, the expressions, (a + b)4 – (a – b)4 can be expanded as
(a + b)4 = 4C0 a4 + 4C1 a3 b + 4C2 a2 b2 + 4C3 a b3 + 4C4 b4
(a - b)4 = 4C0 a4 - 4C1 a3 b + 4C2 a2 b2 - 4C3 a b3 + 4C4 b4
Now, (a + b)4 – (a – b)4 = [4C0 a4 + 4C1 a3 b + 4C2 a2 b2 + 4C3 a b3 + 4C4 b4] – [4C0 a4 - 4C1 a3 b + 4C2 a2
b2 - 4C3 a b3 + 4C4 b4]
= 2[4C1 a3 b + 4C3 a b3]
= 8ab(a2 + b2)
Now, put a = √3 and b = √2, we get
(√3 + √2)4 – (√3 – √2)4 = 8(√3)( √2){( √3)2 + (√2)2}
= 8(√6)(3 + 2)
= 40√6