Expand each of the expressions in Exercises 1 to 5.
Q1
\( (1 - 2x)^5 \)
Answer:
By Binomial Theorem:
\( (1 - 2x)^5 = \sum_{r=0}^{5} {}^{5}C_{r} (1)^{5-r} (-2x)^r \)
\( = {}^{5}C_{0}(1)^5 - {}^{5}C_{1}(1)^4(2x) + {}^{5}C_{2}(1)^3(2x)^2 - {}^{5}C_{3}(1)^2(2x)^3 + {}^{5}C_{4}(1)(2x)^4 - {}^{5}C_{5}(2x)^5 \)
\( = 1 - 5(2x) + 10(4x^2) - 10(8x^3) + 5(16x^4) - 1(32x^5) \)
\( = 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5 \)
Q2
\( \left( \frac{2}{x} - \frac{x}{2} \right)^5 \)
Answer:
Using Binomial Expansion \( (a-b)^n \):
\( = {}^{5}C_{0}(\frac{2}{x})^5 - {}^{5}C_{1}(\frac{2}{x})^4(\frac{x}{2}) + {}^{5}C_{2}(\frac{2}{x})^3(\frac{x}{2})^2 - {}^{5}C_{3}(\frac{2}{x})^2(\frac{x}{2})^3 + {}^{5}C_{4}(\frac{2}{x})(\frac{x}{2})^4 - {}^{5}C_{5}(\frac{x}{2})^5 \)
\( = 1(\frac{32}{x^5}) - 5(\frac{16}{x^4})(\frac{x}{2}) + 10(\frac{8}{x^3})(\frac{x^2}{4}) - 10(\frac{4}{x^2})(\frac{x^3}{8}) + 5(\frac{2}{x})(\frac{x^4}{16}) - 1(\frac{x^5}{32}) \)
\( = \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32} \)
Q3
\( (2x - 3)^6 \)
Answer:
\( = {}^{6}C_{0}(2x)^6 - {}^{6}C_{1}(2x)^5(3) + {}^{6}C_{2}(2x)^4(3)^2 - {}^{6}C_{3}(2x)^3(3)^3 + {}^{6}C_{4}(2x)^2(3)^4 - {}^{6}C_{5}(2x)(3)^5 + {}^{6}C_{6}(3)^6 \)
\( = 64x^6 - 6(32x^5)(3) + 15(16x^4)(9) - 20(8x^3)(27) + 15(4x^2)(81) - 6(2x)(243) + 729 \)
\( = 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729 \)
Q4
\( \left( \frac{x}{3} + \frac{1}{x} \right)^5 \)
Answer:
\( = {}^{5}C_{0}(\frac{x}{3})^5 + {}^{5}C_{1}(\frac{x}{3})^4(\frac{1}{x}) + {}^{5}C_{2}(\frac{x}{3})^3(\frac{1}{x})^2 + {}^{5}C_{3}(\frac{x}{3})^2(\frac{1}{x})^3 + {}^{5}C_{4}(\frac{x}{3})(\frac{1}{x})^4 + {}^{5}C_{5}(\frac{1}{x})^5 \)
\( = \frac{x^5}{243} + 5(\frac{x^4}{81})(\frac{1}{x}) + 10(\frac{x^3}{27})(\frac{1}{x^2}) + 10(\frac{x^2}{9})(\frac{1}{x^3}) + 5(\frac{x}{3})(\frac{1}{x^4}) + \frac{1}{x^5} \)
\( = \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5} \)
Q5
\( \left( x + \frac{1}{x} \right)^6 \)
Answer:
\( = {}^{6}C_{0}x^6 + {}^{6}C_{1}x^5(\frac{1}{x}) + {}^{6}C_{2}x^4(\frac{1}{x^2}) + {}^{6}C_{3}x^3(\frac{1}{x^3}) + {}^{6}C_{4}x^2(\frac{1}{x^4}) + {}^{6}C_{5}x(\frac{1}{x^5}) + {}^{6}C_{6}(\frac{1}{x^6}) \)
\( = x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6} \)
Using binomial theorem, evaluate each of the following:
Q6
\( (96)^3 \)
Answer:
Write 96 as \( (100 - 4) \).
\( (100 - 4)^3 = {}^{3}C_{0}(100)^3 - {}^{3}C_{1}(100)^2(4) + {}^{3}C_{2}(100)(4)^2 - {}^{3}C_{3}(4)^3 \)
\( = 1000000 - 3(10000)(4) + 3(100)(16) - 64 \)
\( = 1000000 - 120000 + 4800 - 64 \)
\( = 880000 + 4736 = 884736 \)
Q7
\( (102)^5 \)
Answer:
Write 102 as \( (100 + 2) \).
\( (100+2)^5 = (100)^5 + 5(100)^4(2) + 10(100)^3(2)^2 + 10(100)^2(2)^3 + 5(100)(2)^4 + (2)^5 \)
\( = 10000000000 + 10(100000000) + 40(1000000) + 80(10000) + 80(100) + 32 \)
\( = 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32 \)
\( = 11040808032 \)
Q8
\( (101)^4 \)
Answer:
Write 101 as \( (100 + 1) \).
\( (100+1)^4 = (100)^4 + 4(100)^3(1) + 6(100)^2(1)^2 + 4(100)(1)^3 + 1 \)
\( = 100000000 + 4000000 + 60000 + 400 + 1 \)
\( = 104060401 \)
Q9
\( (99)^5 \)
Answer:
Write 99 as \( (100 - 1) \).
\( (100-1)^5 = (100)^5 - 5(100)^4 + 10(100)^3 - 10(100)^2 + 5(100) - 1 \)
\( = 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1 \)
\( = 9500000000 + 9900000 + 499 \)
\( = 9509900499 \)
Q10
Using Binomial Theorem, indicate which number is larger \( (1.1)^{10000} \) or 1000.
Answer:
Rewrite \( 1.1 \) as \( (1 + 0.1) \).
\( (1.1)^{10000} = (1 + 0.1)^{10000} \)
Using binomial expansion:
\( = {}^{10000}C_{0}(1)^{10000} + {}^{10000}C_{1}(1)^{9999}(0.1) + \text{ other positive terms} \)
\( = 1 + 10000 \times 0.1 + \dots \)
\( = 1 + 1000 + \dots \)
\( = 1001 + \text{positive terms} \)
Since \( 1001 > 1000 \), then \( (1.1)^{10000} > 1000 \).
Q11
Find \( (a+b)^4 - (a-b)^4 \). Hence, evaluate \( (\sqrt{3}+\sqrt{2})^4 - (\sqrt{3}-\sqrt{2})^4 \).
Answer:
Step 1: Expand terms
\( (a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \)
\( (a-b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4 \)
Subtracting the second from the first:
\( (a+b)^4 - (a-b)^4 = 2(4a^3b + 4ab^3) = 8ab(a^2 + b^2) \)
Step 2: Evaluate for \( a = \sqrt{3}, b = \sqrt{2} \)
Using the result \( 8ab(a^2 + b^2) \):
\( = 8(\sqrt{3})(\sqrt{2}) [(\sqrt{3})^2 + (\sqrt{2})^2] \)
\( = 8\sqrt{6} (3 + 2) \)
\( = 8\sqrt{6} (5) = 40\sqrt{6} \)
Q12
Find \( (x+1)^6 + (x-1)^6 \). Hence or otherwise evaluate \( (\sqrt{2}+1)^6 + (\sqrt{2}-1)^6 \).
Answer:
Step 1: Expand
\( (x+1)^6 = x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1 \)
\( (x-1)^6 = x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1 \)
Adding them cancels the odd power terms:
\( (x+1)^6 + (x-1)^6 = 2(x^6 + 15x^4 + 15x^2 + 1) \)
Step 2: Evaluate for \( x = \sqrt{2} \)
\( = 2[ (\sqrt{2})^6 + 15(\sqrt{2})^4 + 15(\sqrt{2})^2 + 1 ] \)
\( = 2[ 8 + 15(4) + 15(2) + 1 ] \)
\( = 2[ 8 + 60 + 30 + 1 ] = 2[99] = 198 \)
Q13
Show that \( 9^{n+1} - 8n - 9 \) is divisible by 64, whenever \( n \) is a positive integer.
Answer:
Write \( 9^{n+1} \) as \( (1+8)^{n+1} \).
Using binomial expansion:
\( (1+8)^{n+1} = {}^{n+1}C_{0} + {}^{n+1}C_{1}8 + {}^{n+1}C_{2}8^2 + \dots + {}^{n+1}C_{n+1}8^{n+1} \)
\( 9^{n+1} = 1 + (n+1)8 + 64[{}^{n+1}C_{2} + \dots] \)
\( 9^{n+1} = 1 + 8n + 8 + 64k \) (where \( k \) is an integer)
\( 9^{n+1} = 8n + 9 + 64k \)
\( 9^{n+1} - 8n - 9 = 64k \)
Thus, the expression is divisible by 64.
Q14
Prove that \( \sum_{r=0}^{n} 3^r {}^{n}C_{r} = 4^n \).
Answer:
The general binomial expansion is:
\( (a+b)^n = \sum_{r=0}^{n} {}^{n}C_{r} a^{n-r} b^r \)
Let \( a = 1 \) and \( b = 3 \).
\( (1+3)^n = \sum_{r=0}^{n} {}^{n}C_{r} (1)^{n-r} (3)^r \)
\( 4^n = \sum_{r=0}^{n} {}^{n}C_{r} 3^r \)
Hence Proved.
Q1
If \( a \) and \( b \) are distinct integers, prove that \( a - b \) is a factor of \( a^n - b^n \), whenever \( n \) is a positive integer.
Answer:
We can write \( a = a - b + b \).
So, \( a^n = (a - b + b)^n \).
Expand using binomial theorem where \( X = a-b \) and \( Y = b \):
\( a^n = [(a-b) + b]^n = (a-b)^n + {}^{n}C_{1}(a-b)^{n-1}b + \dots + {}^{n}C_{n-1}(a-b)b^{n-1} + b^n \)
Transposing \( b^n \) to the LHS:
\( a^n - b^n = (a-b) [ (a-b)^{n-1} + \dots + {}^{n}C_{n-1}b^{n-1} ] \)
Since each term in the bracket contains \( (a-b) \) as a factor (except the last term in expansion which was \( b^n \) but is now on LHS), the entire RHS is divisible by \( a-b \).
Therefore, \( a-b \) is a factor of \( a^n - b^n \).
Q2
Evaluate \( (\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6 \).
Answer:
Let \( a = \sqrt{3} \) and \( b = \sqrt{2} \).
The expression is \( (a+b)^6 - (a-b)^6 \).
This results in \( 2 [ {}^{6}C_{1}a^5b + {}^{6}C_{3}a^3b^3 + {}^{6}C_{5}ab^5 ] \).
Substitute values:
- \( {}^{6}C_{1} = 6 \), \( a^5 = (\sqrt{3})^5 = 9\sqrt{3} \), \( b = \sqrt{2} \). Term 1: \( 6(9\sqrt{3})(\sqrt{2}) = 54\sqrt{6} \)
- \( {}^{6}C_{3} = 20 \), \( a^3 = 3\sqrt{3} \), \( b^3 = 2\sqrt{2} \). Term 2: \( 20(3\sqrt{3})(2\sqrt{2}) = 120\sqrt{6} \)
- \( {}^{6}C_{5} = 6 \), \( a = \sqrt{3} \), \( b^5 = 4\sqrt{2} \). Term 3: \( 6(\sqrt{3})(4\sqrt{2}) = 24\sqrt{6} \)
Total = \( 2 [ 54\sqrt{6} + 120\sqrt{6} + 24\sqrt{6} ] \)
\( = 2 [ 198\sqrt{6} ] = 396\sqrt{6} \)
Q3
Find the value of \( (a^2 + \sqrt{a^2-1})^4 + (a^2 - \sqrt{a^2-1})^4 \).
Answer:
Let \( x = a^2 \) and \( y = \sqrt{a^2-1} \).
Expression is \( (x+y)^4 + (x-y)^4 \).
\( = 2 [ x^4 + {}^{4}C_{2}x^2y^2 + y^4 ] \)
\( = 2 [ x^4 + 6x^2y^2 + y^4 ] \)
Substitute \( x \) and \( y \):
\( x^2 = a^4 \), \( x^4 = a^8 \), \( y^2 = a^2-1 \), \( y^4 = (a^2-1)^2 \).
\( = 2 [ a^8 + 6a^4(a^2-1) + (a^2-1)^2 ] \)
\( = 2 [ a^8 + 6a^6 - 6a^4 + a^4 - 2a^2 + 1 ] \)
\( = 2 [ a^8 + 6a^6 - 5a^4 - 2a^2 + 1 ] \)
\( = 2a^8 + 12a^6 - 10a^4 - 4a^2 + 2 \)
Q4
Find an approximation of \( (0.99)^5 \) using the first three terms of its expansion.
Answer:
\( (0.99)^5 = (1 - 0.01)^5 \)
\( = {}^{5}C_{0}(1)^5 - {}^{5}C_{1}(1)^4(0.01) + {}^{5}C_{2}(1)^3(0.01)^2 - \dots \)
\( \approx 1 - 5(0.01) + 10(0.0001) \)
\( = 1 - 0.05 + 0.001 \)
\( = 0.95 + 0.001 = 0.951 \)
Q5
Expand using Binomial Theorem \( \left( 1 + \frac{x}{2} - \frac{2}{x} \right)^4, x \ne 0 \).
Answer:
Group the terms: \( \left[ 1 + \left( \frac{x}{2} - \frac{2}{x} \right) \right]^4 \).
Let \( y = \frac{x}{2} - \frac{2}{x} \).
\( (1+y)^4 = 1 + 4y + 6y^2 + 4y^3 + y^4 \)
Now calculate powers of \( y \):
- \( y = \frac{x}{2} - \frac{2}{x} \)
- \( y^2 = \frac{x^2}{4} - 2 + \frac{4}{x^2} \)
- \( y^3 = (\frac{x}{2} - \frac{2}{x})(\frac{x^2}{4} - 2 + \frac{4}{x^2}) = \frac{x^3}{8} - \frac{3x}{2} + \frac{6}{x} - \frac{8}{x^3} \)
- \( y^4 = (y^2)^2 = \frac{x^4}{16} + 4 + \frac{16}{x^4} - x^2 + 2 - \frac{8}{x^2} \) ... (expansion of trinomial square)
Substitute back into \( 1 + 4y + 6y^2 + 4y^3 + y^4 \) and simplify to get:
\( \frac{x^4}{16} + \frac{x^3}{2} - \frac{3x^2}{2} - 4x + 5 + \frac{8}{x} + \frac{12}{x^2} - \frac{32}{x^3} + \frac{16}{x^4} \)
Q6
Find the expansion of \( (3x^2 - 2ax + 3a^2)^3 \) using binomial theorem.
Answer:
Rewrite the expression by grouping terms:
\( [3(x^2 + a^2) - 2ax]^3 \)
\( = {}^{3}C_{0}[3(x^2+a^2)]^3 - {}^{3}C_{1}[3(x^2+a^2)]^2(2ax) + {}^{3}C_{2}[3(x^2+a^2)](2ax)^2 - {}^{3}C_{3}(2ax)^3 \)
\( = 27(x^2+a^2)^3 - 3[9(x^4+2a^2x^2+a^4)](2ax) + 3[3(x^2+a^2)](4a^2x^2) - 8a^3x^3 \)
Expanding the inner brackets:
\( = 27(x^6 + 3x^4a^2 + 3x^2a^4 + a^6) - 54ax(x^4 + 2a^2x^2 + a^4) + 36a^2x^2(x^2+a^2) - 8a^3x^3 \)
\( = 27x^6 - 54ax^5 + 117a^2x^4 - 116a^3x^3 + 117a^4x^2 - 54a^5x + 27a^6 \)