Question 1. 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]
Solutions:
To prove that (a – b) is a factor of (an – bn),
an – bn = k (a – b) where k is some natural number or constant.
a can be written as = a – b + b
an = (a – b + b)n = [(a – b) + b]n
[(a – b) + b]n = nC0 (a – b)n + nC1 (a – b)n-1 b + ……....nCn-1(a – b)bn-1 + nCn bn
an = (a – b)n + n (a – b)n-1 b + ……....nCn-1(a – b)bn-1 + bn
Now, an - bn will be
an – bn = [(a – b)n + n (a – b)n-1 b + ……....nCn-1(a – b)bn-1 + bn] - bn
an – bn = (a – b)n + n (a – b)n-1 b + ……....nCn-1(a – b)bn-1
Taking (a-b) common, we have
an – bn = (a – b) [(a –b)n-1 + n (a – b)n-2 b + …… + nCn-1 bn-1]
an – bn = (a – b) k
Where k = [(a –b)n-1 + n (a – b)n-2 b + …… + nCn-1 bn-1] is a natural number
Hence, it is proved a – b is a factor of an – bn, where n is a positive integer
Question 2. Evaluate (√3+√2)6−(√3−√2)6.
Solutions:
Using binomial theorem the expression (a + b)6 and (a – b)6, can be expanded as follows:
(a + b)6 = 6C0 a6 + 6C1 a5 b + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 a b5 + 6C6 b6
(a – b)6 = 6C0 a6 – 6C1 a5 b + 6C2 a4 b2 – 6C3 a3 b3 + 6C4 a2 b4 – 6C5 a b5 + 6C6 b6
Now adding them,
(a + b)6 – (a – b)6 = 6C0 a6 + 6C1 a5 b + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 a b5 + 6C6 b6 – [6C0 a6 – 6C1 a5 b + 6C2 a4 b2 – 6C3 a3 b3 + 6C4 a2 b4 – 6C5 a b5 + 6C6 b6]
(a + b)6 – (a – b)6 = 2[6C1 a5 b + 6C3 a3 b3 + 6C5 a b5]
Substituting a = √3 and b = √2, we get
(√3 + √2)6 – (√3 – √2)6 = 2 [6 (√3)5 (√2) + 20 (√3)3 (√2)3 + 6 (√3) (√2)5]
= 2 [54(√6) + 120 (√6) + 24 √6]
= 2 (√6) (198)
= 396 √6
Question 3. Find the value of (a^2 + \sqrt{a^2-1})^4 + (a^2 - \sqrt{a^2-1})^4 .
Solutions:
Using binomial theorem the expression (x+y)4 and (x – y)4, can be expanded as follows:
(x + y)4 = 4C0 x4 + 4C1 x3 y + 4C2 x2 y2 + 4C3 x y3 + 4C4 y4
(x – y)4 = 4C0 x4 – 4C1 x3 y + 4C2 x2 y2 – 4C3 x y3 + 4C4 y4
Now adding them,
(x + y)4 + (x - y)4 = 4C0 x4 + 4C1 x3 y + 4C2 x2 y2 + 4C3 x y3 + 4C4 y4 + [4C0 x4 – 4C1 x3 y + 4C2 x2 y2 – 4C3 x y3 + 4C4 y4]
(x + y)4 + (x - y)4 = 2[4C0 x4 + 4C2 x2 y2 + 4C4 y4]
Substituting x = a2 and y =
\sqrt{a^2-1} , we get
(a^2 + \sqrt{a^2-1})^4 + (a^2 - \sqrt{a^2-1})^4 = 2[(a^2)^4 +6 (a^2)^2 (\sqrt{a^2-1})^2 + (\sqrt{a^2-1})^4] = 2[a8 + 6a4 (a2-1) + (a2-1)2]
= 2[a8 + 6a6 - 6a4 + (a4 + 1 - 2(a2)(1))]
= 2[a8 + 6a6 - 6a4 + a4 + 1 - 2a2]
= 2a8 + 12a6 - 10a4 - 4a2 + 2
Question 4. Find an approximation of (0.99)5 using the first three terms of its expansion.
Solutions:
To make 0.99 in binomial form,
0.99 = 1 – 0.01
Now by applying binomial theorem, we get
(0. 99)5 = (1 – 0.01)5
Taking first three terms of its expansion, we have
= 5C0 (1)5 – 5C1 (1)4 (0.01) + 5C2 (1)3 (0.01)2
= 1 – 5 (0.01) + 10 (0.01)2
= 1 – 0.05 + 0.001
= 0.951
Approximation of (0.99)5 = 0.951.
Question 5. Expand using Binomial Theorem (1+\frac{x}{2} - \frac{2}{x})^4 , x≠0.
Solutions:
Grouping
(1+\frac{x}{2} - \frac{2}{x})^4 in binomial form, we have
[(1+\frac{x}{2}) - \frac{2}{x}]^4 Comparing it with (a+b)n,
a =
(1+\frac{x}{2}) , b =\frac{2}{x} and n = 4
[(1+\frac{x}{2}) - \frac{2}{x}]^4 = 4C0(1+\frac{x}{2})^4 – 4C1(1+\frac{x}{2})^3 (\frac{2}{x}) + 4C2(1+\frac{x}{2})^2 (\frac{2}{x})^2 – 4C3(1+\frac{x}{2}) (\frac{2}{x})^3 + 4C4(\frac{2}{x})^4 =
(1+\frac{x}{2})^4 – 4 (1+\frac{x}{2})^3 (\frac{2}{x}) + 6 (1+(\frac{x}{2})^2 (\frac{4}{x^2}) – 4 (1+\frac{x}{2}) (\frac{8}{x^3}) + (\frac{16}{x^4}) =
(1+\frac{x}{2})^4 – (\frac{8}{x}) (1+\frac{x}{2})^3 + (\frac{24}{x^2}) (1+\frac{x}{2})^2 – 4 (\frac{8}{x^3} + (\frac{x}{2})(\frac{8}{x^3})) + (\frac{16}{x^4}) =
(1+\frac{x}{2})^4 – (\frac{8}{x}) (1+\frac{x}{2})^3 + (\frac{24}{x^2}) (1+\frac{x}{2})^2 – 4 (\frac{8}{x^3} + \frac{4}{x^2}) + (\frac{16}{x^4}) Now, lets get the value of
(1+\frac{x}{2})^4, (1+\frac{x}{2})^3 and(1+\frac{x}{2})^2
(1+\frac{x}{2})^2 = 12 + (\frac{x}{2})^2 + 2(1)(\frac{x}{2})
(1+\frac{x}{2})^2 = 1 + \frac{x^2}{4} + x
(1+\frac{x}{2})^3 = 3C0 (1)3 + 3C1 (1)2(\frac{x}{2}) + 3C2 (1)(\frac{x}{2})^2 + 3C3(\frac{x}{2})^3
(1+\frac{x}{2})^3 = 1 + \frac{3x}{2} + \frac{3x^2}{4} + \frac{x^3}{8}
(1+\frac{x}{2})^4 = 4C0 (1)4 + 4C1 (1)3(\frac{x}{2}) + 4C2 (1)2(\frac{x}{2})^2 + 4C3 (1)(\frac{x}{2})^3 + 4C4(\frac{x}{2})^4
(1+\frac{x}{2})^4 = 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} Now, substituting these values in the main equation, we get
=
1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} – (\frac{8}{x}) (1 + \frac{3x}{2} + \frac{3x^2}{4} + \frac{x^3}{8}) + (\frac{24}{x^2}) (1 + \frac{x^2}{4} + x) – 4 (\frac{8}{x^3} + \frac{4}{x^2}) + (\frac{16}{x^4}) =
1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} – (\frac{8}{x} + (\frac{8}{x})(\frac{3x}{2}) + (\frac{8}{x})(\frac{3x^2}{4}) + (\frac{8}{x})(\frac{x^3}{8})) + (\frac{24}{x^2} + (\frac{24}{x^2})(\frac{x^2}{4}) + (\frac{24}{x^2}) (x)) – (\frac{32}{x^3} + \frac{16}{x^2}) + (\frac{16}{x^4}) =
1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} – \frac{8}{x} - 12 - 6x - x^2 + \frac{24}{x^2} + 6+ \frac{24}{x} – \frac{32}{x^3} - \frac{16}{x^2}) + (\frac{16}{x^4}) =
\frac{16}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4} - 4x + \frac{x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} – 5
Question 6. Find the expansion of (3x2– 2ax + 3a2)3 using binomial theorem.
Solutions:
Grouping (3x2– 2ax + 3a2)3 in binomial form, we have
[3x2 + (- 2ax + 3a2)]3
Comparing it with (a+b)n,
a = 3x2, b = -a (2x-3a) and n = 3
[3x2 + (-a (2x-3a))]3
= 3C0 (3x2)3 + 3C1 (3x2)2 (-a (2x-3a)) + 3C2 (3x2) (-a (2x-3a))2 + 3C3 (-a (2x-3a))3
= 27x6 + 3 (9x4) (-a) (2x-3a) + 3 (3x2) (-a)2 (2x-3a)2 + (-a)3 (2x-3a)3
= 27x6 + (-54ax5 + 81a2x4) + 9a2x2 (2x-3a)2 - a3 (2x-3a)3
Now, lets get the value of (2x-3a)2 and (2x-3a)3.
(2x-3a)2 = (2x)2 + (3a)2 - 2(2x)(3a)
(2x-3a)2 = 4x2 + 9a2 -12xa
(2x-3a)3 = (2x)3 - (3a)3 - 3(2x)(3a)(2x-3a)
(2x-3a)3 = 8x3 - 27a3 - 36x2a +54xa2
Now, substituting these values in the main equation, we get
= 27x6 - 54ax5 + 81a2x4 + 9a2x2 (4x2 + 9a2 -12xa) - a3 (8x3 - 27a3 - 36x2a + 54xa2)
= 27x6 - 54ax5 + 81a2x4 + 36a2x4 + 81a4x2 -108x3a3 - (8a3x3 - 27a6 - 36x2a4 + 54xa5)
= 27x6 - 54ax5 + 117a2x4 + 81a4x2 -108x3a3 - 8a3x3 + 27a6 + 36x2a4 - 54xa5
= 27x6 – 54ax5+ 117a2x4 – 116a3x3 + 117a4x2 – 54a5x + 27a6